\displaystyle \textbf{Question 1: }\ \ \text{Differentiate the following functions with respect to } \\ x\text{ from first-principles: } e^{2x}\qquad \qquad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle \ \text{Let }f(x)=e^{2x}.\ \text{Then,}\ f(x+h)=e^{2(x+h)}
\displaystyle \therefore\ \frac{d}{dx}(f(x))=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\displaystyle =\lim_{h\to 0}\frac{e^{2(x+h)}-e^{2x}}{h}=\lim_{h\to 0}\frac{e^{2x}e^{2h}-e^{2x}}{h}
\displaystyle =\lim_{h\to 0}\frac{e^{2x}(e^{2h}-1)}{h}=e^{2x}\lim_{h\to 0}\frac{e^{2h}-1}{h}
\displaystyle =e^{2x}\cdot 2\lim_{h\to 0}\frac{e^{2h}-1}{2h}=2e^{2x}\lim_{y\to 0}\frac{e^{y}-1}{y},\ \text{where }y=2h
\displaystyle =2e^{2x}\cdot 1=2e^{2x}\qquad \left[\because\ \lim_{y\to 0}\frac{e^{y}-1}{y}=1\right]
\displaystyle \therefore\ \frac{d}{dx}(e^{2x})=2e^{2x}

\displaystyle \textbf{Question 2: }\ \ \text{Differentiate the following functions with respect to }x:
\displaystyle  \ \log\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\qquad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \ \text{Let }y=\log\tan\left(\frac{\pi}{4}+\frac{x}{2}\right).\ \text{Putting }\frac{\pi}{4}+\frac{x}{2}=v,\ \tan\left(\frac{\pi}{4}+\frac{x}{2}\right)=\tan v=u,\ \text{we get}
\displaystyle y=\log u,\ u=\tan v\ \text{and}\ v=\frac{\pi}{4}+\frac{x}{2}
\displaystyle \therefore\ \frac{dy}{du}=\frac{1}{u},\ \frac{du}{dv}=\sec^{2}v\ \text{and}\ \frac{dv}{dx}=\frac{1}{2}
\displaystyle \text{Now,}\quad \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dx}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{u}\times\sec^{2}v\times\frac{1}{2}=\frac{1}{\tan v}\sec^{2}v\times\frac{1}{2}\qquad [\because\ u=\tan v]
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{2\sin v\cos v}=\frac{1}{\sin 2v}=\frac{1}{\sin\left(\frac{\pi}{2}+x\right)}=\frac{1}{\cos x}=\sec x\qquad \left[\because\ v=\frac{\pi}{4}+\frac{x}{2}\right]

\displaystyle \textbf{Question 3: }\ \ \text{Differentiate the following functions with respect to }x:
\displaystyle \ \log\left(x+\sqrt{a^{2}+x^{2}}\right) \qquad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{Let }y=\log\left(x+\sqrt{a^{2}+x^{2}}\right).\ \text{Then,}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left\{\log\left(x+\sqrt{a^{2}+x^{2}}\right)\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{x+\sqrt{a^{2}+x^{2}}}\times\frac{d}{dx}\left\{x+\sqrt{a^{2}+x^{2}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{x+\sqrt{a^{2}+x^{2}}}\left\{1+\frac{1}{2}(a^{2}+x^{2})^{-1/2}\times\frac{d}{dx}(a^{2}+x^{2})\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{x+\sqrt{a^{2}+x^{2}}}\left\{1+\frac{1}{2\sqrt{a^{2}+x^{2}}}\times 2x\right\}
\displaystyle =\frac{1}{x+\sqrt{a^{2}+x^{2}}}\times\frac{\sqrt{a^{2}+x^{2}}+x}{\sqrt{a^{2}+x^{2}}}=\frac{1}{\sqrt{a^{2}+x^{2}}}

\displaystyle \textbf{Question 4: }\ \ \text{If }y=\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{n},\ \text{then prove that }\frac{dy}{dx}=\frac{ny}{\sqrt{x^{2}+a^{2}}}.\qquad [\text{CBSE 2005}]
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}\quad y=\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{n}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{d}{dx}\left\{\left(x+\sqrt{x^{2}+a^{2}}\right)^{n}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\frac{d}{dx}\left\{x+\sqrt{x^{2}+a^{2}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\left\{\frac{d}{dx}(x)+\frac{d}{dx}\left(\sqrt{x^{2}+a^{2}}\right)\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\left\{1+\frac{1}{2}(x^{2}+a^{2})^{-1/2}\frac{d}{dx}(x^{2}+a^{2})\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\left\{1+\frac{1}{2\sqrt{x^{2}+a^{2}}}\cdot 2x\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\left\{1+\frac{x}{\sqrt{x^{2}+a^{2}}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n-1}\left\{\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=n\frac{\left\{x+\sqrt{x^{2}+a^{2}}\right\}^{\,n}}{\sqrt{x^{2}+a^{2}}}=\frac{ny}{\sqrt{x^{2}+a^{2}}}

\displaystyle \textbf{Question 5: }\ \ \text{If }y=\sqrt{\frac{1-x}{1+x}},\ \text{prove that }(1-x^{2})\frac{dy}{dx}+y=0.\qquad [\text{CBSE 2004}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{We have,}\quad y=\sqrt{\frac{1-x}{1+x}}
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{-1/2}\frac{d}{dx}\left(\frac{1-x}{1+x}\right)
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\cdot\frac{(1+x)\frac{d}{dx}(1-x)-(1-x)\frac{d}{dx}(1+x)}{(1+x)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\cdot\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\cdot\frac{-1-x-1+x}{(1+x)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=-\sqrt{\frac{1+x}{1-x}}\cdot\frac{1}{(1+x)^{2}}
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}=-(1-x^{2})\sqrt{\frac{1+x}{1-x}}\cdot\frac{1}{(1+x)^{2}}
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}=-\frac{(1-x^{2})}{(1+x)^{2}}\sqrt{\frac{1+x}{1-x}}
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}=-\frac{(1-x)(1+x)}{(1+x)^{2}}\sqrt{\frac{1+x}{1-x}}
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}=-\frac{1-x}{1+x}\sqrt{\frac{1+x}{1-x}}=-\sqrt{\frac{1-x}{1+x}}
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}=-y
\displaystyle \Rightarrow\ (1-x^{2})\frac{dy}{dx}+y=0

\displaystyle \textbf{Question 6: }\ \ \text{If }f(x)=\sqrt{x^{2}+1},\ g(x)=\frac{x+1}{x^{2}+1}\ \text{and} \\ h(x)=2x-3,\ \text{then find }f'\left[h'\{g'(x)\}\right].\qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \ \text{We have,}\quad f(x)=\sqrt{x^{2}+1},\ g(x)=\frac{x+1}{x^{2}+1}\ \text{and}\ h(x)=2x-3
\displaystyle \therefore\ f'(x)=\frac{x}{\sqrt{x^{2}+1}},\quad g'(x)=\frac{1-2x-x^{2}}{(x^{2}+1)^{2}}\ \text{and}\ h'(x)=2\ \text{for all }x\in R
\displaystyle \text{Now,}\quad h'(x)=2\ \text{for all }x\in R
\displaystyle \Rightarrow\ h'\{g'(x)\}=2\ \text{for all }x\in R
\displaystyle \Rightarrow\ f'\left[h'\{g'(x)\}\right]=f'(2)\ \text{for all }x\in R
\displaystyle \Rightarrow\ f'\left[h'\{g'(x)\}\right]=\frac{2}{\sqrt{2^{2}+1}}=\frac{2}{\sqrt{5}}\qquad \left[\because\ f'(x)=\frac{x}{\sqrt{x^{2}+1}}\right]

\displaystyle \textbf{Question 7: }\ \ \text{Differentiate the following functions with respect to }x:
\displaystyle (i)\ \tan^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\},\ x\neq 0
\displaystyle (ii)\ \tan^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\},\ 0<x<\pi. \qquad [\text{CBSE 2004, 2012}]
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let }y=\tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right).\ \text{Putting }x=\tan\theta,\text{ we get}
\displaystyle y=\tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right)=\tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)=\tan^{-1}\left(\frac{2\sin^{2}\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)
\displaystyle \Rightarrow\ y=\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{1}{2}\theta=\frac{1}{2}\tan^{-1}x
\displaystyle \therefore\ \frac{dy}{dx}=\frac{1}{2}\left(\frac{1}{1+x^{2}}\right)
\displaystyle (ii)\ \text{Let }y=\tan^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}
\displaystyle \text{We know that:}
\displaystyle \sqrt{1+\sin x}=\sqrt{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}
\displaystyle =\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2}}=\left|\cos\frac{x}{2}+\sin\frac{x}{2}\right|
\displaystyle \Rightarrow\ \sqrt{1+\sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\ \text{for }0<x<\pi
\displaystyle \text{and,}\quad \sqrt{1-\sin x}=\sqrt{\cos^{2}\frac{x}{2}+\sin^{2}\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}
\displaystyle =\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}}=\left|\cos\frac{x}{2}-\sin\frac{x}{2}\right|
\displaystyle \Rightarrow\ \sqrt{1-\sin x}=\begin{cases} \cos\frac{x}{2}-\sin\frac{x}{2},\ \text{if }0<x<\frac{\pi}{2}\\ -\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right),\ \text{if }\frac{\pi}{2}<x<\pi \end{cases}

\displaystyle \textbf{Question 8: }\ \ \text{If }y=\sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}}\right]\ \text{and }0<x<1, \\ \text{then find }\frac{dy}{dx}.\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle y=\sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^{2}}\right],\ \text{where }0<x<1
\displaystyle \Rightarrow\ y=\sin^{-1}\left[x\sqrt{1-(\sqrt{x})^{2}}-\sqrt{x}\sqrt{1-x^{2}}\right]
\displaystyle \Rightarrow\ y=\sin^{-1}x-\sin^{-1}\sqrt{x}\qquad \left[\text{Using: }\sin^{-1}x-\sin^{-1}y=\sin^{-1}\left\{x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}}\right\}\right]
\displaystyle \text{Differentiating with respect to }x,\text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-(\sqrt{x})^{2}}}\frac{d}{dx}(\sqrt{x})
\displaystyle =\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x}}\times\frac{1}{2\sqrt{x}}

\displaystyle \textbf{Question 9: }\ \ \text{If }y=\tan^{-1}\left\{\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right\},\ -1<x<1, \\ x\neq 0\ \text{find }\frac{dy}{dx} \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle   \text{Putting }x^{2}=\cos 2\theta,\text{ we get}
\displaystyle y=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}\right)
\displaystyle =\tan^{-1}\left(\frac{\sqrt{2\cos^{2}\theta}+\sqrt{2\sin^{2}\theta}}{\sqrt{2\cos^{2}\theta}-\sqrt{2\sin^{2}\theta}}\right)
\displaystyle \Rightarrow\ y=\tan^{-1}\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right) =\tan^{-1}\left(\frac{1+\tan\theta}{1-\tan\theta}\right)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}+\theta\right)\right)
\displaystyle \Rightarrow\ y=\frac{\pi}{4}+\theta\qquad \left[\because\ 0<x^{2}<1\Rightarrow 0<\cos 2\theta<1\Rightarrow 0<2\theta<\frac{\pi}{2}\Rightarrow 0<\theta<\frac{\pi}{4}\Rightarrow \frac{\pi}{4}<\frac{\pi}{4}+\theta<\frac{\pi}{2}\right]
\displaystyle \Rightarrow\ y=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}\qquad \left[\because\ \cos 2\theta=x^{2}\Rightarrow \theta=\frac{1}{2}\cos^{-1}x^{2}\right]
\displaystyle \therefore\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi}{4}\right)+\frac{1}{2}\frac{d}{dx}\left(\cos^{-1}x^{2}\right)
\displaystyle \Rightarrow\ \frac{dy}{dx}=0+\frac{1}{2}\left(\frac{-1}{\sqrt{1-x^{4}}}\right)\frac{d}{dx}(x^{2}) =-\frac{1}{2}\cdot\frac{2x}{\sqrt{1-x^{4}}}=-\frac{x}{\sqrt{1-x^{4}}}

\displaystyle \textbf{Question 10: }\ \ \text{If }x\sqrt{1+y}+y\sqrt{1+x}=0\text{ and }x\neq y,\ \text{prove that } \\ \frac{dy}{dx}=-\frac{1}{(x+1)^{2}}.\qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{We have,}
\displaystyle x\sqrt{1+y}+y\sqrt{1+x}=0
\displaystyle \Rightarrow\ x\sqrt{1+y}=-y\sqrt{1+x}
\displaystyle \Rightarrow\ x^{2}(1+y)=y^{2}(1+x)\qquad [\text{On squaring both sides}]
\displaystyle \Rightarrow\ x^{2}-y^{2}=y^{2}x-x^{2}y
\displaystyle \Rightarrow\ (x+y)(x-y)=-xy(x-y)
\displaystyle \Rightarrow\ x+y=-xy\qquad [\because\ x\neq y]
\displaystyle \Rightarrow\ x=-y-xy
\displaystyle \Rightarrow\ y(1+x)=-x
\displaystyle \Rightarrow\ y=-\frac{x}{1+x}
\displaystyle \Rightarrow\ \frac{dy}{dx}=-\left\{\frac{(1+x)\times 1-x(0+1)}{(1+x)^{2}}\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{1}{(1+x)^{2}}

\displaystyle \textbf{Question 11: }\ \ \text{If }\sin y=x\sin(a+y), \\ \text{prove that }\frac{dy}{dx}=\frac{\sin^{2}(a+y)}{\sin a}.\qquad [\text{CBSE 2009, 2011, 2012}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{Differentiating both sides of the given relation with respect to }x,\text{ we get}
\displaystyle \frac{d}{dx}(\sin y)=\frac{d}{dx}\{x\sin(a+y)\}
\displaystyle \Rightarrow\ \cos y\frac{dy}{dx}=1\cdot\sin(a+y)+x\cos(a+y)\frac{d}{dx}(a+y)
\displaystyle \Rightarrow\ \cos y\frac{dy}{dx}=\sin(a+y)+x\cos(a+y)\frac{dy}{dx}
\displaystyle \Rightarrow\ \cos y\frac{dy}{dx}-x\cos(a+y)\frac{dy}{dx}=\sin(a+y)
\displaystyle \Rightarrow\ \{\cos y-x\cos(a+y)\}\frac{dy}{dx}=\sin(a+y)
\displaystyle \text{Now,}\quad \sin y=x\sin(a+y)\ \Rightarrow\ x=\frac{\sin y}{\sin(a+y)}
\displaystyle \Rightarrow\ \left\{\cos y-\frac{\sin y}{\sin(a+y)}\cos(a+y)\right\}\frac{dy}{dx}=\sin(a+y)
\displaystyle \Rightarrow\ \left\{\frac{\sin(a+y)\cos y-\sin y\cos(a+y)}{\sin(a+y)}\right\}\frac{dy}{dx}=\sin(a+y)
\displaystyle \Rightarrow\ \frac{\sin(a+y-y)}{\sin(a+y)}\frac{dy}{dx}=\sin(a+y)
\displaystyle \Rightarrow\ \frac{\sin a}{\sin(a+y)}\frac{dy}{dx}=\sin(a+y)
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{\sin^{2}(a+y)}{\sin a}

\displaystyle \textbf{Question 12: }\ \ \text{If }y=(\sin x)^{\tan x}+(\cos x)^{\sec x},\ \text{find }\frac{dy}{dx}.\qquad [\text{CBSE 2007}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{We have,}
\displaystyle y=(\sin x)^{\tan x}+(\cos x)^{\sec x}=e^{\tan x\cdot\log\sin x}+e^{\sec x\cdot\log\cos x}
\displaystyle \text{On differentiating both sides with respect to }x,\text{ we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{\tan x\cdot\log\sin x}\right)+\frac{d}{dx}\left(e^{\sec x\cdot\log\cos x}\right)
\displaystyle \Rightarrow\ \frac{dy}{dx}=e^{\tan x\cdot\log\sin x}\frac{d}{dx}(\tan x\cdot\log\sin x)+e^{\sec x\cdot\log\cos x}\frac{d}{dx}(\sec x\cdot\log\cos x)
\displaystyle \Rightarrow\ \frac{dy}{dx}=(\sin x)^{\tan x}\left\{\frac{d}{dx}(\tan x)\cdot\log\sin x+\tan x\frac{d}{dx}(\log\sin x)\right\}
\displaystyle \qquad\qquad\qquad\qquad\qquad +(\cos x)^{\sec x}\left\{\frac{d}{dx}(\sec x)\cdot\log\cos x+\sec x\frac{d}{dx}(\log\cos x)\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=(\sin x)^{\tan x}\left\{\sec^{2}x\cdot\log\sin x+\tan x\cdot\frac{1}{\sin x}\cdot\cos x\right\}
\displaystyle \qquad\qquad\qquad\qquad\qquad +(\cos x)^{\sec x}\left\{\sec x\tan x\cdot\log\cos x+\sec x\left(\frac{1}{\cos x}\right)(-\sin x)\right\}
\displaystyle \Rightarrow\ \frac{dy}{dx}=(\sin x)^{\tan x}\left\{\sec^{2}x\cdot\log\sin x+1\right\}+(\cos x)^{\sec x}\left\{\sec x\tan x\cdot\log\cos x-\sec x\tan x\right\}

\displaystyle \textbf{Question 13: }\ \ \text{Differentiate the following functions with respect to }x:
\displaystyle \ x^{\cot x}+\frac{2x^{2}-3}{x^{2}+x+2} \qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle  \ \text{Let }y=x^{\cot x}+\frac{2x^{2}-3}{x^{2}+x+2}.\ \text{Then,}
\displaystyle y=e^{\cot x\cdot\log x}+\frac{2x^{2}-3}{x^{2}+x+2}\qquad \left[\because\ x^{\cot x}=e^{\log x^{\cot x}}=e^{\cot x\cdot\log x}\right]
\displaystyle \text{On differentiating both sides with respect to }x,\text{ we get}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{d}{dx}\left(e^{\cot x\cdot\log x}\right)+\frac{d}{dx}\left(\frac{2x^{2}-3}{x^{2}+x+2}\right)
\displaystyle \Rightarrow\ \frac{dy}{dx}=e^{\cot x\cdot\log x}\frac{d}{dx}(\cot x\cdot\log x)+\frac{(x^{2}+x+2)\frac{d}{dx}(2x^{2}-3)-(2x^{2}-3)\frac{d}{dx}(x^{2}+x+2)}{(x^{2}+x+2)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=x^{\cot x}\left\{(\log x)\frac{d}{dx}(\cot x)+(\cot x)\frac{d}{dx}(\log x)\right\}+\frac{(x^{2}+x+2)(4x)-(2x^{2}-3)(2x+1)}{(x^{2}+x+2)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=x^{\cot x}\left\{-\mathrm{cosec}^{2}x\cdot\log x+\frac{\cot x}{x}\right\}+\frac{2x^{2}+14x+3}{(x^{2}+x+2)^{2}}

\displaystyle \textbf{Question 14: }\ \ \text{If }x^{y}=e^{x-y},\ \text{prove that } \\ \frac{dy}{dx}=\frac{\log x}{(1+\log x)^{2}}.\qquad [\text{CBSE 2000 C, 2010 C, 2011, 2013}]
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle x^{y}=e^{x-y}
\displaystyle \Rightarrow\ e^{\log x^{y}}=e^{x-y}\qquad \left[\because\ x^{y}=e^{\log x^{y}}=e^{y\log x}\right]
\displaystyle \Rightarrow\ y\log x=x-y
\displaystyle \Rightarrow\ y\log x+y=x
\displaystyle \Rightarrow\ y(1+\log x)=x
\displaystyle \Rightarrow\ y=\frac{x}{1+\log x}
\displaystyle \text{On differentiating both sides with respect to }x,\text{ we get}
\displaystyle \frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{(1+\log x)-1}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}

\displaystyle \textbf{Question 15: }\ \ \text{If }(\cos x)^{y}=(\sin y)^{x},\ \text{find }\frac{dy}{dx}.\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \ \text{We have,}\quad (\cos x)^{y}=(\sin y)^{x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle y\log\cos x=x\log\sin y
\displaystyle \text{Differentiating both sides with respect to }x,\text{ we get}
\displaystyle y\frac{d}{dx}(\log\cos x)+\frac{dy}{dx}\log\cos x=x\frac{d}{dx}(\log\sin y)+(\log\sin y)\cdot 1
\displaystyle \Rightarrow\ y\left(\frac{-\sin x}{\cos x}\right)+\frac{dy}{dx}\log\cos x=x\left(\frac{\cos y}{\sin y}\frac{dy}{dx}\right)+\log\sin y
\displaystyle \Rightarrow\ -y\tan x+\frac{dy}{dx}\log\cos x=\frac{x\cos y}{\sin y}\frac{dy}{dx}+\log\sin y
\displaystyle \Rightarrow\ \frac{dy}{dx}\left(\log\cos x-\frac{x\cos y}{\sin y}\right)=\log\sin y+y\tan x
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{\log\sin y+y\tan x}{\log\cos x-x\cot y}

\displaystyle \textbf{Question 16: }\ \ \text{If }x^{m}y^{n}=(x+y)^{m+n},\ \text{prove that }\frac{dy}{dx}=\frac{y}{x}.\qquad [\text{CBSE 2000, 2014, 2017}]
\displaystyle \text{Answer:}
\displaystyle \ \text{We have,}\quad x^{m}y^{n}=(x+y)^{m+n}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle m\log x+n\log y=(m+n)\log(x+y)
\displaystyle \text{Differentiating both sides with respect to }x,\text{ we get}
\displaystyle m\cdot\frac{1}{x}+n\cdot\frac{1}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\frac{d}{dx}(x+y)
\displaystyle \Rightarrow\ \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)
\displaystyle \Rightarrow\ \left\{\frac{n}{y}-\frac{m+n}{x+y}\right\}\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}
\displaystyle \Rightarrow\ \left\{\frac{nx+ny-my-ny}{y(x+y)}\right\}\frac{dy}{dx}=\left\{\frac{mx+nx-mx-my}{(x+y)x}\right\}
\displaystyle \Rightarrow\ \frac{nx-my}{y(x+y)}\frac{dy}{dx}=\frac{nx-my}{(x+y)x}
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{y}{x}

\displaystyle \textbf{Question 17: }\ \ \text{Find }\frac{dy}{dx}\ \text{in each of the following:}\qquad [\text{CBSE 2011}]
\displaystyle \ x=a\left\{\cos t+\frac{1}{2}\log\tan^{2}\frac{t}{2}\right\}\ \text{and}\ y=a\sin t
\displaystyle \text{Answer:}
\displaystyle  \ x=a\left\{\cos t+\frac{1}{2}\log\tan^{2}\frac{t}{2}\right\}\ \text{and}\ y=a\sin t
\displaystyle \Rightarrow\ x=a\left\{\cos t+\frac{1}{2}\times 2\log\tan\frac{t}{2}\right\}\ \text{and}\ y=a\sin t
\displaystyle \Rightarrow\ x=a\left\{\cos t+\log\tan\frac{t}{2}\right\}\ \text{and}\ y=a\sin t.
\displaystyle \text{Differentiating with respect to }t,\text{ we get}
\displaystyle \frac{dx}{dt}=a\left\{-\sin t+\frac{1}{\tan\frac{t}{2}}\left(\sec^{2}\frac{t}{2}\right)\times\frac{1}{2}\right\}\ \text{and}\ \frac{dy}{dt}=a\cos t
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\left\{-\sin t+\frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right\}\ \text{and}\ \frac{dy}{dt}=a\cos t
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\left\{-\sin t+\frac{1}{\sin t}\right\}\ \text{and}\ \frac{dy}{dt}=a\cos t
\displaystyle \Rightarrow\ \frac{dx}{dt}=a\left\{\frac{-\sin^{2}t+1}{\sin t}\right\}\ \text{and}\ \frac{dy}{dt}=a\cos t
\displaystyle \Rightarrow\ \frac{dx}{dt}=\frac{a\cos^{2}t}{\sin t}\ \text{and}\ \frac{dy}{dt}=a\cos t
\displaystyle \therefore\ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t}{\frac{a\cos^{2}t}{\sin t}}=\tan t

\displaystyle \textbf{Question 18: }\ \ \text{If }x=\sqrt{a^{\sin^{-1}t}},\ y=\sqrt{a^{\cos^{-1}t}},\ a>0\ \text{and }-1<t<1, \\ \text{show that }\frac{dy}{dx}=-\frac{y}{x}.\qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle   \text{We have,}
\displaystyle x=\sqrt{a^{\sin^{-1}t}}\ \text{and}\ y=\sqrt{a^{\cos^{-1}t}}
\displaystyle \Rightarrow\ \frac{dx}{dt}=\frac{1}{2}\left(a^{\sin^{-1}t}\right)^{-1/2}\frac{d}{dt}\left(a^{\sin^{-1}t}\right)\ \text{and}\ \frac{dy}{dt}=\frac{1}{2}\left(a^{\cos^{-1}t}\right)^{-1/2}\frac{d}{dt}\left(a^{\cos^{-1}t}\right)
\displaystyle \Rightarrow\ \frac{dx}{dt}=\frac{1}{2}\left(a^{\sin^{-1}t}\right)^{-1/2}\left(a^{\sin^{-1}t}\log_{e}a\right)\frac{d}{dt}(\sin^{-1}t)
\displaystyle \text{and,}\quad \frac{dy}{dt}=\frac{1}{2}\left(a^{\cos^{-1}t}\right)^{-1/2}\left(a^{\cos^{-1}t}\log_{e}a\right)\frac{d}{dt}(\cos^{-1}t)
\displaystyle \Rightarrow\ \frac{dx}{dt}=\frac{1}{2}\left(a^{\sin^{-1}t}\right)^{1/2}(\log_{e}a)\cdot\frac{1}{\sqrt{1-t^{2}}}=\frac{x\log_{e}a}{2\sqrt{1-t^{2}}}
\displaystyle \text{and,}\quad \frac{dy}{dt}=\frac{1}{2}\left(a^{\cos^{-1}t}\right)^{1/2}(\log_{e}a)\cdot\frac{-1}{\sqrt{1-t^{2}}}=-\frac{y\log_{e}a}{2\sqrt{1-t^{2}}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-\frac{y\log_{e}a}{2\sqrt{1-t^{2}}}}{\frac{x\log_{e}a}{2\sqrt{1-t^{2}}}}=-\frac{y}{x}

\displaystyle \textbf{Question 19: }\ \ \text{If }x\in\left(\frac{1}{\sqrt{2}},1\right),\ \text{differentiate }\tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\ \text{with respect to } \\ \cos^{-1}\left(2x\sqrt{1-x^{2}}\right). \qquad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle   \text{Let }u=\tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\ \text{and}\ v=\cos^{-1}\left(2x\sqrt{1-x^{2}}\right).
\displaystyle \text{Let }x=\sin\theta.\ \text{Then,}
\displaystyle x\in\left(\frac{1}{\sqrt{2}},1\right)\Rightarrow \frac{1}{\sqrt{2}}<\sin\theta<1\Rightarrow \frac{\pi}{4}<\theta<\frac{\pi}{2}\Rightarrow 0<\frac{\pi}{2}-\theta<\frac{\pi}{4}
\displaystyle \text{Now,}\quad u=\tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)
\displaystyle \Rightarrow\ u=\tan^{-1}\left(\frac{\sqrt{1-\sin^{2}\theta}}{\sin\theta}\right)=\tan^{-1}(\cot\theta)=\tan^{-1}\left\{\tan\left(\frac{\pi}{2}-\theta\right)\right\}
\displaystyle \Rightarrow\ u=\frac{\pi}{2}-\theta
\displaystyle \Rightarrow\ u=\frac{\pi}{2}-\sin^{-1}x
\displaystyle \therefore\ \frac{du}{dx}=0-\frac{1}{\sqrt{1-x^{2}}}=-\frac{1}{\sqrt{1-x^{2}}}
\displaystyle v=\cos^{-1}\left(2x\sqrt{1-x^{2}}\right)
\displaystyle \Rightarrow\ v=\frac{\pi}{2}-\sin^{-1}\left(2x\sqrt{1-x^{2}}\right)
\displaystyle \Rightarrow\ v=\frac{\pi}{2}-\sin^{-1}(\sin 2\theta)\qquad [\because\ x=\sin\theta]
\displaystyle \Rightarrow\ v=\frac{\pi}{2}-\sin^{-1}\{\sin(\pi-2\theta)\}
\displaystyle \Rightarrow\ v=\frac{\pi}{2}-(\pi-2\theta)\qquad \left[\because\ \frac{\pi}{4}<\theta<\frac{\pi}{2}\Rightarrow 0<\pi-2\theta<\frac{\pi}{2}\right]
\displaystyle \Rightarrow\ v=-\frac{\pi}{2}+2\theta=-\frac{\pi}{2}+2\sin^{-1}x
\displaystyle \therefore\ \frac{dv}{dx}=\frac{2}{\sqrt{1-x^{2}}}
\displaystyle \therefore\ \frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{-\frac{1}{\sqrt{1-x^{2}}}}{\frac{2}{\sqrt{1-x^{2}}}}=-\frac{1}{2}

\displaystyle \textbf{Question 20. }\text{The function }f(x)=|x|\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) continuous and differentiable everywhere.}
\displaystyle \text{(b) continuous and differentiable nowhere.}
\displaystyle \text{(c) continuous everywhere, but differentiable everywhere }   \text{except at }x=0
\displaystyle \text{(d) continuous everywhere, but differentiable nowhere.}
\displaystyle \text{Answer:}
\displaystyle \text{(c)} 
\displaystyle \text{We know that, modulus function is everywhere continuous.}
\displaystyle \text{Now, check differentiability of }f(x)\text{ at }x=0
\displaystyle Lf'(0)=\lim_{h\to 0}\frac{f(0-h)-f(0)}{-h}
\displaystyle =\lim_{h\to 0}\frac{f(-h)-f(0)}{-h}
\displaystyle =\lim_{h\to 0}\frac{|-h|-|0|}{-h}=\lim_{h\to 0}\frac{h}{-h}
\displaystyle =\lim_{h\to 0}(-1)=-1
\displaystyle Rf'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{|h|-|0|}{h}=\lim_{h\to 0}\frac{h}{h}=\lim_{h\to 0}(1)=1
\displaystyle \therefore Lf'(0)\neq Rf'(0)
\displaystyle \Rightarrow f\text{ is not differentiable at }x=0
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\displaystyle \textbf{Question 21. }\text{The function }f(x)=x|x|\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) continuous and differentiable at }x=0
\displaystyle \text{(b) continuous but not differentiable at }x=0
\displaystyle \text{(c) differentiable but not continuous at }x=0
\displaystyle \text{(d) neither differentiable nor continuous at }x=0
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }f(x)=x|x|=\begin{cases}x^2, & x\geq 0\\-x^2, & x<0\end{cases} 
\displaystyle Lf'(0)=\lim_{h\to 0}\frac{f(0-h)-f(0)}{-h}
\displaystyle =\lim_{h\to 0}\frac{f(-h)-f(0)}{-h}=\lim_{h\to 0}\left(\frac{-h^2-0}{-h}\right)
\displaystyle =\lim_{h\to 0}(h)=0
\displaystyle Rf'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\left(\frac{h^2-0}{h}\right)
\displaystyle =\lim_{h\to 0}(h)=0
\displaystyle \therefore \text{LHD}=\text{RHD}=0
\displaystyle \Rightarrow f(x)\text{ is differentiable at }x=0
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\displaystyle \textbf{Question 22. }\text{The value of }k\text{ for which function }f(x)=\begin{cases}x^2,\ x\geq0\\kx,\ x<0\end{cases}
\displaystyle \text{is differentiable at }x=0\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }1 \qquad \text{(b) }2  \qquad  \text{(c) any real number} \qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{(d) Since, }f(x)\text{ is differentiable at }x=0 
\displaystyle Lf'(0)=\lim_{h\to 0}\frac{f(0-h)-f(0)}{-h}
\displaystyle =\lim_{h\to 0}\frac{f(-h)-f(0)}{-h}
\displaystyle =\lim_{h\to 0}\frac{k(-h)-0}{-h}=k\qquad\ldots\text{(i)}
\displaystyle Rf'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{f(h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{h^2-0}{h}=\lim_{h\to 0}(h)=0\qquad\ldots\text{(ii)}
\displaystyle \text{Since }f(x)\text{ is differentiable at }x=0
\displaystyle \therefore Lf'(0)=Rf'(0)
\displaystyle \Rightarrow k=0\qquad\text{[using Eqs. (i) and (ii)]}
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\displaystyle \textbf{Question 23. }\text{If }f(x)=2|x|+3|\sin x|+6,\text{ then the right hand } \text{derivative of }
\displaystyle f(x)\text{ at }x=0\text{ is}    \hspace{2.2cm} \text{[CBSE 2023, CBSE 2020]}
\displaystyle \text{(a) }6 \qquad \text{(b) }5 \qquad \text{(c) }3 \qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, }f(x)=2|x|+3|\sin x|+6 
\displaystyle Rf'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{f(h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{(2|h|+3|\sin h|+6)-(2|0|+3|\sin 0|+6)}{h}
\displaystyle =\lim_{h\to 0}\frac{(2h+3\sin h+6)-(6)}{h}
\displaystyle =\lim_{h\to 0}\frac{2h+3\sin h}{h}
\displaystyle =\lim_{h\to 0}\left(2+3\cdot\frac{\sin h}{h}\right)
\displaystyle =2+\lim_{h\to 0}\left(\frac{3\sin h}{h}\right)=2+3=5
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\displaystyle \textbf{Question 24. }\text{If }y=\frac{\cos x-\sin x}{\cos x+\sin x},\text{ then }\frac{dy}{dx}\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }-\sec^2\left(\frac{\pi}{4}-x\right) \qquad \text{(b) }\sec^2\left(\frac{\pi}{4}-x\right)
\displaystyle \text{(c) }\log\left|\sec\left(\frac{\pi}{4}-x\right)\right| \qquad \text{(d) }-\log\left|\sec\left(\frac{\pi}{4}-x\right)\right|
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }y=\frac{\cos x-\sin x}{\cos x+\sin x} 
\displaystyle =\frac{1-\tan x}{1+\tan x}\qquad\text{(divide each term by }\cos x\text{)}
\displaystyle =\frac{\tan\pi/4-\tan x}{1+\tan\pi/4\tan x}\Rightarrow y=\tan\left(\frac{\pi}{4}-x\right)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get} 
\displaystyle \frac{dy}{dx}=\sec^2\left(\frac{\pi}{4}-x\right)\cdot(-1)
\displaystyle \Rightarrow \frac{dy}{dx}=-\sec^2\left(\frac{\pi}{4}-x\right)
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\displaystyle \textbf{Question 25. }\text{If }y=\sin^2(x^3),\text{ then }\frac{dy}{dx}\text{ is equal to}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }2\sin x^3\cos x^3 \qquad \text{(b) }3x^3\sin x^3\cos x^3
\displaystyle \text{(c) }6x^2\sin x^3\cos x^3 \qquad \text{(d) }2x^2\sin^2(x^3)
\displaystyle \text{Answer:}
\displaystyle \text{(c) We have, }y=\sin^2(x^3) 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=2\sin(x^3)\cdot\frac{d}{dx}\sin(x^3)
\displaystyle =2\sin(x^3)\cos(x^3)\cdot\frac{d}{dx}(x^3)
\displaystyle =2\sin(x^3)\cos(x^3)(3x^2)=6x^2\sin x^3\cos x^3
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\displaystyle \textbf{Question 26. }\text{If }y=\log(\sin e^x),\text{ then }\frac{dy}{dx}\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\cot e^x \qquad \text{(b) }\mathrm{cosec}\ e^x
\displaystyle \text{(c) }e^x\cot e^x \qquad \text{(d) }e^x\mathrm{cosec}\ e^x
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }y=\log(\sin e^x) 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{\sin e^x}\cdot\frac{d}{dx}(\sin e^x)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\sin e^x}\cdot\cos e^x\cdot e^x\Rightarrow \frac{dy}{dx}=\cot e^x\cdot e^x
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\displaystyle \textbf{Question 27. }\text{If }f(x)=|\cos x|,\text{ then }f'\left(\frac{3\pi}{4}\right)\text{ is}    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }1 \qquad \text{(b) }-1 \qquad \text{(c) }-\frac{1}{\sqrt2} \qquad \text{(d) }\frac{1}{\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }f(x)=|\cos x| 
\displaystyle \Rightarrow f(x)=-\cos x,\text{ if }\frac{\pi}{2}<x<\pi
\displaystyle \Rightarrow f'(x)=\sin x
\displaystyle \therefore f'\left(\frac{3\pi}{4}\right)=\sin\left(\frac{3\pi}{4}\right)=\sin\left(\pi-\frac{\pi}{4}\right)=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}
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\displaystyle \textbf{Question 28. }\text{If }(a+bx)e^{\frac{y}{x}}=x,\text{ then prove that }x^2\frac{d^2y}{dx^2}=\left(\frac{a}{a+bx}\right)^2.    \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(a+bx)e^{y/x}=x\Rightarrow e^{y/x}=\frac{x}{a+bx} 
\displaystyle \text{Taking }\log_e\text{ both sides, we get}
\displaystyle \log_e e^{y/x}=\log_e\left(\frac{x}{a+bx}\right)
\displaystyle \Rightarrow \frac{y}{x}\log_e e=\log_e\left(\frac{x}{a+bx}\right)
\displaystyle \Rightarrow \frac{y}{x}=\log_e\frac{x}{a+bx}\qquad[\because\log_e e=1]
\displaystyle \Rightarrow \frac{y}{x}=\log_e x-\log_e(a+bx)
\displaystyle \text{Differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{x\frac{dy}{dx}-y}{x^2}=\frac{1}{x}-\frac{b}{a+bx}
\displaystyle \Rightarrow x\frac{dy}{dx}-y=x^2\left(\frac{1}{x}-\frac{b}{a+bx}\right)=\frac{ax}{a+bx}
\displaystyle \text{Differentiating again w.r.t. }x\text{, we get}
\displaystyle \Rightarrow x\frac{d^2y}{dx^2}+\frac{dy}{dx}-\frac{dy}{dx}=\frac{(a+bx)a-ax(b)}{(a+bx)^2}
\displaystyle \Rightarrow x\frac{d^2y}{dx^2}=\frac{a^2}{(a+bx)^2}=\left(\frac{a}{a+bx}\right)^2
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\displaystyle \textbf{Question 29. }\text{If }f(x)=\begin{cases}x^2,\text{ if }x\geq1\\x,\text{ if }x<1\end{cases},\text{ then show that }f\text{ is not } \text{differentiable at }
\displaystyle x=1.    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}x^2, & \text{if }x\geq 1\\x, & \text{if }x<1\end{cases} 
\displaystyle Lf'(1)=\lim_{h\to 0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h\to 0}\frac{(1-h)-(1)}{-h}=\lim_{h\to 0}\frac{1-h-1}{-h}
\displaystyle =\lim_{h\to 0}\left(\frac{-h}{-h}\right)=\lim_{h\to 0}(1)=1
\displaystyle Rf'(1)=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to 0}\frac{(1+h)^2-(1)^2}{h}
\displaystyle =\lim_{h\to 0}\frac{(1+h^2+2h)-1}{h}
\displaystyle =\lim_{h\to 0}\frac{h^2+2h}{h}=\lim_{h\to 0}(h+2)=0+2=2
\displaystyle \text{Since, }Lf'(1)\neq Rf'(1)
\displaystyle \text{Hence, }f(x)\text{ is not differentiable at }x=1\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 30. }\text{If }(x^2+y^2)^2=xy,\text{ then find }\frac{dy}{dx}.    \hspace{2.2cm} \text{[CBSE 2023, CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }(x^2+y^2)^2=xy 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle 2(x^2+y^2)\left[2x+2y\frac{dy}{dx}\right]=\left[x\frac{dy}{dx}+y\right]
\displaystyle \Rightarrow 4(x^2+y^2)\left(x+y\frac{dy}{dx}\right)=\left(y+x\frac{dy}{dx}\right)
\displaystyle \Rightarrow 4(x^2+y^2)x+4(x^2+y^2)y\frac{dy}{dx}=y+x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}\left[4(x^2+y^2)y-x\right]=y-4x(x^2+y^2)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4(x^2+y^2)y-x}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-[y-4x(x^2+y^2)]}{[x-4y(x^2+y^2)]}
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\displaystyle \textbf{Question 31. }\text{If }y=(x+\sqrt{x^2-1})^2,\text{ then show that } (x^2-1)\left(\frac{dy}{dx}\right)^2=4y^2.
\displaystyle     \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=(x+\sqrt{x^2-1})^2 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=2(x+\sqrt{x^2-1})\cdot\frac{d}{dx}(x+\sqrt{x^2-1})
\displaystyle \Rightarrow \frac{dy}{dx}=2(x+\sqrt{x^2-1})\left(1+\frac{2x}{2\sqrt{x^2-1}}\right)
\displaystyle =2(x+\sqrt{x^2-1})\left(\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}\right)
\displaystyle =\frac{2(x+\sqrt{x^2-1})^2}{\sqrt{x^2-1}}=\frac{2y}{\sqrt{x^2-1}}
\displaystyle [\because y=(x+\sqrt{x^2-1})^2]
\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)^2=\frac{4y^2}{x^2-1}
\displaystyle \Rightarrow (x^2-1)\left(\frac{dy}{dx}\right)^2=4y^2\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 32. }\text{If }y=x^{\frac{1}{x}},\text{ then find }\frac{dy}{dx}\text{ at }x=1.    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=x^{1/x} 
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \log y=\frac{1}{x}\log x
\displaystyle \text{On differentiating both sides, we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\cdot\frac{d}{dx}(\log x)+\log x\cdot\frac{d}{dx}\left(\frac{1}{x}\right)
\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\cdot\frac{1}{x}+\log x\left(\frac{-1}{x^2}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=y\left(\frac{1}{x^2}-\frac{\log x}{x^2}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=x^{1/x}\left(\frac{1-\log x}{x^2}\right)
\displaystyle \therefore \text{At }x=1
\displaystyle \frac{dy}{dx}=1\cdot\frac{(1-\log 1)}{1}=1-0=1
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\displaystyle \textbf{Question 33. }\text{If }x=a\sin2t,\ y=a(\cos2t+\log\tan t),\text{ then find }\frac{dy}{dx}. \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a\sin 2t 
\displaystyle \therefore \frac{dx}{dt}=a\cos 2t\cdot(2)=2a\cos 2t
\displaystyle \text{and }y=a(\cos 2t+\log\tan t)
\displaystyle \therefore \frac{dy}{dt}=a\left(-2\sin 2t+\frac{1}{\tan t}\cdot\sec^2 t\right)
\displaystyle =a\left(-2\sin 2t+\frac{2}{2\sin t\cos t}\right)
\displaystyle =a\left(-2\sin 2t+\frac{2}{\sin 2t}\right)
\displaystyle =2a\left(\frac{-\sin^2 2t+1}{\sin 2t}\right)=2a\cdot\frac{\cos^2 2t}{\sin 2t}
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a\cos^2 2t/\sin 2t}{2a\cos 2t}=\cot 2t
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\displaystyle \textbf{Question 34. }\text{If }xy=e^{x-y},\text{ then show that }\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}.   \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }xy=e^{x-y} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d}{dx}(xy)=\frac{d}{dx}(e^{x-y})
\displaystyle \Rightarrow x\frac{dy}{dx}+y\frac{d}{dx}(x)=e^{x-y}\cdot\frac{d}{dx}(x-y)
\displaystyle \Rightarrow x\frac{dy}{dx}+y=e^{x-y}\left(1-\frac{dy}{dx}\right)
\displaystyle \Rightarrow (x+e^{x-y})\frac{dy}{dx}=e^{x-y}-y\Rightarrow \frac{dy}{dx}=\frac{e^{x-y}-y}{x+e^{x-y}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{xy-y}{x+xy}\qquad[\because xy=e^{x-y}]
\displaystyle =\frac{y(x-1)}{x(1+y)}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 35. }\text{If }x=a\cos t\text{ and }y=b\sin t,\text{ then find }\frac{d^2y}{dx^2}.    \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a\cos t,\ y=b\sin t 
\displaystyle \text{On differentiating w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=a(-\sin t)=-a\sin t
\displaystyle \text{and }\frac{dy}{dt}=b\cos t
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{b\cos t}{-a\sin t}=\frac{-b}{a}\cot t
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{-b}{a}\cot t\right)=\frac{-b}{a}\cdot\frac{d}{dt}(\cot t)\cdot\frac{dt}{dx}
\displaystyle =\frac{-b}{a}(-\text{cosec}^2 t)\left(\frac{-1}{a\sin t}\right)=\frac{-b}{a^2}\text{cosec}^3 t
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\displaystyle \textbf{Question 36. }\text{If }y=\sqrt{ax+b},\text{ prove that }y\left(\frac{d^2y}{dx^2}\right)+\left(\frac{dy}{dx}\right)^2=0.   \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\sqrt{ax+b} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{ax+b}}\cdot\frac{d}{dx}(ax+b)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{a}{2\sqrt{ax+b}}=\frac{a}{2y}\qquad[\because y=\sqrt{ax+b}]
\displaystyle \Rightarrow y\frac{dy}{dx}=\frac{a}{2}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle y\frac{d^2y}{dx^2}+\frac{dy}{dx}\cdot\frac{dy}{dx}=0
\displaystyle \Rightarrow y\left(\frac{d^2y}{dx^2}\right)+\left(\frac{dy}{dx}\right)^2=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 37. }\text{Differentiate }\sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\text{ with respect to } \sin^{-1}\left[2x\sqrt{1-x^2}\right].   
\displaystyle   \hspace{2.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) 
\displaystyle \text{On putting }x=\sin\theta
\displaystyle \Rightarrow \theta=\sin^{-1}x
\displaystyle \therefore u=\sec^{-1}\left(\frac{1}{\sqrt{1-\sin^2\theta}}\right)=\sec^{-1}\left(\frac{1}{\sqrt{\cos^2\theta}}\right)
\displaystyle =\sec^{-1}(\sec\theta)=\theta=\sin^{-1}x
\displaystyle \therefore \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}
\displaystyle \text{and let }v=\sin^{-1}(2x\sqrt{1-x^2})
\displaystyle \text{On putting }x=\sin\theta
\displaystyle \Rightarrow \theta=\sin^{-1}x
\displaystyle \therefore v=\sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})
\displaystyle =\sin^{-1}(2\sin\theta\cdot\cos\theta)
\displaystyle =\sin^{-1}(\sin 2\theta)=2\theta=2\sin^{-1}x
\displaystyle \therefore \frac{dv}{dx}=\frac{d}{dx}(2\sin^{-1}x)=\frac{2}{\sqrt{1-x^2}}
\displaystyle \therefore \frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{1/\sqrt{1-x^2}}{2/\sqrt{1-x^2}}\Rightarrow \frac{du}{dv}=\frac{1}{2}
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\displaystyle \textbf{Question 38. }\text{If }y=\tan x+\sec x,\text{ then prove that }\frac{d^2y}{dx^2}=\frac{\cos x}{(1-\sin x)^2}.    \hspace{0.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\tan x+\sec x 
\displaystyle \Rightarrow y=\frac{\sin x}{\cos x}+\frac{1}{\cos x}=\frac{\sin x+1}{\cos x}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{\cos x\cdot\frac{d}{dx}(\sin x+1)-(\sin x+1)\cdot\frac{d}{dx}(\cos x)}{(\cos x)^2}
\displaystyle =\frac{\cos x(\cos x)-(\sin x+1)(-\sin x)}{\cos^2 x}
\displaystyle =\frac{\cos^2x+\sin^2x+\sin x}{\cos^2x}=\frac{1+\sin x}{\cos^2x}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1+\sin x}{1-\sin^2x}=\frac{1}{1-\sin x}
\displaystyle \text{Again, differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{(1-\sin x)(0)-1(0-\cos x)}{(1-\sin x)^2}=\frac{\cos x}{(1-\sin x)^2}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 39. }\text{If }y=Ae^{5x}+Be^{-5x},\text{ then }\frac{d^2y}{dx^2}\text{ is equal to}     \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{(a) }25y \qquad \text{(b) }5y \qquad \text{(c) }-25y \qquad \text{(d) }15y
\displaystyle \text{Answer:}
\displaystyle \text{(a) We have, }y=Ae^{5x}+Be^{-5x} 
\displaystyle \Rightarrow \frac{dy}{dx}=Ae^{5x}\cdot(5)+Be^{-5x}(-5)
\displaystyle =5Ae^{5x}-5Be^{-5x}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=5Ae^{5x}(5)-5Be^{-5x}(-5)
\displaystyle =25Ae^{5x}+25Be^{-5x}
\displaystyle \therefore \frac{d^2y}{dx^2}=25(Ae^{5x}+Be^{-5x})=25y
\\

\displaystyle \textbf{Question 40. }\text{If }f(x)=x|x|,\text{ then find }f'(x).    \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=x|x|=\begin{cases}-x^2, & x<0\\x^2, & x\geq 0\end{cases} 
\displaystyle f'(x)=\begin{cases}-2x, & x<0\\2x, & x>0\end{cases}
\displaystyle \therefore f'(x)=2|x|
\\

\displaystyle \textbf{Question 41. }\text{Find the value of }\frac{dy}{dx}\text{ at }\theta=\frac{\pi}{3},\text{ if}
\displaystyle x=\cos\theta-\cos2\theta,\ y=\sin\theta-\sin2\theta.    \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x=\cos\theta-\cos 2\theta,\ y=\sin\theta-\sin 2\theta 
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=-\sin\theta+2\sin 2\theta
\displaystyle \frac{dy}{d\theta}=\cos\theta-2\cos 2\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{\cos\theta-2\cos 2\theta}{-\sin\theta+2\sin 2\theta}
\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{3}}=\frac{\cos\frac{\pi}{3}-2\cos\frac{2\pi}{3}}{-\sin\frac{\pi}{3}+2\sin\frac{2\pi}{3}}
\displaystyle =\frac{\dfrac{1}{2}+2\left(\dfrac{1}{2}\right)}{-\dfrac{\sqrt{3}}{2}+2\left(\dfrac{\sqrt{3}}{2}\right)}=\frac{\dfrac{3}{2}}{\dfrac{\sqrt{3}}{2}}=\frac{3}{\sqrt{3}}=\sqrt{3}
\\

\displaystyle \textbf{Question 42. }\text{Find the differential of }\sin^2x\text{ with respect to }e^{\cos x}.    \hspace{0.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\sin^2x\text{ and }z=e^{\cos x} 
\displaystyle \text{On differentiating }y\text{ and }z\text{ w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=2\sin x\cos x
\displaystyle \text{and }\frac{dz}{dx}=e^{\cos x}(-\sin x)=-(\sin x)e^{\cos x}
\displaystyle \text{Now, }\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{2\sin x\cos x}{-(\sin x)e^{\cos x}}=\frac{-2\cos x}{e^{\cos x}}
\\

\displaystyle \textbf{Question 43. }\text{If }y=\sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right),\text{ then show that } \frac{dy}{dx}=\frac{-1}{2\sqrt{1-x^2}}. 
\displaystyle    \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }y=\sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) 
\displaystyle \text{On putting }x=\sin\theta\text{, we get}
\displaystyle y=\sin^{-1}\left(\frac{\sqrt{1+\sin\theta}+\sqrt{1-\sin\theta}}{2}\right)
\displaystyle \left[\because 1\pm\sin\theta=\left(\cos\frac{\theta}{2}\pm\sin\frac{\theta}{2}\right)^2\right]
\displaystyle \Rightarrow y=\sin^{-1}\left[\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}+\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{2}\right]
\displaystyle \Rightarrow y=\sin^{-1}\left[\frac{2\cos\frac{\theta}{2}}{2}\right]
\displaystyle \Rightarrow y=\sin^{-1}\left(\cos\frac{\theta}{2}\right)=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right)
\displaystyle \Rightarrow y=\frac{\pi}{2}-\frac{\theta}{2}\Rightarrow y=\frac{\pi}{2}-\frac{1}{2}\sin^{-1}x
\displaystyle \text{On differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=-\frac{1}{2\sqrt{1-x^2}}\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 44. }\text{If }x=a\cos\theta\text{ and }y=b\sin\theta,\text{ then find }\frac{d^2y}{dx^2}.    \hspace{2.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x=a\cos\theta\text{ and }y=b\sin\theta 
\displaystyle \therefore \frac{dx}{d\theta}=-a\sin\theta,\ \frac{dy}{d\theta}=b\cos\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{b\cos\theta}{-a\sin\theta}=\frac{-b}{a}\cot\theta
\displaystyle \text{Again, }\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)
\displaystyle =\frac{d}{dx}\left(\frac{-b}{a}\cot\theta\right)=\frac{-b}{a}(-\text{cosec}^2\theta)\cdot\frac{d\theta}{dx}
\displaystyle =\frac{b}{a}\text{cosec}^2\theta\times\frac{1}{(-a\sin\theta)}\qquad\left[\because\frac{d\theta}{dx}=\frac{1}{dx/d\theta}\right]
\displaystyle =\frac{-b}{a^2}\text{cosec}^3\theta
\\

\displaystyle \textbf{Question 45. }\text{Differentiate }e^{\sqrt{3x}},\text{ with respect to }x.    \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=e^{\sqrt{3x}} 
\displaystyle \text{Then, }\frac{dy}{dx}=\frac{d(e^{\sqrt{3x}})}{dx}=\frac{3\cdot e^{\sqrt{3x}}}{2\cdot\sqrt{3x}}=\frac{3e^{\sqrt{3x}}}{2\sqrt{3x}}
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\displaystyle \textbf{Question 46. }\text{If }y=\cos(\sqrt{3x}),\text{ then find }\frac{dy}{dx}.    \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\cos(\sqrt{3x}) 
\displaystyle \text{Differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\{\cos(\sqrt{3x})\}=-\sin(\sqrt{3x})\cdot\frac{d}{dx}(\sqrt{3x})
\displaystyle =-\sin(\sqrt{3x})\cdot\frac{1}{2\sqrt{3x}}\cdot3=-\frac{3\sin\sqrt{3x}}{2\sqrt{3x}}
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\displaystyle \textbf{Question 47. }\text{If }y=x|x|,\text{ find }\frac{dy}{dx}\text{ for }x<0.    \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }y=x|x| 
\displaystyle \text{When }x<0\text{, then }|x|=-x
\displaystyle \therefore y=x(-x)=-x^2\Rightarrow \frac{dy}{dx}=-2x
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\displaystyle \textbf{Question 48. }\text{If }(\cos x)^y=(\cos y)^x,\text{ then find }\frac{dy}{dx}.    \hspace{2.2cm} \text{[CBSE 2019, CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(\cos x)^y=(\cos y)^x 
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log(\cos x)^y=\log(\cos y)^x
\displaystyle \Rightarrow y\log(\cos x)=x\log(\cos y)\qquad[\because\log x^n=n\log x]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle y\cdot\frac{d}{dx}\log(\cos x)+\log\cos x\cdot\frac{d}{dx}(y)=x\cdot\frac{d}{dx}\log(\cos y)+\log(\cos y)\cdot\frac{d}{dx}(x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow y\cdot\frac{1}{\cos x}\cdot\frac{d}{dx}(\cos x)+\log(\cos x)\frac{dy}{dx}=x\cdot\frac{1}{\cos y}\cdot\frac{d}{dx}(\cos y)+\log\cos y\cdot1
\displaystyle \Rightarrow y\cdot\frac{1}{\cos x}(-\sin x)+\log(\cos x)\cdot\frac{dy}{dx}=x\cdot\frac{1}{\cos y}(-\sin y)\cdot\frac{dy}{dx}+\log(\cos y)\cdot1
\displaystyle \Rightarrow -y\tan x+\log(\cos x)\frac{dy}{dx}=-x\tan y\frac{dy}{dx}+\log(\cos y)
\displaystyle \Rightarrow [x\tan y+\log(\cos x)]\frac{dy}{dx}=\log(\cos y)+y\tan x
\displaystyle \therefore \frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{x\tan y+\log(\cos x)}
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\displaystyle \textbf{Question 49. }\text{If }x\sqrt{1+y}+y\sqrt{1+x}=0\text{, }(x\neq y)\text{, then prove that}
\displaystyle \dfrac{dy}{dx}=-\dfrac{1}{(1+x)^2}    \hspace{2.2cm} \text{[CBSE 2019; CBSE 2012; CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given equation is }x\sqrt{1+y}+y\sqrt{1+x}=0\text{, where }x\neq y 
\displaystyle \text{we first convert the given equation into }y=f(x)\text{ form.}
\displaystyle \text{Clearly, }x\sqrt{1+y}=-y\sqrt{1+x}
\displaystyle \text{On squaring both sides, we get}
\displaystyle x^2(1+y)=y^2(1+x)
\displaystyle \Rightarrow x^2+x^2y=y^2+y^2x
\displaystyle \Rightarrow x^2-y^2=y^2x-x^2y
\displaystyle \Rightarrow (x-y)(x+y)=-xy(x-y)
\displaystyle [\because a^2-b^2=(a-b)(a+b)]
\displaystyle \Rightarrow (x-y)(x+y)+xy(x-y)=0
\displaystyle \Rightarrow (x-y)(x+y+xy)=0
\displaystyle \therefore \text{Either }x-y=0\text{ or }x+y+xy=0
\displaystyle \text{Now, }x-y=0\Rightarrow x=y
\displaystyle \text{But it is given that }x\neq y\text{. So, it is a contradiction.}
\displaystyle \therefore x-y=0\text{ is rejected.}
\displaystyle \text{Now, consider }y+xy+x=0
\displaystyle \Rightarrow y(1+x)=-x
\displaystyle \Rightarrow y=\frac{-x}{1+x}\qquad\ldots\text{(i)}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{(1+x)\cdot\frac{d}{dx}(-x)-(-x)\cdot\frac{d}{dx}(1+x)}{(1+x)^2}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{(1+x)(-1)+x(1)}{(1+x)^2}\Rightarrow \frac{dy}{dx}=\frac{-1-x+x}{(1+x)^2}
\displaystyle \therefore \frac{dy}{dx}=\frac{-1}{(1+x)^2}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 50. }\text{If }y=(\sin^{-1}x)^2\text{, prove that }   (1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-2=0    \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=(\sin^{-1}x)^2\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^2}}
\displaystyle \text{Again, differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{(\sqrt{1-x^2})\left(\dfrac{2}{\sqrt{1-x^2}}\right)-\left(\dfrac{1}{2}\cdot\dfrac{(-2x)}{\sqrt{1-x^2}}\right)\cdot(2\sin^{-1}x)}{(\sqrt{1-x^2})^2}
\displaystyle =\frac{2+\dfrac{2x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}
\displaystyle \text{Now consider, }(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-2
\displaystyle =(1-x^2)\left(\frac{2+\dfrac{2x\sin^{-1}x}{\sqrt{1-x^2}}}{1-x^2}\right)-x\left(\frac{2\sin^{-1}x}{\sqrt{1-x^2}}\right)-2
\displaystyle =2+\frac{2x\sin^{-1}x}{\sqrt{1-x^2}}-\frac{2x\sin^{-1}x}{\sqrt{1-x^2}}-2
\displaystyle =0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 51. }\text{If }(x-a)^2+(y-b)^2=c^2\text{, for some }c>0\text{, prove that}
\displaystyle \dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}}\text{ is a constant independent of }a\text{ and }b.   \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(x-a)^2+(y-b)^2=c^2 
\displaystyle \text{Differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{d\{(x-a)^2\}}{dx}+\frac{d\{(y-b)^2\}}{dx}=\frac{d(c^2)}{dx}
\displaystyle \Rightarrow 2(x-a)\cdot\frac{d(x-a)}{dx}+2(y-b)\cdot\frac{d(y-b)}{dx}=0
\displaystyle \Rightarrow 2(x-a)\cdot(1-0)+2(y-b)\cdot\left(\frac{dy}{dx}-0\right)=0
\displaystyle \Rightarrow 2(y-b)\cdot\frac{dy}{dx}=-2(x-a)
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{(x-a)}{(y-b)}
\displaystyle \text{Again, differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left\{-\frac{(x-a)}{(y-b)}\right\}
\displaystyle \frac{d^2y}{dx^2}=-\frac{\left\{(y-b)\frac{d(x-a)}{dx}-(x-a)\frac{d(y-b)}{dx}\right\}}{(y-b)^2}
\displaystyle =-\frac{\left\{(y-b)\cdot(1-0)-(x-a)\left(\frac{dy}{dx}-0\right)\right\}}{(y-b)^2}
\displaystyle =-\frac{\left\{(y-b)-\frac{dy}{dx}(x-a)\right\}}{(y-b)^2}
\displaystyle =-\frac{\left\{(y-b)+\frac{(x-a)}{(y-b)}(x-a)\right\}}{(y-b)^2}
\displaystyle =-\frac{\{(y-b)^2+(x-a)^2\}}{(y-b)^3}
\displaystyle =-\frac{c^2}{(y-b)^3}\qquad[\because(x-a)^2+(y-b)^2=c^2]
\displaystyle \text{Now, }\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}=\frac{\left[1+\left\{-\frac{(x-a)}{(y-b)}\right\}^2\right]^{3/2}}{\frac{-c^2}{(y-b)^3}}
\displaystyle =\frac{\left[\frac{(y-b)^2+(x-a)^2}{(y-b)^2}\right]^{3/2}}{\frac{-c^2}{(y-b)^3}}
\displaystyle =\frac{\left[\frac{c^2}{(y-b)^2}\right]^{3/2}}{\frac{-c^2}{(y-b)^3}}=-\left[\frac{c^2}{(y-b)^2}\right]^{3/2}\times\frac{(y-b)^3}{c^2}
\displaystyle =-\left(\frac{c}{y-b}\right)^{2\times\frac{3}{2}}\times\frac{(y-b)^3}{c^2}=-\frac{c^3}{c^2}\times\frac{(y-b)^3}{(y-b)^3}
\displaystyle =-c
\displaystyle \text{which is constant independent of }a\text{ and }b.\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 52. }\text{If }x=ae^t(\sin t+\cos t)\text{ and }y=ae^t(\sin t-\cos t)\text{, then prove that}
\displaystyle \dfrac{dy}{dx}=\dfrac{x+y}{x-y}    \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=ae^t(\sin t+\cos t) 
\displaystyle \text{and }y=ae^t(\sin t-\cos t)
\displaystyle \therefore \frac{dx}{dt}=a\left[e^t(\cos t-\sin t)+e^t(\sin t+\cos t)\right]
\displaystyle =-y+x=x-y
\displaystyle \text{and }\frac{dy}{dt}=a\left[e^t(\sin t-\cos t)+e^t(\sin t+\cos t)\right]
\displaystyle =y+x=x+y
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 53. }\text{Differentiate }x^{\sin x}+(\sin x)^{\cos x}\text{ with respect to }x.    \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=x^{\sin x}+(\sin x)^{\cos x} 
\displaystyle \Rightarrow y=e^{\sin x\log x}+e^{\cos x\log\sin x}
\displaystyle [\because a^b=e^{\log a^b}=e^{b\log a}]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=e^{\sin x\log x}\cdot\frac{d}{dx}(\sin x\cdot\log x)+e^{\cos x\log\sin x}\cdot\frac{d}{dx}(\cos x\cdot\log\sin x)
\displaystyle =x^{\sin x}\left\{\log x\cdot\frac{d}{dx}(\sin x)+\sin x\cdot\frac{d}{dx}(\log x)\right\}
\displaystyle +(\sin x)^{\cos x}\left\{\log\sin x\cdot\frac{d}{dx}(\cos x)+\cos x\cdot\frac{d}{dx}(\log\sin x)\right\}
\displaystyle =x^{\sin x}\left\{\cos x\cdot\log x+\frac{\sin x}{x}\right\}+(\sin x)^{\cos x}\left\{-\sin x\log(\sin x)+\cos x\times\frac{1}{\sin x}\times\cos x\right\}
\displaystyle =x^{\sin x}\left\{\cos x\cdot\log x+\frac{\sin x}{x}\right\}+(\sin x)^{\cos x}\left\{-\sin x\log(\sin x)+\frac{\cos^2 x}{\sin x}\right\}
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\displaystyle \textbf{Question 54. }\text{If }\log(x^2+y^2)=2\tan^{-1}\left(\dfrac{y}{x}\right)\text{, then show that}   \dfrac{dy}{dx}=\dfrac{x+y}{x-y}   \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \log(x^2+y^2)=2\tan^{-1}\left(\frac{y}{x}\right) 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{x^2+y^2}\left(2x+2y\frac{dy}{dx}\right)=2\cdot\frac{1}{1+\frac{y^2}{x^2}}\cdot\left(\frac{y'\cdot x-y}{x^2}\right)
\displaystyle \Rightarrow \frac{2(x+y\cdot y')}{x^2+y^2}=\frac{2x^2}{x^2+y^2}\left(\frac{y'x-y}{x^2}\right)\qquad\left[\because y'=\frac{dy}{dx}\right]
\displaystyle \Rightarrow x+y\cdot y'=y'x-y
\displaystyle \Rightarrow y'(x-y)=x+y
\displaystyle \Rightarrow y'=\frac{x+y}{x-y}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 55. }\text{If }x^y-y^x=a^b\text{, find }\dfrac{dy}{dx}.  \hspace{2.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x^y-y^x=a^b 
\displaystyle \text{Let }x^y=u\text{ and }y^x=v
\displaystyle \text{Then, }u-v=a^b\Rightarrow \frac{du}{dx}-\frac{dv}{dx}=0\qquad\ldots\text{(i)}
\displaystyle \text{Now, }u=x^y\Rightarrow \log u=y\log x
\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\log x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{du}{dx}=y\cdot x^{y-1}+x^y\cdot\log x\frac{dy}{dx}
\displaystyle \text{and }v=y^x\Rightarrow \log v=x\log y
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=\frac{x}{y}\frac{dy}{dx}+\log y
\displaystyle \Rightarrow \frac{dv}{dx}=xy^{x-1}\frac{dy}{dx}+y^x\log y
\displaystyle \text{Now, Eq. (i) becomes,}
\displaystyle y\cdot x^{y-1}+x^y\log x\frac{dy}{dx}-xy^{x-1}\frac{dy}{dx}-y^x\log y=0
\displaystyle \Rightarrow \frac{dy}{dx}(x^y\log x-xy^{x-1})=y^x\log y-y\cdot x^{y-1}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^x\cdot\log y-y\cdot x^{y-1}}{x^y\cdot\log x-x\cdot y^{x-1}}
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\displaystyle \textbf{Question 56. }\text{If }x=\cos t+\log\left(\tan\left(\dfrac{t}{2}\right)\right)\text{, }y=\sin t\text{, then find the values of }
\displaystyle  \dfrac{d^2y}{dt^2}\text{ and }\dfrac{d^2y}{dx^2}\text{ at }t=\dfrac{\pi}{4}. \hspace{2.2cm} \text{[CBSE 2019; CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given that, }y=\sin t 
\displaystyle \Rightarrow \frac{dy}{dt}=\cos t\qquad\text{[differentiate w.r.t. }t\text{]}
\displaystyle \Rightarrow \frac{d^2y}{dt^2}=-\sin t\qquad\text{[differentiate w.r.t. }t\text{]}
\displaystyle \left[\frac{d^2y}{dt^2}\right]_{t=\frac{\pi}{4}}=-\sin\frac{\pi}{4}=-\frac{1}{\sqrt{2}}
\displaystyle \text{Again, }x=\cos t+\log\left(\tan\left(\frac{t}{2}\right)\right)
\displaystyle \Rightarrow \frac{dx}{dt}=-\sin t+\frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\qquad\text{[differentiate w.r.t. }t\text{]}
\displaystyle =-\sin t+\frac{\cos\frac{t}{2}}{2\cdot\sin\frac{t}{2}\cdot\cos^2\frac{t}{2}}
\displaystyle =-\sin t+\frac{1}{\sin\left(2\times\frac{t}{2}\right)}\qquad[\because 2\sin a\cos a=\sin 2a]
\displaystyle =-\sin t+\text{cosec }t\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\cos t}{\text{cosec }t-\sin t}\qquad\text{[using Eqs. (i) and (ii)]}
\displaystyle =\frac{\cos t}{\frac{1-\sin^2 t}{\sin t}}\cdot\sin t=\frac{\sin t\cdot\cos t}{\cos^2 t}
\displaystyle \Rightarrow \frac{dy}{dx}=\tan t
\displaystyle \therefore \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}=\frac{\sec^2 t}{\text{cosec }t-\sin t}
\displaystyle =\frac{\sec^2 t\cdot\sin t}{\cos^2 t}=\sec^3 t\cdot\tan t
\displaystyle \Rightarrow \left[\frac{d^2y}{dx^2}\right]_{t=\frac{\pi}{4}}=\sec^3\frac{\pi}{4}\cdot\tan\frac{\pi}{4}=2\sqrt{2}\times1=2\sqrt{2}
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\displaystyle \textbf{Question 57. }\text{Differentiate }\tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)\text{ with respect to }x.    \hspace{1.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)=\tan^{-1}\left(\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\right) 
\displaystyle \left[\because 1+\cos A=2\cos^2\frac{A}{2}\text{ and }\sin A=2\sin\frac{A}{2}\cos\frac{A}{2}\right]
\displaystyle =\tan^{-1}\left(\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right)=\tan^{-1}\left(\cot\frac{x}{2}\right)
\displaystyle =\tan^{-1}\left(\tan\left(\frac{\pi}{2}-\frac{x}{2}\right)\right)
\displaystyle =\frac{\pi}{2}-\frac{x}{2}\qquad[\because\tan^{-1}(\tan\theta)=\theta]
\displaystyle \text{Now, on differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=0-\frac{1}{2}=-\frac{1}{2}
\\

\displaystyle \textbf{Question 58. }\text{Differentiate }\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\text{ with respect to }x.    \hspace{1.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=\tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan^{-1}\left(\frac{1-\tan x}{1+\tan x}\right) 
\displaystyle =\tan^{-1}\left(\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right)
\displaystyle =\frac{\pi}{4}-x\qquad[\because\tan^{-1}(\tan\theta)=\theta]
\displaystyle \text{On differentiating w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=-1
\\

\displaystyle \textbf{Question 59. }\text{If }y=\sin(\sin x)\text{, prove that}   \dfrac{d^2y}{dx^2}+\tan x\dfrac{dy}{dx}+y\cos^2 x=0    \hspace{1.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\sin(\sin x)\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\cos(\sin x)\cdot\cos x\qquad\ldots\text{(ii)}
\displaystyle \text{Again, on differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\cos(\sin x)\cdot(-\sin x)+\cos x\cdot(-\sin(\sin x))\cdot\cos x
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{1}{\cos x}\cdot\left(\frac{dy}{dx}\right)(-\sin x)-y\cos^2 x\qquad\text{[using Eqs. (i) and (ii)]}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\tan x\frac{dy}{dx}-y\cos^2x
\displaystyle \Rightarrow \frac{d^2y}{dx^2}+\tan x\frac{dy}{dx}+y\cos^2x=0\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 60. }\text{If }x=a(2\theta-\sin 2\theta)\text{ and }y=a(1-\cos 2\theta)\text{, find }
\displaystyle  \dfrac{dy}{dx}\text{ when }\theta=\dfrac{\pi}{3}.  \hspace{2.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }x=a(2\theta-\sin 2\theta) 
\displaystyle \text{and }y=a(1-\cos 2\theta)
\displaystyle \text{Clearly, }\frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta}\right)}{\left(\frac{dx}{d\theta}\right)}\qquad\ldots\text{(i)}
\displaystyle \text{Here, }\frac{dx}{d\theta}=a(2-\cos 2\theta\cdot2)=2a(1-\cos 2\theta)
\displaystyle \text{and }\frac{dy}{d\theta}=a(0+2\sin 2\theta)=2a\sin 2\theta
\displaystyle \text{From Eq. (i), we get}
\displaystyle \frac{dy}{dx}=\frac{2a\sin 2\theta}{2a(1-\cos 2\theta)}
\displaystyle =\frac{\sin 2\theta}{1-\cos 2\theta}=\frac{2\sin\theta\cos\theta}{2\sin^2\theta}
\displaystyle [\because\cos 2A=1-2\sin^2A]
\displaystyle =\cot\theta
\displaystyle \text{Now, }\left(\frac{dy}{dx}\right)_{\theta=\pi/3}=\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}
\\

\displaystyle \textbf{Question 61. }\text{If }\sin y=x\cos(a+y)\text{, then show that }   \dfrac{dy}{dx}=\dfrac{\cos^2(a+y)}{\cos a}
\displaystyle \text{Also, show that }\dfrac{dy}{dx}=\cos a\text{, when }x=0.   \hspace{2.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin y=x\cos(a+y)\qquad\ldots\text{(i)} 
\displaystyle \Rightarrow x=\frac{\sin y}{\cos(a+y)}
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dx}{dy}=\frac{\cos(a+y)\frac{d}{dy}(\sin y)-\sin y\frac{d}{dy}\cos(a+y)}{\cos^2(a+y)}
\displaystyle \text{[using quotient rule of derivative]}
\displaystyle \frac{dx}{dy}=\frac{\cos(a+y)\cos y+\sin y\sin(a+y)}{\cos^2(a+y)}
\displaystyle =\frac{\cos(a+y-y)}{\cos^2(a+y)}
\displaystyle [\because\cos A\cos B+\sin A\sin B=\cos(A-B)]
\displaystyle =\frac{\cos a}{\cos^2(a+y)}\Rightarrow \frac{dy}{dx}=\frac{\cos^2(a+y)}{\cos a}
\displaystyle \text{Put }x=0\text{ in Eq. (i), we get }y=0
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{\cos^2(a+0)}{\cos a}=\frac{\cos^2 a}{\cos a}=\cos a\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 62. }\text{If }x=a\sec^3\theta\text{ and }y=a\tan^3\theta\text{, find }\dfrac{d^2y}{dx^2}\text{ at }\theta=\dfrac{\pi}{3}.   \hspace{1.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=a\tan^3\theta\text{ and }x=a\sec^3\theta 
\displaystyle \text{On differentiating w.r.t. }\theta\text{, we get}
\displaystyle \frac{dy}{d\theta}=3a\tan^2\theta\cdot\frac{d}{d\theta}(\tan\theta)
\displaystyle \Rightarrow \frac{dy}{d\theta}=3a\tan^2\theta\sec^2\theta
\displaystyle \text{and }\frac{dx}{d\theta}=3a\sec^2\theta\cdot\frac{d}{d\theta}(\sec\theta)=3a\sec^2\theta\sec\theta\tan\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{3a\tan^2\theta\sec^2\theta}{3a\sec^2\theta\sec\theta\tan\theta}
\displaystyle \frac{dy}{dx}=\frac{\tan\theta}{\sec\theta}=\sin\theta
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(\sin\theta)=\frac{d}{d\theta}(\sin\theta)\cdot\frac{d\theta}{dx}
\displaystyle =\frac{\cos\theta}{3a\sec^3\theta\tan\theta}=\frac{\cos^5\theta}{3a\sin\theta}
\displaystyle \text{At }\theta=\frac{\pi}{3}\text{,}
\displaystyle \frac{d^2y}{dx^2}=\frac{\cos^5\frac{\pi}{3}}{3a\sin\frac{\pi}{3}}=\frac{\left(\frac{1}{2}\right)^5}{3a\cdot\frac{\sqrt{3}}{2}}=\frac{\frac{1}{32}}{\frac{3a\sqrt{3}}{2}}=\frac{1}{2^5\times3a\sqrt{3}}=\frac{1}{48\sqrt{3}\,a}
\\

\displaystyle \textbf{Question 63. }\text{If }y=e^{\tan^{-1}x}\text{, prove that }(1+x^2)\dfrac{d^2y}{dx^2}+(2x-1)\dfrac{dy}{dx}=0.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }y=e^{\tan^{-1}x} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=e^{\tan^{-1}x}\cdot\frac{d}{dx}(\tan^{-1}x)
\displaystyle \Rightarrow \frac{dy}{dx}=e^{\tan^{-1}x}\times\frac{1}{(1+x^2)}
\displaystyle \Rightarrow (1+x^2)\frac{dy}{dx}=e^{\tan^{-1}x}\qquad\ldots\text{(i)}
\displaystyle  \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle (1+x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}=e^{\tan^{-1}x}\times\frac{1}{(1+x^2)}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+(2x)\frac{dy}{dx}=\frac{dy}{dx}\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+(2x-1)\frac{dy}{dx}=0\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 64. }\text{Find }\frac{dy}{dx}\text{ at }x=1,\ y=\frac{\pi}{4},\text{ if }\sin^2y+\cos xy=K.    \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\sin^2y+\cos xy=K 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d}{dx}(\sin^2y+\cos xy)=\frac{d}{dx}(K)
\displaystyle \Rightarrow \frac{d}{dx}(\sin^2y)+\frac{d}{dx}(\cos xy)=0
\displaystyle \Rightarrow 2\sin y\cos y\frac{dy}{dx}+(-\sin xy)\cdot\frac{d}{dx}(xy)=0
\displaystyle \Rightarrow \sin 2y\frac{dy}{dx}-\sin xy\left(x\frac{dy}{dx}+y\cdot1\right)=0
\displaystyle \Rightarrow \sin 2y\frac{dy}{dx}-x\sin xy\frac{dy}{dx}=y\sin xy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y\sin(xy)}{\sin 2y-x\sin(xy)}
\displaystyle \therefore \left(\frac{dy}{dx}\right)\text{ at }x=1,\ y=\frac{\pi}{4}=\frac{\frac{\pi}{4}\sin\left(1\cdot\frac{\pi}{4}\right)}{\sin\left(2\cdot\frac{\pi}{4}\right)-1\cdot\sin\left(1\cdot\frac{\pi}{4}\right)}
\displaystyle =\frac{\frac{\pi}{4}\sin\left(\frac{\pi}{4}\right)}{\sin\left(\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{4}\right)}=\frac{\frac{\pi}{4}\cdot\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}
\displaystyle =\frac{\frac{\pi}{4\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}}=\frac{\pi}{4\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}-1}=\frac{\pi}{4(\sqrt{2}-1)}
\\

\displaystyle \textbf{Question 65. }\text{If }y=\sin^{-1}(6x\sqrt{1-9x^2}),\ -\frac{1}{3\sqrt2}<x<\frac{1}{3\sqrt2}, \text{ then find }\frac{dy}{dx}.
\displaystyle     \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\sin^{-1}(6x\sqrt{1-9x^2}) 
\displaystyle \Rightarrow y=\sin^{-1}(2\cdot3x\sqrt{1-(3x)^2})
\displaystyle \text{On putting }3x=\sin\theta\text{, then}
\displaystyle y=\sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})
\displaystyle \Rightarrow y=\sin^{-1}(2\sin\theta\cdot\cos\theta)
\displaystyle \Rightarrow y=\sin^{-1}(\sin 2\theta)\Rightarrow y=2\theta
\displaystyle \Rightarrow y=2\sin^{-1}(3x)\qquad[\because\theta=\sin^{-1}(3x)]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2}{\sqrt{1-9x^2}}\cdot(3)\Rightarrow \frac{dy}{dx}=\frac{6}{\sqrt{1-9x^2}}
\\

\displaystyle \textbf{Question 66. }\text{If }x^y+y^x=a^b\text{, then find }\dfrac{dy}{dx}.    \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{If }x^y+y^x=a^b\text{, then find }\frac{dy}{dx} 
\displaystyle \text{Let }u=x^y\text{ and }v=y^x
\displaystyle \text{Then, }u+v=a^b\Rightarrow \frac{du}{dx}+\frac{dv}{dx}=0\qquad\ldots\text{(i)}
\displaystyle \text{Now, }u=x^y\Rightarrow \log u=y\log x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\log x\frac{dy}{dx}
\displaystyle \Rightarrow \frac{du}{dx}=x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)\qquad\ldots\text{(ii)}
\displaystyle \text{And }v=y^x\Rightarrow \log v=x\log y
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{v}\frac{dv}{dx}=\frac{x}{y}\frac{dy}{dx}+\log y
\displaystyle \Rightarrow \frac{dv}{dx}=y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)\qquad\ldots\text{(iii)}
\displaystyle \text{From Eqs. (i), (ii) and (iii), we get}
\displaystyle x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)+y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}\left(x^y\log x+xy^{x-1}\right)=-\left(yx^{y-1}+y^x\log y\right)
\displaystyle \therefore \frac{dy}{dx}=-\frac{yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}
\\

\displaystyle \textbf{Question 67. }\text{If }e^y(x+1)=1\text{, then show that }\dfrac{d^2y}{dx^2}=\left(\dfrac{dy}{dx}\right)^2.    \hspace{2.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }e^y(x+1)=1 
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log[e^y(x+1)]=\log 1
\displaystyle \Rightarrow \log e^y+\log(x+1)=\log 1
\displaystyle \Rightarrow y+\log(x+1)=\log 1\qquad[\because\log e^y=y]
\displaystyle \frac{dy}{dx}+\frac{1}{x+1}=0\qquad\ldots\text{(i)}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}-\frac{1}{(x+1)^2}=0
\displaystyle \Rightarrow \frac{d^2y}{dx^2}-\left(-\frac{dy}{dx}\right)^2=0\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}-\left(\frac{dy}{dx}\right)^2=0
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 68. }\text{If }y=x^x\text{, then prove that}   \dfrac{d^2y}{dx^2}-\dfrac{1}{y}\left(\dfrac{dy}{dx}\right)^2-\dfrac{y}{x}=0.   \hspace{1.2cm} \text{[CBSE 2016, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=x^x 
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log y=\log x^x\Rightarrow \log y=x\log x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=x\cdot\frac{d}{dx}(\log x)+\log x\cdot\frac{d}{dx}(x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\times\frac{1}{x}+\log x\cdot1
\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=(1+\log x)
\displaystyle \Rightarrow \frac{dy}{dx}=y(1+\log x)\qquad\ldots\text{(i)}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=y\cdot\frac{d}{dx}(1+\log x)+(1+\log x)\frac{dy}{dx}
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=y\times\frac{1}{x}+(1+\log x)\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{y}{x}+\frac{1}{y}\left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)\qquad\text{[using Eq. (i)]}
\displaystyle \therefore \frac{d^2y}{dx^2}-\frac{1}{y}\left(\frac{dy}{dx}\right)^2-\frac{y}{x}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 69. }\text{Differentiate }\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)\text{ with respect to }
\displaystyle  \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)\text{, when }x\neq 0. \hspace{2.2cm} \text{[CBSE 2016, 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }u=\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) 
\displaystyle \text{Put }x=\tan\theta\Rightarrow\theta=\tan^{-1}x\text{, then}
\displaystyle u=\tan^{-1}\left[\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right]=\tan^{-1}\left[\frac{\sqrt{\sec^2\theta}-1}{\tan\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{\sec\theta-1}{\tan\theta}\right]=\tan^{-1}\left[\frac{1-\cos\theta}{\sin\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{2\sin^2\theta/2}{2\sin\theta/2\cdot\cos\theta/2}\right]=\tan^{-1}[\tan\theta/2]
\displaystyle \Rightarrow u=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x\qquad[\because\tan^{-1}(\tan\theta)=\theta]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{du}{dx}=\frac{1}{2(1+x^2)}\qquad\left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\right]
\displaystyle \text{Also, let }v=\sin^{-1}\left(\frac{2x}{1+x^2}\right)
\displaystyle \text{Put }x=\tan\theta\Rightarrow\theta=\tan^{-1}x\text{, then we get}
\displaystyle v=\sin^{-1}\left[\frac{2\tan\theta}{1+\tan^2\theta}\right]
\displaystyle \Rightarrow v=\sin^{-1}[\sin 2\theta]
\displaystyle \Rightarrow v=2\theta\Rightarrow v=2\tan^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=\frac{2}{1+x^2}\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}=\frac{1}{2(1+x^2)}\times\frac{(1+x^2)}{2}
\displaystyle \text{[from Eqs. (i) and (ii)]}
\displaystyle \therefore \frac{du}{dv}=\frac{1}{4}
\\

\displaystyle \textbf{Question 70. }\text{If }x=a\sin 2t(1+\cos 2t)\text{ and }y=b\cos 2t(1-\cos 2t), \\ \text{then find the values of }\dfrac{dy}{dx}\text{ at }t=\dfrac{\pi}{4}\text{ and }t=\dfrac{\pi}{3}.
\displaystyle \text{If }x=a\sin 2t(1+\cos 2t)\text{ and }y=b\cos 2t(1-\cos 2t)\text{, then show that at } \\ t=\dfrac{\pi}{4}\text{, }\dfrac{dy}{dx}=\dfrac{b}{a}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2016; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a\sin2t(1+\cos2t) 
\displaystyle \text{and }y=b\cos2t(1-\cos2t)
\displaystyle \text{On differentiating }x\text{ and }y\text{ separately w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=a\left[\sin2t\cdot\frac{d}{dt}(1+\cos2t)+(1+\cos2t)\cdot\frac{d}{dt}(\sin2t)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =a[\sin2t\times(0-2\sin2t)+(1+\cos2t)(2\cos2t)]
\displaystyle =a(-2\sin^22t+2\cos2t+2\cos^22t)
\displaystyle =a[2(\cos^22t-\sin^22t)+2\cos2t]
\displaystyle =a(2\cos4t+2\cos2t)\qquad[\because\cos^2\theta-\sin^2\theta=\cos2\theta]
\displaystyle =2a\left[2\cos\left(\frac{4t+2t}{2}\right)\cdot\cos\left(\frac{4t-2t}{2}\right)\right]
\displaystyle \left[\because\cos x+\cos y=2\cos\left(\frac{x+y}{2}\right)\cdot\cos\left(\frac{x-y}{2}\right)\right]
\displaystyle =4a\cos3t\cos t
\displaystyle \text{and }\frac{dy}{dt}=b\left[\cos2t\cdot\frac{d}{dt}(1-\cos2t)+(1-\cos2t)\cdot\frac{d}{dt}(\cos2t)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =b[\cos2t\times(0+2\sin2t)+(1-\cos2t)(-2\sin2t)]
\displaystyle =b(2\sin2t\cos2t-2\sin2t+2\sin2t\cos2t)
\displaystyle =2b(2\sin2t\cos2t-\sin2t)
\displaystyle =2b(\sin4t-\sin2t)\qquad[\because2\sin2\theta\cos2\theta=\sin4\theta]
\displaystyle =2b\left[2\cos\left(\frac{4t+2t}{2}\right)\sin\left(\frac{4t-2t}{2}\right)\right]
\displaystyle \left[\because\sin x-\sin y=2\cos\left(\frac{x+y}{2}\right)\cdot\sin\left(\frac{x-y}{2}\right)\right]
\displaystyle =4b\cos3t\sin t
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4b\cos3t\sin t}{4a\cos3t\cdot\cos t}=\frac{b}{a}\tan t
\displaystyle \text{At }t=\frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}\tan\frac{\pi}{4}=\frac{b}{a}
\displaystyle \text{At }t=\frac{\pi}{3},\ \frac{dy}{dx}=\frac{b}{a}\tan\frac{\pi}{3}=\frac{\sqrt{3}\,b}{a}
\\

\displaystyle \textbf{Question 71. }\text{If }x\cos(a+y)=\cos y\text{, then prove that }\dfrac{dy}{dx}=\dfrac{\cos^2(a+y)}{\sin a}\text{. Hence, show that}
\displaystyle \sin a\dfrac{d^2y}{dx^2}+\sin 2(a+y)\dfrac{dy}{dx}=0.
\displaystyle \text{If }\cos y=x\cos(a+y)\text{, where }\cos a\neq\pm 1\text{, prove that }\dfrac{dy}{dx}=\dfrac{\cos^2(a+y)}{\sin a}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2016; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x\cos(a+y)=\cos y\Rightarrow x=\frac{\cos y}{\cos(a+y)} 
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dx}{dy}=\frac{\cos(a+y)\cdot\frac{d}{dy}\cos y-\cos y\cdot\frac{d}{dy}\cos(a+y)}{\cos^2(a+y)}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =\frac{\cos(a+y)\times(-\sin y)+\cos y\times\sin(a+y)}{\cos^2(a+y)}
\displaystyle =\frac{\sin(a+y)\cos y-\cos(a+y)\sin y}{\cos^2(a+y)}
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{\sin(a+y-y)}{\cos^2(a+y)}=\frac{\sin a}{\cos^2(a+y)}
\displaystyle [\because\sin A\cos B-\cos A\sin B=\sin(A-B)]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}\qquad\ldots\text{(i)}
\displaystyle \text{Again, on differentiating both sides of Eq. (i) w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{1}{\sin a}\cdot\frac{d}{dx}\cos^2(a+y)
\displaystyle =\frac{1}{\sin a}\times\frac{d}{dy}\cos^2(a+y)\times\frac{dy}{dx}
\displaystyle =\frac{1}{\sin a}\times2\cos(a+y)[-\sin(a+y)]\times\frac{dy}{dx}
\displaystyle =-\frac{2\sin(a+y)\cos(a+y)}{\sin a}\times\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\frac{\sin2(a+y)}{\sin a}\cdot\frac{dy}{dx}\qquad[\because2\sin\theta\cos\theta=\sin2\theta]
\displaystyle \therefore \sin a\cdot\frac{d^2y}{dx^2}+\sin2(a+y)\cdot\frac{dy}{dx}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 72. }\text{Find }\dfrac{dy}{dx}\text{, if }y=\sin^{-1}\left[\dfrac{6x-4\sqrt{1-4x^2}}{5}\right].   \hspace{2.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\sin^{-1}\left(\frac{6x-4\sqrt{1-4x^2}}{5}\right) 
\displaystyle \text{On putting }x=\frac{1}{2}\sin\theta\text{, we have}
\displaystyle \therefore y=\sin^{-1}\left[\frac{6\times\frac{\sin\theta}{2}-4\sqrt{1-4\times\left(\frac{\sin\theta}{2}\right)^2}}{5}\right]
\displaystyle =\sin^{-1}\left(\frac{3\sin\theta-4\sqrt{1-\sin^2\theta}}{5}\right)
\displaystyle =\sin^{-1}\left(\frac{3\sin\theta-4\cos\theta}{5}\right)
\displaystyle =\sin^{-1}\left(\frac{3}{5}\sin\theta-\frac{4}{5}\cos\theta\right)\qquad\ldots\text{(i)}
\displaystyle \text{Let }\cos\phi=\frac{3}{5}\text{, then }\sin\phi=\sqrt{1-\cos^2\phi}=\sqrt{1-\left(\frac{3}{5}\right)^2}
\displaystyle =\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
\displaystyle \text{Now, Eq. (i) becomes}
\displaystyle y=\sin^{-1}(\cos\phi\sin\theta-\sin\phi\cos\theta)
\displaystyle =\sin^{-1}[\sin(\theta-\phi)]=\theta-\phi
\displaystyle \Rightarrow y=\sin^{-1}(2x)-\cos^{-1}\left(\frac{3}{5}\right)
\displaystyle \left[\because x=\frac{1}{2}\sin\theta\Rightarrow\sin\theta=2x\Rightarrow\theta=\sin^{-1}(2x)\right.
\displaystyle \left.\text{and }\cos\phi=\frac{3}{5}\Rightarrow\phi=\cos^{-1}\frac{3}{5}\right]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-(2x)^2}}\cdot\frac{d}{dx}(2x)-0=\frac{2}{\sqrt{1-4x^2}}
\\

\displaystyle \textbf{Question 73. }\text{Find the values of }a\text{ and }b\text{, if the function }f\text{ defined by}
\displaystyle f(x)=\begin{cases} x^2+3x+a, & x\leq 1 \\ bx+2, & x>1 \end{cases}\text{ is differentiable at }x=1.   \hspace{2.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}x^2+3x+a, & x\leq1\\bx+2, & x>1\end{cases}\text{ is differentiable at }x=1   
\displaystyle \therefore Lf'(1)=Rf'(1)\qquad\ldots\text{(i)}
\displaystyle \text{Here, }Lf'(1)=\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h\to0}\frac{(1-h)^2+3(1-h)+a-(4+a)}{-h}
\displaystyle =\lim_{h\to0}\frac{1+h^2-2h+3-3h+a-4-a}{-h}
\displaystyle =\lim_{h\to0}\frac{h^2-5h}{-h}
\displaystyle =\lim_{h\to0}5-h=5
\displaystyle \text{and }Rf'(1)=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to0}\frac{b(1+h)+2-(4+a)}{h}
\displaystyle =\lim_{h\to0}\frac{b+bh+2-4-a}{h}
\displaystyle =\lim_{h\to0}\frac{bh+b-a-2}{h}
\displaystyle \text{Clearly, for }Rf'(1)\text{ to be exist }b-a-2\text{ should be equal to }0\text{, i.e.}
\displaystyle b-a-2=0\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }Rf'(1)=\lim_{h\to0}\frac{bh}{h}=\lim_{h\to0}b=b
\displaystyle \text{From Eq. (i), we have}
\displaystyle Lf'(1)=Rf'(1)
\displaystyle \Rightarrow 5=b\Rightarrow b=5
\displaystyle \text{Now, on substituting }b=5\text{ in Eq. (ii), we get}
\displaystyle 5-a-2=0\Rightarrow a=3
\displaystyle \text{Hence, }a=3\text{ and }b=5
\\

\displaystyle \textbf{Question 74. }\text{If }x=\sin t\text{ and }y=\sin pt\text{, then prove that}
\displaystyle (1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+p^2y=0.    \hspace{2.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=\sin t\text{ and }y=\sin pt 
\displaystyle \text{On differentiating }x\text{ and }y\text{ separately w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=\cos t\text{ and }\frac{dy}{dt}=\cos pt\cdot p\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos pt\cdot p}{\cos t}
\displaystyle \text{Now, on differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=p\cdot\frac{[\cos t(-\sin pt\cdot p)-\cos pt(-\sin t)]}{\cos^2 t}\cdot\frac{dt}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{p[\cos pt\cdot\sin t-\cos t\sin pt\cdot p]}{\cos^2 t}\cdot\frac{1}{\cos t}
\displaystyle \Rightarrow \cos^2 t\frac{d^2y}{dx^2}=\frac{p\cos pt\cdot\sin t-p^2\cos t\sin pt}{\cos t}
\displaystyle \Rightarrow (1-\sin^2 t)\frac{d^2y}{dx^2}=\frac{p\cos pt}{\cos t}\cdot\sin t-p^2\sin pt
\displaystyle \Rightarrow (1-x^2)\frac{d^2y}{dx^2}=\frac{dy}{dx}\cdot x-p^2y
\displaystyle \left[\because\frac{dy}{dx}=\frac{p\cos pt}{\cos t},\ x=\sin t\text{ and }y=\sin pt\right]
\displaystyle \therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 75. }\text{If }y=\tan^{-1}\left(\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\text{, }x^2\leq 1\text{, then find }dy/dx.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=\tan^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) 
\displaystyle \text{On putting }x^2=\sin\theta
\displaystyle \Rightarrow \theta=\sin^{-1}x^2
\displaystyle \therefore y=\tan^{-1}\left(\frac{\sqrt{1+\sin\theta}+\sqrt{1-\sin\theta}}{\sqrt{1+\sin\theta}-\sqrt{1-\sin\theta}}\right)
\displaystyle =\tan^{-1}\left(\frac{\sqrt{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}+\sqrt{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}}{\sqrt{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}-\sqrt{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}}\right)
\displaystyle =\tan^{-1}\left(\frac{\sqrt{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}+\sqrt{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}}{\sqrt{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}-\sqrt{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}}\right)
\displaystyle =\tan^{-1}\left(\frac{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)+\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)-\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}\right)
\displaystyle =\tan^{-1}\left(\frac{2\cos\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\right)=\tan^{-1}\left(\cot\frac{\theta}{2}\right)
\displaystyle =\tan^{-1}\left(\tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right)\qquad\left[\because\cot\theta=\tan\left(\frac{\pi}{2}-\theta\right)\right]
\displaystyle =\frac{\pi}{2}-\frac{\theta}{2}\qquad[\because\tan^{-1}(\tan\theta)=\theta]
\displaystyle \Rightarrow y=\frac{\pi}{2}-\frac{1}{2}\sin^{-1}x^2\qquad[\because\theta=\sin^{-1}x^2]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{\sqrt{1-(x^2)^2}}\cdot(2x)=\frac{-x}{\sqrt{1-x^4}}
\\

\displaystyle \textbf{Question 76. }\text{If }x=a\cos\theta+b\sin\theta\text{, }y=a\sin\theta-b\cos\theta\text{, then show that}
\displaystyle y^2\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+y=0.   \hspace{2.2cm} \text{[CBSE 2015; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a\cos\theta+b\sin\theta\qquad\ldots\text{(i)} 
\displaystyle \text{and }y=a\sin\theta-b\cos\theta\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eqs. (i) and (ii) w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=-a\sin\theta+b\cos\theta\qquad\ldots\text{(iii)}
\displaystyle \text{and }\frac{dy}{d\theta}=a\cos\theta+b\sin\theta\qquad\ldots\text{(iv)}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\cos\theta+b\sin\theta}{b\cos\theta-a\sin\theta}
\displaystyle \text{[dividing Eq. (iv) by Eq. (iii)]}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x}{-y}\qquad\text{[from Eqs. (i) and (ii)]}\qquad\ldots\text{(v)}
\displaystyle \text{Again, differentiating both sides of Eq. (v) w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=-\left[\frac{y\cdot1-x\cdot\frac{dy}{dx}}{y^2}\right]
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle \Rightarrow y^2\frac{d^2y}{dx^2}=-\left[y-x\frac{dy}{dx}\right]
\displaystyle \therefore y^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 77. }\text{Show that the function }f(x)=|x+1|+|x-1|\text{, for all }x\in R\text{, is not differentiable at the points }x=-1\text{ and }x=1.    \hspace{2.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=|x+1|+|x-1|,\ \forall x\in R 
\displaystyle \text{It can be rewritten as}
\displaystyle f(x)=\begin{cases}-(x+1)-(x-1), & x<-1\\(x+1)-(x-1), & -1\leq x<1\\(x+1)+(x-1), & x\geq1\end{cases}
\displaystyle =\begin{cases}-2x, & x<-1\\2, & -1\leq x<1\\2x, & x\geq1\end{cases}
\displaystyle \textbf{Differentiability at }x=-1
\displaystyle Lf'(-1)=\lim_{h\to0}\frac{f(-1-h)-f(-1)}{-h}
\displaystyle =\lim_{h\to0}\frac{-2(-1-h)-2}{-h}
\displaystyle =\lim_{h\to0}\frac{2+2h-2}{-h}
\displaystyle =\lim_{h\to0}(-2)=-2
\displaystyle Rf'(-1)=\lim_{h\to0}\frac{f(-1+h)-f(-1)}{h}
\displaystyle =\lim_{h\to0}\frac{2-2}{h}=\lim_{h\to0}\frac{0}{h}=0
\displaystyle \therefore Lf'(-1)\neq Rf'(-1)
\displaystyle \therefore f\text{ is not differentiable at }x=-1
\displaystyle \textbf{Differentiability at }x=1
\displaystyle Lf'(1)=\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h\to0}\frac{2-2}{-h}=\lim_{h\to0}\frac{0}{-h}=0
\displaystyle \text{and }Rf'(1)=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to0}\frac{2(1+h)-2}{h}
\displaystyle =\lim_{h\to0}\frac{2+2h-2}{h}
\displaystyle =\lim_{h\to0}\frac{2h}{h}=2
\displaystyle \therefore Lf'(1)\neq Rf'(1)
\displaystyle \therefore f\text{ is not differentiable at }x=1
\displaystyle \text{Hence, }f\text{ is not differentiable at }x=1\text{ and }-1
\\

\displaystyle \textbf{Question 78. }\text{If }y=e^{m\sin^{-1}x}\text{, then show that}   (1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-m^2y=0.    \hspace{1.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=e^{m\sin^{-1}x}\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides of Eq. (i) w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=e^{m\sin^{-1}x}\cdot\frac{d}{dx}(m\sin^{-1}x)
\displaystyle \text{[by using chain rule of derivative]}
\displaystyle =e^{m\sin^{-1}x}\cdot m\cdot\frac{1}{\sqrt{1-x^2}}
\displaystyle \Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=my\qquad\text{[from Eq. (i)]}
\displaystyle \text{Now, on squaring both sides, we get}
\displaystyle (1-x^2)\left(\frac{dy}{dx}\right)^2=m^2y^2\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eq. (ii) w.r.t. }x\text{, we get}
\displaystyle (1-x^2)\cdot2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2(-2x)=2m^2y\left(\frac{dy}{dx}\right)
\displaystyle \Rightarrow (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=m^2y
\displaystyle \left[\text{dividing both sides by }2\left(\frac{dy}{dx}\right)\right]
\displaystyle \therefore (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-m^2y=0\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 79. }\text{If }y=\left(x+\sqrt{1+x^2}\right)^n\text{, then show that}   (1+x^2)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=n^2y.    \hspace{0.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=(x+\sqrt{1+x^2})^n\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=n(x+\sqrt{1+x^2})^{n-1}\left(1+\frac{2x}{2\sqrt{1+x^2}}\right)
\displaystyle \text{[by using chain rule of derivative]}
\displaystyle \Rightarrow \frac{dy}{dx}=n(x+\sqrt{1+x^2})^{n-1}\left(\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{n(x+\sqrt{1+x^2})^n}{\sqrt{1+x^2}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{ny}{\sqrt{1+x^2}}\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow \sqrt{1+x^2}\frac{dy}{dx}=ny\qquad\ldots\text{(ii)}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \sqrt{1+x^2}\frac{d^2y}{dx^2}+\frac{2x}{2\sqrt{1+x^2}}\cdot\frac{dy}{dx}=n\cdot\frac{dy}{dx}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=n\cdot\sqrt{1+x^2}\frac{dy}{dx}
\displaystyle \text{[multiplying both sides by }\sqrt{1+x^2}\text{]}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=n\sqrt{1+x^2}\cdot\frac{ny}{\sqrt{1+x^2}}
\displaystyle \text{[from Eq. (ii)]}
\displaystyle \therefore (1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=n^2y\qquad \text{Hence proved.}
\\

\displaystyle \textbf{Question 80. }\text{Find whether the following function is differentiable at }
\displaystyle x=1\text{ and }x=2\text{ or not. } f(x)=\begin{cases} x, & x<1 \\ 2-x, & 1\leq x\leq 2 \\ -2+3x-x^2, & x>2 \end{cases}    \hspace{2.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }f(x)=\begin{cases}x, & x<1\\2-x, & 1\leq x\leq2\\-2+3x-x^2, & x>2\end{cases} 
\displaystyle \textbf{Differentiability at }x=1
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h\to0}\frac{(1-h)-[2-(1)]}{-h}=\lim_{h\to0}\frac{-h}{-h}=1
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to0}\frac{2-(1+h)-(2-1)}{h}
\displaystyle =\lim_{h\to0}\frac{-h}{h}=-1
\displaystyle \therefore \text{LHD}\neq\text{RHD}
\displaystyle \text{So, }f(x)\text{ is not differentiable at }x=1
\displaystyle \textbf{Differentiability at }x=2
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(2-h)-f(2)}{-h}
\displaystyle =\lim_{h\to0}\frac{2-(2-h)-(2-2)}{-h}
\displaystyle =\lim_{h\to0}\frac{h}{-h}=-1
\displaystyle \text{RHD}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}
\displaystyle =\lim_{h\to0}\frac{-2+3(2+h)-(2+h)^2-(2-2)}{h}
\displaystyle =\lim_{h\to0}\frac{-2+6+3h-(4+h^2+4h)-0}{h}
\displaystyle =\lim_{h\to0}\frac{-h^2-h}{h}
\displaystyle =\lim_{h\to0}\frac{-h(h+1)}{h}=-(0+1)=-1
\displaystyle \therefore \text{LHD}=\text{RHD}
\displaystyle \text{So, }f(x)\text{ is differentiable at }x=2
\displaystyle \text{Hence, }f(x)\text{ is not differentiable at }x=1\text{, but it is differentiable at }x=2
\\

\displaystyle \textbf{Question 81. }\text{For what value of }\lambda\text{, function defined by}
\displaystyle f(x)=\begin{cases} \lambda(x^2+2), & \text{if }x\leq 0 \\ 4x+6, & \text{if }x>0 \end{cases}\text{ is continuous at }x=0\text{?}
\displaystyle \text{Hence, check the differentiability of }f(x)\text{ at }x=0.    \hspace{2.2cm} \text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, function is }f(x)=\begin{cases}\lambda(x^2+2), & \text{if }x\leq0\\4x+6, & \text{if }x>0\end{cases} 
\displaystyle \text{Let }f(x)\text{ is continuous at }x=0
\displaystyle \text{Then, LHL}=\text{RHL}=f(0)\qquad\ldots\text{(i)}
\displaystyle \text{Here, RHL}=\lim_{x\to0^+}f(x)=\lim_{x\to0^+}(4x+6)
\displaystyle =\lim_{h\to0}[4(0+h)+6]
\displaystyle \text{[put }x=0+h\text{; when }x\to0^+\text{, then }h\to0\text{]}
\displaystyle =\lim_{h\to0}(4h+6)=4\times0+6=6
\displaystyle \text{and }f(0)=\lambda(0^2+2)=2\lambda
\displaystyle \text{From Eq. (i), RHL}=f(0)
\displaystyle \Rightarrow 2\lambda=6\Rightarrow\lambda=3
\displaystyle \text{Now, given function becomes}
\displaystyle f(x)=\begin{cases}3(x^2+2), & \text{if }x\leq0\\4x+6, & \text{if }x>0\end{cases}
\displaystyle \text{Now, let us check the differentiability at }x=0
\displaystyle \text{LHD}=\lim_{h\to0}\frac{f(0-h)-f(0)}{-h}
\displaystyle =\lim_{h\to0}\frac{3[(0-h)^2+2]-3(0+2)}{-h}
\displaystyle =\lim_{h\to0}\frac{3[h^2+2]-6}{-h}=\lim_{h\to0}(-3h)=0
\displaystyle \text{and RHD}=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim_{h\to0}\frac{[4(0+h)+6]-3(0+2)}{h}=\lim_{h\to0}\frac{4h}{h}=4
\displaystyle \therefore \text{LHD}\neq\text{RHD}
\displaystyle \therefore f(x)\text{ is not differentiable at }x=0
\\

\displaystyle \textbf{Question 82. }\text{If }y=(\sin x)^x+\sin^{-1}\sqrt{x}\text{, then find }\dfrac{dy}{dx}.    \hspace{2.2cm} \text{[CBSE 2015C, 2013C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=(\sin x)^x+\sin^{-1}\sqrt{x}\qquad\ldots\text{(i)} 
\displaystyle \text{Let }u=(\sin x)^x\qquad\ldots\text{(ii)}
\displaystyle \text{Then, Eq. (i) becomes, }y=u+\sin^{-1}\sqrt{x}\qquad\ldots\text{(iii)}
\displaystyle \text{On taking log both sides of Eq. (ii), we get}
\displaystyle \log u=x\log(\sin x)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=x\cdot\frac{d}{dx}(\log(\sin x))+\log(\sin x)\cdot\frac{d}{dx}(x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{du}{dx}=u\left[x\times\frac{1}{\sin x}\frac{d}{dx}(\sin x)+\log(\sin x)(1)\right]
\displaystyle \Rightarrow \frac{du}{dx}=(\sin x)^x\left[\frac{x}{\sin x}\times\cos x+\log(\sin x)\right]
\displaystyle \text{[from Eq. (ii)]}
\displaystyle \Rightarrow \frac{du}{dx}=(\sin x)^x[x\cot x+\log(\sin x)]\qquad\ldots\text{(iv)}
\displaystyle \text{On differentiating both sides of Eq. (iii) w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot\frac{d}{dx}(\sqrt{x})
\displaystyle \therefore \frac{dy}{dx}=(\sin x)^x[x\cot x+\log(\sin x)]+\frac{1}{\sqrt{1-x}}\times\frac{1}{2\sqrt{x}}\qquad\text{[from Eq. (iv)]}
\\

\displaystyle \textbf{Question 83. }\text{If }y=\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}-\log\sqrt{1-x^2}\text{, then prove that}   \dfrac{dy}{dx}=\dfrac{\cos^{-1}x}{(1-x^2)^{3/2}}.    \hspace{0.2cm} \text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }y=\frac{x\cos^{-1}x}{\sqrt{1-x^2}}-\log\sqrt{1-x^2}\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides of Eq. (i) w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\right)-\frac{d}{dx}(\log\sqrt{1-x^2})
\displaystyle \sqrt{1-x^2}\left[x\cdot\frac{(-1)}{\sqrt{1-x^2}}+\cos^{-1}x\right]
\displaystyle =\frac{-x\cos^{-1}x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot(-2x)}{(\sqrt{1-x^2})^2}
\displaystyle -\frac{1}{\sqrt{1-x^2}}\cdot\frac{(-2x)}{2\sqrt{1-x^2}}
\displaystyle =\frac{-x+\sqrt{1-x^2}\cos^{-1}x+\frac{x^2\cos^{-1}x}{\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}+\frac{x}{(\sqrt{1-x^2})^2}
\displaystyle =\frac{-x+\sqrt{1-x^2}\cos^{-1}x+\frac{x^2\cos^{-1}x}{\sqrt{1-x^2}}+x}{(\sqrt{1-x^2})^2}
\displaystyle =\frac{(1-x^2)\cos^{-1}x+x^2\cos^{-1}x}{(\sqrt{1-x^2})^3}=\frac{\cos^{-1}x}{(1-x^2)^{3/2}}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 84. }\text{Write the derivative of }\sin x\text{ with respect to }\cos x.    \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }u=\sin x 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{du}{dx}=\cos x\qquad\ldots\text{(i)}
\displaystyle \text{Also, let }v=\cos x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=-\sin x\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}=-\frac{\cos x}{\sin x}\qquad\text{[from Eqs. (i) and (ii)]}
\displaystyle \therefore \frac{du}{dv}=-\cot x
\\

\displaystyle \textbf{Question 85. }\text{If }y=\sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\right]\text{ and }0<x<1\text{, then find }\dfrac{dy}{dx}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2014C; CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=\sin^{-1}\left[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\right] 
\displaystyle \text{Above equation can be rewritten as}
\displaystyle y=\sin^{-1}\left[x\sqrt{1-(\sqrt{x})^2}-\sqrt{x}\sqrt{1-x^2}\right]
\displaystyle \text{Now, put }\sqrt{x}=\sin\theta\text{ and }x=\sin\phi\text{, so that}
\displaystyle y=\sin^{-1}[\sin\phi\sqrt{1-\sin^2\theta}-\sin\theta\sqrt{1-\sin^2\phi}]
\displaystyle \Rightarrow y=\sin^{-1}[\sin\phi\cos\theta-\sin\theta\cos\phi]
\displaystyle [\because\sqrt{1-\sin^2x}=\cos x]
\displaystyle \Rightarrow y=\sin^{-1}[\sin(\phi-\theta)]
\displaystyle [\because\sin\phi\cos\theta-\cos\phi\sin\theta=\sin(\phi-\theta)]
\displaystyle \Rightarrow y=\phi-\theta\qquad[\because\sin^{-1}\sin\theta=\theta]
\displaystyle \Rightarrow y=\sin^{-1}x-\sin^{-1}\sqrt{x}
\displaystyle [\because\phi=\sin^{-1}x\text{ and }\theta=\sin^{-1}\sqrt{x}]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-(\sqrt{x})^2}}\times\frac{d}{dx}(\sqrt{x})
\displaystyle \left[\because\frac{d}{d\theta}(\sin^{-1}\theta)=\frac{1}{\sqrt{1-\theta^2}}\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}
\displaystyle \text{Hence, }\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x-x^2}}
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\displaystyle \textbf{Question 86. }\text{If }e^x+e^y=e^{x+y}\text{, prove that }\dfrac{dy}{dx}+e^{y-x}=0.    \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }e^x+e^y=e^{x+y}\qquad\ldots\text{(i)} 
\displaystyle \text{On dividing Eq. (i) by }e^{x+y}\text{, we get}
\displaystyle e^{-y}+e^{-x}=1\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eq. (ii) w.r.t. }x\text{, we get}
\displaystyle e^{-y}\cdot\left(\frac{-dy}{dx}\right)+e^{-x}(-1)=0
\displaystyle \Rightarrow -e^{-y}\frac{dy}{dx}-e^{-x}=0\Rightarrow -e^{-y}\frac{dy}{dx}=e^{-x}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{e^{-x}}{e^{-y}}\Rightarrow \frac{dy}{dx}=-e^{(y-x)}
\displaystyle \therefore \frac{dy}{dx}+e^{(y-x)}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 87. }\text{Find the value of }\dfrac{dy}{dx}\text{ at }\theta=\dfrac{\pi}{4}\text{, if }
\displaystyle  x=ae^\theta(\sin\theta-\cos\theta)\text{ and }y=ae^\theta(\sin\theta+\cos\theta). \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x=ae^\theta(\sin\theta-\cos\theta) 
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=a\frac{d}{d\theta}[e^\theta\sin\theta-e^\theta\cos\theta]
\displaystyle =a\left[\frac{d}{d\theta}(e^\theta\sin\theta)-\frac{d}{d\theta}(e^\theta\cos\theta)\right]
\displaystyle =a\left[e^\theta\frac{d}{d\theta}(\sin\theta)+\sin\theta\frac{d}{d\theta}(e^\theta)-e^\theta\frac{d}{d\theta}(\cos\theta)-\cos\theta\frac{d}{d\theta}(e^\theta)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =a[e^\theta\cos\theta+e^\theta\sin\theta-e^\theta(-\sin\theta)-e^\theta\cos\theta]
\displaystyle =a[e^\theta\cos\theta+e^\theta\sin\theta+e^\theta\sin\theta-e^\theta\cos\theta]
\displaystyle \Rightarrow \frac{dx}{d\theta}=a[2e^\theta\sin\theta]=2ae^\theta\sin\theta\qquad\ldots\text{(i)}
\displaystyle \text{Also, we have }y=ae^\theta(\sin\theta+\cos\theta)
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dy}{d\theta}=a\left[\frac{d}{d\theta}(e^\theta\sin\theta)+\frac{d}{d\theta}(e^\theta\cos\theta)\right]
\displaystyle =a\left[e^\theta\frac{d}{d\theta}(\sin\theta)+\sin\theta\frac{d}{d\theta}(e^\theta)+e^\theta\frac{d}{d\theta}(\cos\theta)+\cos\theta\frac{d}{d\theta}(e^\theta)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =a[e^\theta\cos\theta+e^\theta\sin\theta-e^\theta\sin\theta+e^\theta\cos\theta]
\displaystyle =a[2e^\theta\cos\theta]\Rightarrow \frac{dy}{d\theta}=2ae^\theta\cos\theta\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy}{d\theta}\times\frac{d\theta}{dx}=\frac{2ae^\theta\cos\theta}{2ae^\theta\sin\theta}
\displaystyle \text{[from Eqs. (i) and (ii)]}
\displaystyle =\cot\theta
\displaystyle \text{At }\theta=\frac{\pi}{4},\ \frac{dy}{dx}=\cot\frac{\pi}{4}=1\qquad\left[\because\cot\frac{\pi}{4}=1\right]
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\displaystyle \textbf{Question 88. }\text{If }x=a\left(\cos t+\log\tan\dfrac{t}{2}\right)\text{, }y=a\sin t\text{, then evaluate }
\displaystyle  \dfrac{d^2y}{dx^2}\text{ at }t=\dfrac{\pi}{3}.  \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }x=a\left(\cos t+\log\tan\frac{t}{2}\right),\ y=a\sin t\text{, then evaluate }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{3} 
\displaystyle \frac{dx}{dt}=-\sin t+\frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}
\displaystyle =-\sin t+\frac{\cos\frac{t}{2}}{2\sin\frac{t}{2}\cdot\cos^2\frac{t}{2}}=-\sin t+\frac{1}{\sin t\cos\frac{t}{2}\cdot\frac{1}{\cos\frac{t}{2}}}
\displaystyle =-\sin t+\frac{1}{\sin t}=\frac{-\sin^2t+1}{\sin t}=\frac{\cos^2t}{\sin t}
\displaystyle \frac{dy}{dt}=a\cos t
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t}{\frac{\cos^2t}{\sin t}}=\frac{a\sin t}{\cos t}=a\tan t
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(a\tan t)\cdot\frac{dt}{dx}
\displaystyle =a\sec^2t\cdot\frac{\sin t}{\cos^2t}=\frac{a\sin t}{\cos^4t}=\frac{a\sin t\sec^4t}{1}
\displaystyle \text{At }t=\frac{\pi}{3}\text{:}
\displaystyle \left(\frac{d^2y}{dx^2}\right)_{t=\frac{\pi}{3}}=\frac{a\sin\frac{\pi}{3}}{\cos^4\frac{\pi}{3}}=\frac{a\cdot\frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^4}=\frac{\frac{a\sqrt{3}}{2}}{\frac{1}{16}}=8\sqrt{3}\,a
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\displaystyle \textbf{Question 89. }\text{If }x^my^n=(x+y)^{m+n}\text{, prove that }\dfrac{dy}{dx}=\dfrac{y}{x}.   \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x^my^n=(x+y)^{m+n} 
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log(x^my^n)=\log(x+y)^{m+n}
\displaystyle \Rightarrow \log(x^m)+\log(y^n)=(m+n)\log(x+y)
\displaystyle \Rightarrow m\log x+n\log y=(m+n)\log(x+y)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)
\displaystyle \Rightarrow \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{m}{x}-\frac{m+n}{x+y}=\frac{m+n}{x+y}\frac{dy}{dx}-\frac{n}{y}\frac{dy}{dx}
\displaystyle \Rightarrow \left[\frac{my+ny-nx-ny}{y(x+y)}\right]\frac{dy}{dx}=\frac{mx+my-mx-nx}{x(x+y)}
\displaystyle \Rightarrow \frac{dy}{dx}\left[\frac{my-nx}{y}\right]=\frac{my-nx}{x}
\displaystyle \text{Hence, }\frac{dy}{dx}=\frac{y}{x}
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\displaystyle \textbf{Question 90. }\text{Differentiate }\tan^{-1}\left(\dfrac{\sqrt{1-x^2}}{x}\right)\text{ with respect to }
\displaystyle  \cos^{-1}(2x\sqrt{1-x^2})\text{, when }x\neq 0. \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }u=\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) 
\displaystyle \text{On putting }x=\cos\theta\Rightarrow\theta=\cos^{-1}x
\displaystyle \text{Then, }u=\tan^{-1}\left(\frac{\sqrt{1-\cos^2\theta}}{\cos\theta}\right)=\tan^{-1}\left(\frac{\sqrt{\sin^2\theta}}{\cos\theta}\right)
\displaystyle [\because\cos^2\theta+\sin^2\theta=1\Rightarrow\sin^2\theta=1-\cos^2\theta]
\displaystyle =\tan^{-1}\left[\frac{\sin\theta}{\cos\theta}\right]=\tan^{-1}[\tan\theta]=\theta
\displaystyle \Rightarrow u=\cos^{-1}x\qquad[\because x=\cos\theta]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}
\displaystyle \text{Again, let }v=\cos^{-1}(2x\sqrt{1-x^2})
\displaystyle \text{On putting }x=\cos\theta\Rightarrow\theta=\cos^{-1}x
\displaystyle \text{Then, }v=\cos^{-1}[2\cos\theta\sqrt{1-\cos^2\theta}]
\displaystyle =\cos^{-1}[2\cos\theta\sin\theta]\qquad\left[\because\sin\theta=\sqrt{1-\cos^2\theta}\Rightarrow\sin^2\theta=1-\cos^2\theta\right]
\displaystyle =\cos^{-1}[\sin2\theta]
\displaystyle =\cos^{-1}\left[\cos\left(\frac{\pi}{2}-2\theta\right)\right]=\frac{\pi}{2}-2\theta
\displaystyle \Rightarrow v=\frac{\pi}{2}-2\cos^{-1}x\qquad[\because\theta=\cos^{-1}x]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}
\displaystyle \text{Now, }\frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}=-\frac{1}{\sqrt{1-x^2}}\times\frac{\sqrt{1-x^2}}{2}=-\frac{1}{2}
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\displaystyle \textbf{Question 91. }\text{Differentiate }\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\text{ with respect to }\sin^{-1}(2x\sqrt{1-x^2}).
\displaystyle  \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }u=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) 
\displaystyle \text{On putting }x=\sin\theta\Rightarrow\theta=\sin^{-1}x\text{, then}
\displaystyle u=\tan^{-1}\left[\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\right]
\displaystyle \Rightarrow u=\tan^{-1}\left[\frac{\sin\theta}{\cos\theta}\right]\qquad\left[\because\sin^2A+\cos^2A=1\Rightarrow\cos A=\sqrt{(1-\sin^2A)}\right]
\displaystyle \Rightarrow u=\tan^{-1}(\tan\theta)\Rightarrow u=\theta
\displaystyle \Rightarrow u=\sin^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\qquad\ldots\text{(i)}
\displaystyle \text{Again, let }v=\sin^{-1}(2x\sqrt{1-x^2})
\displaystyle \text{On putting }x=\sin\theta\Rightarrow\theta=\sin^{-1}x\text{, then}
\displaystyle v=\sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})
\displaystyle \Rightarrow v=\sin^{-1}(2\sin\theta\cos\theta)
\displaystyle \Rightarrow v=\sin^{-1}(\sin2\theta)\Rightarrow v=2\theta
\displaystyle \Rightarrow v=2\sin^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}\Rightarrow\frac{du}{dv}=\frac{1}{\sqrt{1-x^2}}\times\frac{\sqrt{1-x^2}}{2}
\displaystyle \therefore \frac{du}{dv}=\frac{1}{2}
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\displaystyle \textbf{Question 92. }\text{If }y=Pe^{ax}+Qe^{bx}\text{, then show that }   \dfrac{d^2y}{dx^2}-(a+b)\dfrac{dy}{dx}+aby=0.    \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=Pe^{ax}+Qe^{bx}\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=P\frac{d}{dx}(e^{ax})+Q\frac{d}{dx}(e^{bx})
\displaystyle \Rightarrow \frac{dy}{dx}=Pae^{ax}+Qbe^{bx}\qquad\ldots\text{(ii)}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=Pa\frac{d}{dx}(e^{ax})+Qb\frac{d}{dx}(e^{bx})
\displaystyle =Pa(ae^{ax})+Qb(be^{bx})
\displaystyle =a^2Pe^{ax}+b^2Qe^{bx}\qquad\ldots\text{(iii)}
\displaystyle \text{Now, LHS}=\frac{d^2y}{dx^2}-(a+b)\frac{dy}{dx}+aby
\displaystyle \text{On putting the values from Eqs. (i), (ii) and (iii), we get}
\displaystyle \text{LHS}=a^2Pe^{ax}+b^2Qe^{bx}-(a+b)(aPe^{ax}+bQe^{bx})+ab(Pe^{ax}+Qe^{bx})
\displaystyle =a^2Pe^{ax}+b^2Qe^{bx}-a^2Pe^{ax}-abQe^{bx}-abPe^{ax}-b^2Qe^{bx}+abPe^{ax}+abQe^{bx}
\displaystyle =0=\text{RHS}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 93. }\text{If }x=\cos t(3-2\cos^2 t)\text{ and }y=\sin t(3-2\sin^2 t)
\displaystyle  \text{, then find the value of }\dfrac{dy}{dx}\text{ at }t=\dfrac{\pi}{4}. \hspace{2.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x=\cos t(3-2\cos^2t) 
\displaystyle \Rightarrow x=3\cos t-2\cos^3t
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=3(-\sin t)-2(3)\cos^2t(-\sin t)
\displaystyle \Rightarrow \frac{dx}{dt}=-3\sin t+6\cos^2t\sin t\qquad\ldots\text{(i)}
\displaystyle \text{Also, }y=\sin t(3-2\sin^2t)\Rightarrow y=3\sin t-2\sin^3t
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dy}{dt}=3\cos t-2\times3\times\sin^2t\cos t
\displaystyle \Rightarrow \frac{dy}{dt}=3\cos t-6\sin^2t\cos t\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=\frac{3\cos t-6\sin^2t\cos t}{-3\sin t+6\cos^2t\sin t}
\displaystyle \text{[from Eqs. (i) and (ii)]}
\displaystyle =\frac{\cos t-2\cos t\sin^2t}{-\sin t+2\cos^2t\sin t}
\displaystyle =\frac{\cos t(1-2\sin^2t)}{\sin t(2\cos^2t-1)}=\frac{\cos t\cdot\cos2t}{\sin t\cdot\cos2t}
\displaystyle =\cot(t)
\displaystyle \therefore \frac{dy}{dx}\bigg|_{t=\frac{\pi}{4}}=\cot\left(\frac{\pi}{4}\right)=1
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\displaystyle \textbf{Question 94. }\text{If }(x-y)e^{\frac{x}{x-y}}=a\text{, prove that }y\dfrac{dy}{dx}+x=2y.    \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }(x-y)\cdot e^{\frac{x}{x-y}}=a 
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log\left[(x-y)\cdot e^{\frac{x}{x-y}}\right]=\log a
\displaystyle \Rightarrow \log(x-y)+\log e^{\frac{x}{x-y}}=\log a\qquad[\because\log(mn)=\log m+\log n]
\displaystyle \Rightarrow \log(x-y)+\frac{x}{x-y}\log_e e=\log a
\displaystyle \Rightarrow \log(x-y)+\frac{x}{x-y}=\log a\qquad[\because\log_e e=1]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d}{dx}[\log(x-y)]+\frac{d}{dx}\left(\frac{x}{x-y}\right)=\frac{d}{dx}(\log a)
\displaystyle \Rightarrow \frac{1}{x-y}\frac{d}{dx}(x-y)+\frac{(x-y)\frac{d}{dx}(x)-x\frac{d}{dx}(x-y)}{(x-y)^2}=0
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle \Rightarrow \frac{1}{x-y}\cdot(1-y')+\frac{(x-y)-x(1-y')}{(x-y)^2}=0
\displaystyle \text{where }y'=dy/dx
\displaystyle \Rightarrow (x-y)(1-y')+x-y-x(1-y')=0
\displaystyle \Rightarrow yy'+x-2y=0
\displaystyle \therefore y\frac{dy}{dx}+x=2y\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 95. }\text{If }x=a(\cos t+t\sin t)\text{ and }y=a(\sin t-t\cos t)
\displaystyle  \text{, then find the value of }\dfrac{d^2y}{dx^2}\text{ at }t=\dfrac{\pi}{4}. \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }x=a(\cos t+t\sin t)\text{ and }y=a(\sin t-t\cos t)\text{, then find the value of }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{4} 
\displaystyle \frac{dx}{dt}=a\left[-\sin t+\sin t+t\cos t\right]=at\cos t
\displaystyle \frac{dy}{dt}=a\left[\cos t-\cos t+t\sin t\right]=at\sin t
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{at\sin t}{at\cos t}=\tan t
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(\tan t)=\frac{d}{dt}(\tan t)\cdot\frac{dt}{dx}=\sec^2t\cdot\frac{1}{at\cos t}=\frac{\sec^3t}{at}
\displaystyle \text{At }t=\frac{\pi}{4}\text{:}
\displaystyle \left(\frac{d^2y}{dx^2}\right)_{t=\frac{\pi}{4}}=\frac{\sec^3\frac{\pi}{4}}{a\cdot\frac{\pi}{4}}=\frac{(\sqrt{2})^3}{\frac{a\pi}{4}}=\frac{2\sqrt{2}\cdot4}{a\pi}=\frac{8\sqrt{2}}{a\pi}
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\displaystyle \textbf{Question 96. }\text{If }y=\tan^{-1}\left(\dfrac{a}{x}\right)+\log\sqrt{\dfrac{x-a}{x+a}}\text{, prove that}   \dfrac{dy}{dx}=\dfrac{2a^3}{x^4-a^4}.    \hspace{0.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=\tan^{-1}\left(\frac{a}{x}\right)+\log\sqrt{\frac{x-a}{x+a}} 
\displaystyle =\tan^{-1}\left(\frac{a}{x}\right)+\log\left(\frac{x-a}{x+a}\right)^{1/2}
\displaystyle \Rightarrow y=\tan^{-1}\left(\frac{a}{x}\right)+\frac{1}{2}[\log(x-a)-\log(x+a)]
\displaystyle \left[\because\log\frac{m}{n}=\log m-\log n\right]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{1+\frac{a^2}{x^2}}\cdot\left(\frac{-a}{x^2}\right)+\frac{1}{2}\left[\frac{1}{x-a}-\frac{1}{x+a}\right]
\displaystyle \left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\text{ and }\frac{d}{dx}(\log x)=\frac{1}{x}\right]
\displaystyle =\frac{-a}{x^2+a^2}+\frac{1}{2}\left[\frac{x+a-x+a}{(x-a)(x+a)}\right]
\displaystyle =\frac{-a}{x^2+a^2}+\frac{a}{x^2-a^2}
\displaystyle =\frac{-x^2a+a^3+x^2a+a^3}{(x^2+a^2)(x^2-a^2)}
\displaystyle \therefore \frac{dy}{dx}=\frac{2a^3}{x^4-a^4}\qquad[\because(a+b)(a-b)=a^2-b^2]\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 97. }\text{If }(\tan^{-1}x)^y+y^{\cot x}=1\text{, then find }dy/dx.    \hspace{2.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }u=(\tan^{-1}x)^y\text{ and }v=y^{\cot x} 
\displaystyle \text{Then, given equation becomes }u+v=1
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{du}{dx}+\frac{dv}{dx}=0\qquad\ldots\text{(i)}
\displaystyle \text{Now, }u=(\tan^{-1}x)^y
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log u=y\log(\tan^{-1}x)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(y)\cdot\log(\tan^{-1}x)+y\frac{d}{dx}(\log\tan^{-1}x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{1}{u}\frac{du}{dx}=\frac{dy}{dx}\log(\tan^{-1}x)+\frac{y}{(\tan^{-1}x)(1+x^2)}
\displaystyle \Rightarrow \frac{du}{dx}=(\tan^{-1}x)^y\left[\frac{dy}{dx}\log(\tan^{-1}x)+\frac{y}{(\tan^{-1}x)(1+x^2)}\right]\qquad\ldots\text{(ii)}
\displaystyle \text{Also, }v=y^{\cot x}
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log v=\cot x\log y
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(\cot x)\cdot\log y+\cot x\cdot\frac{d}{dx}(\log y)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{1}{v}\frac{dv}{dx}=-\text{cosec}^2x\log y+\frac{\cot x}{y}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dv}{dx}=y^{\cot x}\left[-\text{cosec}^2x\log y+\frac{\cot x}{y}\frac{dy}{dx}\right]\qquad\ldots\text{(iii)}
\displaystyle \text{On putting values from Eqs. (ii) and (iii) in Eq. (i), we get} 
\displaystyle (\tan^{-1}x)^y\left[\frac{dy}{dx}\log(\tan^{-1}x)+\frac{y}{(\tan^{-1}x)(1+x^2)}\right]+y^{\cot x}\left[-\text{cosec}^2x\log y+\frac{\cot x}{y}\frac{dy}{dx}\right]=0
\displaystyle \Rightarrow \frac{dy}{dx}\left[(\tan^{-1}x)^y\log(\tan^{-1}x)+\cot x\cdot y^{\cot x-1}\right]
\displaystyle =-\left[\frac{y}{1+x^2}(\tan^{-1}x)^{y-1}-y^{\cot x}\text{cosec}^2x\log y\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-\left[\frac{y}{1+x^2}(\tan^{-1}x)^{y-1}-y^{\cot x}\text{cosec}^2x\log y\right]}{[(\tan^{-1}x)^y\log(\tan^{-1}x)+\cot x\cdot y^{\cot x-1}]}
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\displaystyle \textbf{Question 98. }\text{If }x=2\cos\theta-\cos 2\theta\text{ and }y=2\sin\theta-\sin 2\theta\text{, then prove that }
\displaystyle  \dfrac{dy}{dx}=\tan\left(\dfrac{3\theta}{2}\right).  \hspace{2.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x=2\cos\theta-\cos2\theta 
\displaystyle \text{and }y=2\sin\theta-\sin2\theta
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=-2\sin\theta+2\sin2\theta
\displaystyle \text{and }\frac{dy}{d\theta}=2\cos\theta-2\cos2\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{2(\cos\theta-\cos2\theta)}{2(-\sin\theta+\sin2\theta)}
\displaystyle =\frac{2\sin\left(\frac{\theta+2\theta}{2}\right)\sin\left(\frac{2\theta-\theta}{2}\right)}{2\cos\left(\frac{2\theta+\theta}{2}\right)\sin\left(\frac{2\theta-\theta}{2}\right)}
\displaystyle \left[\because\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\right]
\displaystyle \left[\text{and }\sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\right]
\displaystyle =\frac{\sin\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}=\tan\left(\frac{3\theta}{2}\right)\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 99. }\text{If }y=x\log\left(\dfrac{x}{a+bx}\right)\text{, then prove that }   x^3\dfrac{d^2y}{dx^2}=\left(x\dfrac{dy}{dx}-y\right)^2.    \hspace{0.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=x\log\left(\frac{x}{a+bx}\right)\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=x\cdot\frac{d}{dx}\log\left(\frac{x}{a+bx}\right)+\log\left(\frac{x}{a+bx}\right)\cdot\frac{d}{dx}(x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =x\left(\frac{1}{\frac{x}{a+bx}}\right)\frac{d}{dx}\left(\frac{x}{a+bx}\right)+\log\left(\frac{x}{a+bx}\right)\cdot1
\displaystyle \left[\because\frac{d}{dx}(\log x)=\frac{1}{x}\right]
\displaystyle =(a+bx)\left[\frac{(a+bx)(1)-x(b)}{(a+bx)^2}\right]+\log\left(\frac{x}{a+bx}\right)
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =(a+bx)\left[\frac{a}{(a+bx)^2}\right]+\log\left(\frac{x}{a+bx}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{a}{a+bx}+\log\left(\frac{x}{a+bx}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{a}{a+bx}+\frac{y}{x}\qquad\text{[using Eq. (i)]}
\displaystyle \Rightarrow \frac{dy}{dx}-\frac{y}{x}=\frac{a}{a+bx}
\displaystyle \Rightarrow x\frac{dy}{dx}-y=\frac{ax}{a+bx}\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \Rightarrow x\frac{d^2y}{dx^2}+\frac{dy}{dx}(1)-\frac{dy}{dx}=a\left[\frac{(a+bx)\cdot(1)-x(b)}{(a+bx)^2}\right]
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle \Rightarrow x\frac{d^2y}{dx^2}=\frac{a^2}{(a+bx)^2}
\displaystyle \Rightarrow x^3\frac{d^2y}{dx^2}=\frac{a^2x^2}{(a+bx)^2}=\left[\frac{ax}{a+bx}\right]^2
\displaystyle \text{[multiplying both sides by }x^2\text{]}
\displaystyle \Rightarrow x^3\frac{d^2y}{dx^2}=\left(x\frac{dy}{dx}-y\right)^2\qquad\text{[using Eq. (ii)]}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 100. }\text{If }x=\cos\theta\text{ and }y=\sin^3\theta\text{, then prove that}
\displaystyle y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=3\sin^2\theta(5\cos^2\theta-1).   \hspace{2.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x=\cos\theta\qquad\ldots\text{(i)} 
\displaystyle \text{and }y=\sin^3\theta\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eqs. (i) and (ii) w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=-\sin\theta\text{ and }\frac{dy}{d\theta}=3\sin^2\theta\cdot\cos\theta
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}
\displaystyle =-3\sin\theta\cos\theta=\frac{-3}{2}(2\sin\theta\cos\theta)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-3}{2}\sin2\theta\qquad\ldots\text{(iii)}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{-3}{2}\cdot2\cos2\theta\cdot\frac{d\theta}{dx}=-3\cos2\theta\cdot\left(\frac{-1}{\sin\theta}\right)
\displaystyle \left[\because\frac{d\theta}{dx}=\frac{1}{dx/d\theta}=\frac{-1}{\sin\theta}\right]
\displaystyle =\frac{3\cos2\theta}{\sin\theta}\qquad\ldots\text{(iv)}
\displaystyle \text{Now, consider LHS}=y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2
\displaystyle =\sin^3\theta\left(\frac{3\cos2\theta}{\sin\theta}\right)+\left(\frac{-3}{2}\sin2\theta\right)^2
\displaystyle =\sin^2\theta\cdot3\cos2\theta+\frac{9}{4}\sin^22\theta
\displaystyle =3\sin^2\theta(2\cos^2\theta-1)+\frac{9}{4}(4\sin^2\theta\cos^2\theta)
\displaystyle =6\sin^2\theta\cos^2\theta-3\sin^2\theta+9\sin^2\theta\cos^2\theta
\displaystyle =15\sin^2\theta\cos^2\theta-3\sin^2\theta
\displaystyle =3\sin^2\theta(5\cos^2\theta-1)\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 101. }\text{Differentiate the following function with respect to }x.   (\log x)^x+x^{\log x}   \hspace{0.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }y=(\log x)^x+x^{\log x} 
\displaystyle \text{Also, let }u=(\log x)^x\text{ and }v=x^{\log x}\text{, then }y=u+v
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\qquad\ldots\text{(i)}
\displaystyle \text{Now, consider }u=(\log x)^x
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log u=\log(\log x)^x=x\log(\log x)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=x\cdot\frac{d}{dx}\log(\log x)+\log(\log x)\cdot\frac{d}{dx}(x)
\displaystyle =\frac{x}{\log x}\cdot\frac{1}{x}+\log(\log x)
\displaystyle \frac{du}{dx}=u\left[\frac{1}{\log x}+\log(\log x)\right]
\displaystyle \Rightarrow \frac{du}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right]\qquad\ldots\text{(ii)}
\displaystyle \text{Since, }v=x^{\log x}
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log v=\log(x^{\log x})=(\log x)(\log x)=(\log x)^2
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{v}\frac{dv}{dx}=2\log x\cdot\frac{1}{x}\Rightarrow \frac{dv}{dx}=v\left[\frac{2\log x}{x}\right]
\displaystyle \Rightarrow \frac{dv}{dx}=x^{\log x}\left[\frac{2\log x}{x}\right]\qquad[\because v=x^{\log x}]\qquad\ldots\text{(iii)}
\displaystyle \text{From Eqs. (i), (ii) and (iii), we get}
\displaystyle \frac{dy}{dx}=(\log x)^x\left\{\frac{1}{\log x}+\log(\log x)\right\}+2\left(\frac{\log x}{x}\right)x^{\log x}
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\displaystyle \textbf{Question 102. }\text{If }y=\log\left[x+\sqrt{x^2+a^2}\right]\text{, then show that }   (x^2+a^2)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=0.    \hspace{0.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }y=\log\left[x+\sqrt{x^2+a^2}\right] 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\cdot\frac{d}{dx}(x+\sqrt{x^2+a^2})
\displaystyle \left[\because\frac{d}{dx}(\log f(x))=\frac{1}{f(x)}\cdot\frac{d}{dx}f(x)\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\left(1+\frac{2x}{2\sqrt{x^2+a^2}}\right)
\displaystyle \left[\because\frac{d}{dx}(\sqrt{x^2+a^2})=\frac{1}{2\sqrt{x^2+a^2}}\times2x\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+a^2}}\left(\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}\right)
\displaystyle \Rightarrow \frac{dy}{dx}\left(\sqrt{x^2+a^2}\right)=1
\displaystyle \text{Again, on differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \sqrt{x^2+a^2}\cdot\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\cdot\frac{d}{dx}(\sqrt{x^2+a^2})=\frac{d(1)}{dx}
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}(\sqrt{x^2+a^2})+\frac{dy}{dx}\cdot\frac{1\cdot2x}{2\sqrt{x^2+a^2}}=0
\displaystyle \text{On multiplying both sides by }\sqrt{x^2+a^2}\text{, we get}
\displaystyle \frac{d^2y}{dx^2}(\sqrt{x^2+a^2})^2+\frac{dy}{dx}\times\frac{x\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}=0
\displaystyle \therefore (x^2+a^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 103. }\text{Show that the function }f(x)=|x-3|\text{, }x\in R
\displaystyle  \text{, is continuous but not differentiable at }x=3. \hspace{2.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }f(x)=|x-3| 
\displaystyle \text{First, we check the continuity of }f(x)\text{ at }x=3
\displaystyle \text{Here, LHL}=\lim_{x\to3^-}|x-3|=\lim_{h\to0}|3-h-3|
\displaystyle \text{[put }x=3-h\text{; when }x\to3^-\text{, then }h\to0\text{]}
\displaystyle =\lim_{h\to0}|-h|=0
\displaystyle \text{RHL}=\lim_{x\to3^+}|x-3|=\lim_{h\to0}|3+h-3|=\lim_{h\to0}|h|=0
\displaystyle \text{[put }x=3+h\text{; when }x\to3^+\text{, then }h\to0\text{]}
\displaystyle \text{and }f(3)=|3-3|=0
\displaystyle \text{Thus, LHL}=\text{RHL}=f(3)
\displaystyle \text{Hence, }f\text{ is continuous at }x=3
\displaystyle \text{Now, let us check the differentiability of }f(x)\text{ at }x=3
\displaystyle \text{LHD}=f'(3^-)=\lim_{h\to0}\frac{f(3-h)-f(3)}{-h}
\displaystyle \left[\because Lf'(a)=\lim_{h\to0}\frac{f(a-h)-f(a)}{-h}\right]
\displaystyle =\lim_{h\to0}\frac{|3-h-3|-|3-3|}{-h}
\displaystyle =\lim_{h\to0}\frac{|-h|}{-h}=\lim_{h\to0}\frac{h}{-h}=-1
\displaystyle [\because|-x|=x,\text{ if }x>0]
\displaystyle \text{RHD}=f'(3^+)=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}
\displaystyle \left[\because Rf'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\right]
\displaystyle =\lim_{h\to0}\frac{|3+h-3|-|3-3|}{h}
\displaystyle =\lim_{h\to0}\frac{|h|}{h}=\lim_{h\to0}\frac{h}{h}=1
\displaystyle [\because|x|=x,\text{ if }x>0]
\displaystyle \text{Since, LHD}\neq\text{RHD at }x=3
\displaystyle \text{So, }f\text{ is not differentiable.}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 104. }\text{If }x=a\sin t\text{ and }y=a\left[\cos t+\log\tan\left(\dfrac{t}{2}\right)\right]\text{, then find }\dfrac{d^2y}{dx^2}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }x=a\sin t\text{ and }y=a[\cos t+\log\tan(t/2)]\text{, then find }\frac{d^2y}{dx^2} 
\displaystyle \frac{dx}{dt}=a\cos t
\displaystyle \frac{dy}{dt}=a\left[-\sin t+\frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\right]
\displaystyle =a\left[-\sin t+\frac{\cos\frac{t}{2}}{2\sin\frac{t}{2}\cdot\cos^2\frac{t}{2}}\right]
\displaystyle =a\left[-\sin t+\frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right]
\displaystyle =a\left[-\sin t+\frac{1}{\sin t}\right]=a\cdot\frac{\cos^2t}{\sin t}
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cdot\frac{\cos^2t}{\sin t}}{a\cos t}=\frac{\cos t}{\sin t}=\cot t
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(\cot t)=\frac{d}{dt}(\cot t)\cdot\frac{dt}{dx}=-\text{cosec}^2t\cdot\frac{1}{a\cos t}
\displaystyle \therefore \frac{d^2y}{dx^2}=\frac{-\text{cosec}^2t}{a\cos t}=\frac{-\text{cosec}^2t\cdot\sec t}{a}
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\displaystyle \textbf{Question 105. }\text{Differentiate the following with respect to }x. \  \sin^{-1}\left[\dfrac{2^{x+1}\cdot 3^x}{1+(36)^x}\right]    \hspace{0.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }y=\sin^{-1}\left(\frac{2^{x+1}\cdot3^x}{1+(36)^x}\right) 
\displaystyle =\sin^{-1}\left(\frac{2\cdot2^x\cdot3^x}{1+(6^2)^x}\right)=\sin^{-1}\left(\frac{2\cdot6^x}{1+(6^x)^2}\right)
\displaystyle \text{On putting }6^x=\tan\theta\Rightarrow\theta=\tan^{-1}(6^x)
\displaystyle \text{Then, }y=\sin^{-1}\left(\frac{2\cdot\tan\theta}{1+\tan^2\theta}\right)
\displaystyle =\sin^{-1}(\sin2\theta)\qquad\left[\because\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\right]
\displaystyle =2\theta
\displaystyle \Rightarrow y=2\tan^{-1}(6^x)\qquad[\because\theta=\tan^{-1}(6^x)]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{2}{1+(6^x)^2}\cdot\frac{d}{dx}(6^x)\qquad\left[\because\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}\right]
\displaystyle \therefore \frac{dy}{dx}=\frac{2}{1+(6^x)^2}\cdot6^x\cdot\log6=\frac{2^{x+1}\cdot3^x}{1+(36)^x}\log6
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\displaystyle \textbf{Question 106. }\text{If }x=a\cos^3\theta\text{ and }y=a\sin^3\theta\text{, then find the value of }
\displaystyle  \dfrac{d^2y}{dx^2}\text{ at }\theta=\dfrac{\pi}{6}.  \hspace{2.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x=a\cos^3\theta\text{ and }y=a\sin^3\theta 
\displaystyle \text{On differentiating both sides of }x\text{ and }y\text{ w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=3a\cos^2\theta\cdot\frac{d}{d\theta}(\cos\theta)=3a\cos^2\theta\cdot(-\sin\theta)
\displaystyle =-3a\cos^2\theta\cdot\sin\theta
\displaystyle \text{and }\frac{dy}{d\theta}=3a\sin^2\theta\cdot\frac{d}{d\theta}(\sin\theta)
\displaystyle =3a\sin^2\theta\cdot(\cos\theta)=3a\sin^2\theta\cdot\cos\theta
\displaystyle \text{Now, }\frac{dy}{dx}=\left(\frac{dy/d\theta}{dx/d\theta}\right)
\displaystyle =\frac{3a\sin^2\theta\cdot\cos\theta}{-3a\cos^2\theta\cdot\sin\theta}=-\tan\theta
\displaystyle \text{Again, on differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(-\tan\theta)=-\frac{d}{d\theta}(\tan\theta)\cdot\frac{d\theta}{dx}
\displaystyle =-\sec^2\theta\cdot\frac{d\theta}{dx}
\displaystyle =-\sec^2\theta\cdot\left(\frac{-1}{3a\cos^2\theta\sin\theta}\right)
\displaystyle \left[\because\frac{d\theta}{dx}=\frac{1}{dx/d\theta}=\frac{-1}{3a\cos^2\theta\sin\theta}\right]
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{1}{3a\cos^4\theta\cdot\sin\theta}
\displaystyle  \text{At }\theta=\frac{\pi}{6}\text{,}
\displaystyle \left(\frac{d^2y}{dx^2}\right)_{\theta=\frac{\pi}{6}}=\frac{1}{3a\left(\cos\frac{\pi}{6}\right)^4\left(\sin\frac{\pi}{6}\right)}
\displaystyle =\frac{1}{3a\left(\frac{\sqrt{3}}{2}\right)^4\left(\frac{1}{2}\right)}
\displaystyle =\frac{1}{3a\cdot\frac{9}{16}\cdot\frac{1}{2}}=\frac{32}{27a}
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\displaystyle \textbf{Question 107. }\text{If }x\sin(a+y)+\sin a\cos(a+y)=0\text{, then prove that}
\displaystyle \dfrac{dy}{dx}=\dfrac{\sin^2(a+y)}{\sin a}.    \hspace{2.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x\sin(a+y)+\sin a\cos(a+y)=0 
\displaystyle \Rightarrow x=\frac{-\sin a\cos(a+y)}{\sin(a+y)}
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dx}{dy}=\frac{\sin(a+y)\cdot\frac{d}{dy}\{\sin a\cos(a+y)\}-[-\sin a\cos(a+y)\cdot\frac{d}{dy}\{\sin(a+y)\}]}{\sin^2(a+y)}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =\frac{\left\{\sin(a+y)\cdot\sin a\sin(a+y)+\sin a\cos(a+y)\cos(a+y)\right\}}{\sin^2(a+y)}
\displaystyle =\frac{\sin a}{\sin^2(a+y)}\{\sin^2(a+y)+\cos^2(a+y)\}
\displaystyle =\frac{\sin a}{\sin^2(a+y)}\cdot1\qquad[\because\sin^2\theta+\cos^2\theta=1]
\displaystyle \therefore \frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 108. }\text{If }x^y=e^{x-y}\text{, then prove that }\dfrac{dy}{dx}=\dfrac{\log x}{(1+\log x)^2}.
\displaystyle \text{Or If }x^y=e^{x-y}\text{, then prove that }\dfrac{dy}{dx}=\dfrac{\log x}{\{\log(xe)\}^2}.    \hspace{0.2cm} \text{[CBSE 2013; CBSE 2010; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }x^y=e^{x-y} 
\displaystyle \text{On taking log both sides, we get}
\displaystyle y\log_e x=(x-y)\log_e e
\displaystyle \Rightarrow y\log_e x=x-y\qquad[\because\log_e e=1]
\displaystyle \Rightarrow y(1+\log x)=x\Rightarrow y=\frac{x}{1+\log x}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=\frac{(1+\log x)\frac{d}{dx}(x)-x\frac{d}{dx}(1+\log x)}{(1+\log x)^2}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =\frac{1+\log x-x\cdot\frac{1}{x}}{(1+\log x)^2}=\frac{1+\log x-1}{(1+\log x)^2}
\displaystyle \text{Hence, }\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}
\displaystyle \text{Also, it can be written as}
\displaystyle \frac{dy}{dx}=\frac{\log x}{(\log_e e+\log x)^2}\qquad[\log_e e=1]
\displaystyle \therefore \frac{dy}{dx}=\frac{\log x}{\{\log(ex)\}^2}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 109. }\text{If }y^x=e^{y-x}\text{, then prove that }\dfrac{dy}{dx}=\dfrac{(1+\log y)^2}{\log y}.    \hspace{1.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }y^x=e^{y-x}\text{, then prove that }\frac{dy}{dx}=\frac{(1+\log y)^2}{\log y} 
\displaystyle \text{Taking log both sides, we get}
\displaystyle x\log y=y-x
\displaystyle \Rightarrow x\log y+x=y
\displaystyle \Rightarrow x(1+\log y)=y\Rightarrow x=\frac{y}{1+\log y}
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dx}{dy}=\frac{(1+\log y)\cdot1-y\cdot\frac{1}{y}}{(1+\log y)^2}=\frac{1+\log y-1}{(1+\log y)^2}=\frac{\log y}{(1+\log y)^2}
\displaystyle \therefore \frac{dy}{dx}=\frac{(1+\log y)^2}{\log y}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 110. }\text{If }\sin y=x\sin(a+y)\text{, then prove that }\dfrac{dy}{dx}=\dfrac{\sin^2(a+y)}{\sin a}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }\sin y=x\sin(a+y)\text{, then prove that }\frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a} 
\displaystyle \text{Given, }\sin y=x\sin(a+y)\Rightarrow x=\frac{\sin y}{\sin(a+y)}
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dx}{dy}=\frac{\sin(a+y)\cos y-\sin y\cos(a+y)}{\sin^2(a+y)}
\displaystyle =\frac{\sin(a+y-y)}{\sin^2(a+y)}=\frac{\sin a}{\sin^2(a+y)}
\displaystyle \therefore \frac{dy}{dx}=\frac{\sin^2(a+y)}{\sin a}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 111. }\text{If }y=\sin^{-1}x\text{, show that }(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}=0.    \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{If }y=\sin^{-1}x\text{, show that }(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0 
\displaystyle \text{Given, }y=\sin^{-1}x
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}
\displaystyle \Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=1
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \sqrt{1-x^2}\frac{d^2y}{dx^2}+\frac{dy}{dx}\cdot\frac{-2x}{2\sqrt{1-x^2}}=0
\displaystyle \text{Multiplying both sides by }\sqrt{1-x^2}\text{:}
\displaystyle (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 112. }\text{If }x=\sqrt{a^{\sin^{-1}t}}\text{ and }y=\sqrt{a^{\cos^{-1}t}}
\displaystyle  \text{, then show that }\dfrac{dy}{dx}=\dfrac{-y}{x}.  \hspace{2.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=\sqrt{a^{\sin^{-1}t}}\text{ and }y=\sqrt{a^{\cos^{-1}t}} 
\displaystyle \text{Consider, }x=(a^{\sin^{-1}t})^{1/2}
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=\frac{1}{2}(a^{\sin^{-1}t})^{-1/2}\cdot\frac{d}{dt}(a^{\sin^{-1}t})
\displaystyle \text{[by using chain rule of derivative]}
\displaystyle =\frac{1}{2}(a^{\sin^{-1}t})^{-1/2}\cdot a^{\sin^{-1}t}\log a\cdot\frac{d}{dt}(\sin^{-1}t)
\displaystyle \left[\because\frac{d}{dx}(a^x)=a^x\log a\right]
\displaystyle =\frac{1}{2}(a^{\sin^{-1}t})^{1/2}\cdot\log a\cdot\frac{1}{\sqrt{1-t^2}}
\displaystyle \Rightarrow \frac{dx}{dt}=\frac{\frac{1}{2}\sqrt{a^{\sin^{-1}t}}\cdot\log a}{\sqrt{1-t^2}}\qquad\ldots\text{(i)}
\displaystyle \text{Now, consider }y=(a^{\cos^{-1}t})^{1/2}
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dy}{dt}=\frac{1}{2}(a^{\cos^{-1}t})^{-1/2}\cdot\frac{d}{dt}(a^{\cos^{-1}t})
\displaystyle \text{[by using chain rule of derivative]}
\displaystyle \left[\because\frac{d}{dx}(a^x)=a^x\log a\right]
\displaystyle =\frac{1}{2}(a^{\cos^{-1}t})^{1/2}\log a\cdot\frac{(-1)}{\sqrt{1-t^2}}
\displaystyle \Rightarrow \frac{dy}{dt}=\frac{-\frac{1}{2}\sqrt{a^{\cos^{-1}t}}\cdot\log a}{\sqrt{1-t^2}}\qquad\ldots\text{(ii)}
\displaystyle \text{On dividing Eq. (ii) by Eq. (i), we get}
\displaystyle \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{\dfrac{-\frac{1}{2}\sqrt{a^{\cos^{-1}t}}\cdot\log a}{\sqrt{1-t^2}}}{\dfrac{\frac{1}{2}\sqrt{a^{\sin^{-1}t}}\cdot\log a}{\sqrt{1-t^2}}}
\displaystyle =-\frac{\sqrt{a^{\cos^{-1}t}}}{\sqrt{a^{\sin^{-1}t}}}=-\frac{y}{x}
\displaystyle \left[\text{given, }\sqrt{a^{\cos^{-1}t}}=y\text{ and }\sqrt{a^{\sin^{-1}t}}=x\right]\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 113. }\text{If }y=x^{\sin x-\cos x}+\dfrac{x^2-1}{x^2+1}\text{, then find }\dfrac{dy}{dx}.   \hspace{2.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=x^{\sin x-\cos x}+\frac{x^2-1}{x^2+1} 
\displaystyle \text{Let }u=x^{\sin x-\cos x}\text{ and }v=\frac{x^2-1}{x^2+1}
\displaystyle \text{Then, the given equation becomes}
\displaystyle y=u+v\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\qquad\ldots\text{(i)}
\displaystyle \text{Consider, }u=x^{\sin x-\cos x}
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log u=(\sin x-\cos x)\cdot\log x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=(\sin x-\cos x)\cdot\frac{1}{x}+\log x\cdot(\cos x+\sin x)
\displaystyle \Rightarrow \frac{du}{dx}=x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x\cdot(\cos x+\sin x)\right]\qquad\ldots\text{(ii)}
\displaystyle \text{Now, consider }v=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=0-\frac{(x^2+1)\frac{d}{dx}(2)-2\frac{d}{dx}(x^2+1)}{(x^2+1)^2}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle \Rightarrow \frac{dv}{dx}=-\left[\frac{0-2\cdot2x}{(x^2+1)^2}\right]=\frac{4x}{(x^2+1)^2}\qquad\ldots\text{(iii)}
\displaystyle \text{On substituting the values from Eqs. (ii) and (iii) into Eq. (i), we get}
\displaystyle \frac{dy}{dx}=x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x(\cos x+\sin x)\right]+\frac{4x}{(x^2+1)^2}
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\displaystyle \textbf{Question 114. }\text{If }x=a(\cos t+t\sin t)\text{ and }y=a(\sin t-t\cos t)
\displaystyle  \text{, then find }\dfrac{d^2x}{dt^2}\text{, }\dfrac{d^2y}{dt^2}\text{ and }\dfrac{d^2y}{dx^2}.  \hspace{2.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a(\cos t+t\sin t) 
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dx}{dt}=a\left[-\sin t+\frac{d}{dt}(t)\cdot\sin t+t\cdot\frac{d}{dt}(\sin t)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow \frac{dx}{dt}=a(-\sin t+1\cdot\sin t+t\cos t)=at\cos t\qquad\ldots\text{(i)}
\displaystyle \text{Also given, }y=a(\sin t-t\cos t)
\displaystyle \text{On differentiating both sides w.r.t. }t\text{, we get}
\displaystyle \frac{dy}{dt}=a\left[\cos t-\frac{d}{dt}(t)\cos t-t\frac{d}{dt}(\cos t)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \frac{dy}{dt}=a(\cos t-\cos t\cdot1+t\sin t)=at\sin t\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{at\sin t}{at\cos t}=\tan t
\displaystyle \text{[from Eqs. (i) and (ii)]}
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(\tan t)\cdot\frac{dt}{dx}=\sec^2t\cdot\frac{1}{dx/dt}
\displaystyle =\frac{\sec^2t}{at\cos t}=\frac{\sec^3t}{at}\qquad\text{[from Eq. (i)]}
\displaystyle \text{Also, }\frac{d^2x}{dt^2}=\frac{d}{dt}(at\cos t)
\displaystyle =a\left[\frac{d}{dt}(t)\cdot\cos t+t\cdot\frac{d}{dt}(\cos t)\right]
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =a[\cos t-t\sin t]
\displaystyle \text{and }\frac{d^2y}{dt^2}=\frac{d}{dt}\left(\frac{dy}{dt}\right)=\frac{d}{dt}(at\sin t)
\displaystyle =a(\sin t+t\cos t)
\\

\displaystyle \textbf{Question 115. }\text{If }x=a\left(\cos t+\log\tan\dfrac{t}{2}\right)\text{ and }y=a\sin t\text{, find }\dfrac{d^2y}{dt^2}\text{ and }\dfrac{d^2y}{dx^2}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{If }x=a\left(\cos t+\log\tan\frac{t}{2}\right)\text{ and }y=a\sin t\text{, find }\frac{d^2y}{dt^2}\text{ and }\frac{d^2y}{dx^2} 
\displaystyle \frac{dy}{dt}=a\cos t
\displaystyle \frac{d^2y}{dt^2}=-a\sin t
\displaystyle \frac{dx}{dt}=a\left(-\sin t+\frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\right)=a\cdot\frac{\cos^2t}{\sin t}
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t}{a\cdot\frac{\cos^2t}{\sin t}}=\frac{\sin t}{\cos t}=\tan t
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dt}(\tan t)\cdot\frac{dt}{dx}=\sec^2t\cdot\frac{\sin t}{a\cos^2t}=\frac{\sin t\sec^4t}{a}
\\

\displaystyle \textbf{Question 116. }\text{Differentiate }\tan^{-1}\left[\dfrac{\sqrt{1+x^2}-1}{x}\right]\text{ with respect to }x.   \hspace{0.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Differentiate }\tan^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]\text{ with respect to }x 
\displaystyle \text{Let }y=\tan^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]
\displaystyle \text{Put }x=\tan\theta\Rightarrow\theta=\tan^{-1}x
\displaystyle y=\tan^{-1}\left[\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right]=\tan^{-1}\left[\frac{\sec\theta-1}{\tan\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{1-\cos\theta}{\sin\theta}\right]=\tan^{-1}\left[\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right]
\displaystyle =\tan^{-1}\left[\tan\frac{\theta}{2}\right]=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x
\displaystyle \therefore \frac{dy}{dx}=\frac{1}{2}\cdot\frac{1}{1+x^2}=\frac{1}{2(1+x^2)}
\\

\displaystyle \textbf{Question 117. }\text{If }y=(\tan^{-1}x)^2\text{, then show that}
\displaystyle (x^2+1)^2\dfrac{d^2y}{dx^2}+2x(x^2+1)\dfrac{dy}{dx}=2.   \hspace{2.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=(\tan^{-1}x)^2 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=2\tan^{-1}x\cdot\frac{1}{1+x^2}\qquad\left[\because\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}\right]
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\tan^{-1}x}{1+x^2}
\displaystyle \Rightarrow (1+x^2)\frac{dy}{dx}=2\tan^{-1}x
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle (1+x^2)\cdot\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\cdot\frac{d}{dx}(1+x^2)=\frac{d}{dx}(2\tan^{-1}x)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow (1+x^2)\cdot\frac{d^2y}{dx^2}+\frac{dy}{dx}\cdot2x=\frac{2}{1+x^2}
\displaystyle \left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\right]
\displaystyle \text{On multiplying both sides by }(1+x^2)\text{, we get}
\displaystyle (1+x^2)^2\frac{d^2y}{dx^2}+\frac{dy}{dx}\cdot2x\cdot(1+x^2)=\frac{2}{1+x^2}\cdot(1+x^2)
\displaystyle \Rightarrow (1+x^2)^2\frac{d^2y}{dx^2}+2x(1+x^2)\frac{dy}{dx}=2\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 118. }\text{Find }\dfrac{dy}{dx}\text{, when }y=x^{\cot x}+\dfrac{2x^2-3}{x^2+x+2}.    \hspace{2.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }y=x^{\cot x}+\frac{2x^2-3}{x^2+x+2} 
\displaystyle \text{Let }u=x^{\cot x}\text{ and }v=\frac{2x^2-3}{x^2+x+2}
\displaystyle \text{Then, given equation becomes }y=u+v
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\qquad\text{[differentiate w.r.t. }x\text{]}
\displaystyle \text{Consider, }u=x^{\cot x}
\displaystyle \text{On taking log both sides, we get}
\displaystyle \log u=\cot x\log x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{u}\frac{du}{dx}=\cot x\cdot\frac{1}{x}-\text{cosec}^2x\cdot\log x
\displaystyle \Rightarrow \frac{du}{dx}=u\left(\frac{\cot x}{x}-\text{cosec}^2x\cdot\log x\right)
\displaystyle =x^{\cot x}\left(\frac{\cot x}{x}-\text{cosec}^2x\cdot\log x\right)\qquad\ldots\text{(ii)}
\displaystyle \text{Now, consider }v=\frac{2x^2-3}{x^2+x+2}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dv}{dx}=\frac{(x^2+x+2)(4x)-(2x^2-3)(2x+1)}{(x^2+x+2)^2}
\displaystyle \text{[by using quotient rule of derivative]}
\displaystyle =\frac{4x^3+4x^2+8x-4x^3-2x^2+6x+3}{(x^2+x+2)^2}
\displaystyle =\frac{2x^2+14x+3}{(x^2+x+2)^2}\qquad\ldots\text{(iii)}
\displaystyle \text{On substituting the values from Eqs. (ii) and (iii) into Eq. (i), we get}
\displaystyle \frac{dy}{dx}=x^{\cot x}\left(\frac{\cot x}{x}-\text{cosec}^2x\cdot\log x\right)+\frac{2x^2+14x+3}{(x^2+x+2)^2}
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\displaystyle \textbf{Question 119. }\text{If }x=a(\theta-\sin\theta)\text{, }y=a(1+\cos\theta)\text{, then find }\dfrac{d^2y}{dx^2}.   \hspace{0.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a(\theta-\sin\theta)\text{ and }y=a(1+\cos\theta) 
\displaystyle \text{On differentiating both sides of }x\text{ and }y\text{ w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=a(1-\cos\theta)\text{ and }\frac{dy}{d\theta}=-a\sin\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta}\right)}{\left(\frac{dx}{d\theta}\right)}=\frac{-a\sin\theta}{a(1-\cos\theta)}=\frac{-\sin\theta}{1-\cos\theta}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}=-\cot\frac{\theta}{2}\qquad\ldots\text{(i)}
\displaystyle \left[\because\sin A=2\sin\frac{A}{2}\cos\frac{A}{2}\text{ and }1-\cos A=2\sin^2\frac{A}{2}\right]
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(-\cot\frac{\theta}{2}\right)=\frac{d}{d\theta}\left(-\cot\frac{\theta}{2}\right)\times\frac{d\theta}{dx}
\displaystyle \left[\because\frac{d}{dx}[f(\theta)]=\frac{d}{d\theta}f(\theta)\times\frac{d\theta}{dx}\right]
\displaystyle =\frac{1}{2}\text{cosec}^2\frac{\theta}{2}\times\frac{1}{a(1-\cos\theta)}
\displaystyle \left[\because\frac{d}{d\theta}(\cot A)=-\text{cosec}^2A\right]
\displaystyle =\frac{1}{2a}\text{cosec}^2\frac{\theta}{2}\times\frac{1}{2\sin^2\frac{\theta}{2}}=\frac{1}{4a}\text{cosec}^4\frac{\theta}{2}
\\

\displaystyle \textbf{Question 120. }\text{Prove that }\dfrac{d}{dx}\left[\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x}{a}\right)\right]=\sqrt{a^2-x^2}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, LHS}=\frac{d}{dx}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]   
\displaystyle =\left[\frac{x}{2}\times\frac{d}{dx}\sqrt{a^2-x^2}+\sqrt{a^2-x^2}\times\frac{d}{dx}\left(\frac{x}{2}\right)\right]
\displaystyle +\frac{a^2}{2}\times\frac{d}{dx}\sin^{-1}\frac{x}{a}
\displaystyle \text{[by using product rule of derivative]}
\displaystyle =\frac{x}{2}\cdot\frac{1}{2\sqrt{a^2-x^2}}\cdot\frac{d}{dx}(a^2-x^2)+\sqrt{a^2-x^2}\cdot\frac{1}{2}
\displaystyle +\frac{a^2}{2}\times\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\cdot\frac{d}{dx}\left(\frac{x}{a}\right)
\displaystyle \text{[by using chain rule of derivative]}
\displaystyle =\left[\frac{x}{2}\cdot\frac{-2x}{2\sqrt{a^2-x^2}}+\sqrt{a^2-x^2}\cdot\frac{1}{2}+\frac{a^2}{2}\times\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\cdot\frac{1}{a}\right]
\displaystyle =\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{\sqrt{a^2-x^2}}{2}+\frac{a^2}{2a}\cdot\frac{1}{\sqrt{\frac{a^2-x^2}{a^2}}}
\displaystyle =\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{\sqrt{a^2-x^2}}{2}+\frac{a^2}{2a}\times\frac{a}{\sqrt{a^2-x^2}}
\displaystyle =\frac{-x^2}{2\sqrt{a^2-x^2}}+\frac{\sqrt{a^2-x^2}}{2}+\frac{a^2}{2\sqrt{a^2-x^2}}
\displaystyle =\frac{-x^2+(a^2-x^2)+a^2}{2\sqrt{a^2-x^2}}
\displaystyle =\frac{2a^2-2x^2}{2\sqrt{a^2-x^2}}=\frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}}=\sqrt{a^2-x^2}
\displaystyle =\text{RHS}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 121. }\text{If }y=\log\left[x+\sqrt{x^2+1}\right]\text{, then prove that}
\displaystyle (x^2+1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=0.    \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{If }y=\log[x+\sqrt{x^2+1}]\text{, then prove that }(x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0 
\displaystyle \text{Given, }y=\log[x+\sqrt{x^2+1}]
\displaystyle \frac{dy}{dx}=\frac{1}{x+\sqrt{x^2+1}}\cdot\left(1+\frac{x}{\sqrt{x^2+1}}\right)
\displaystyle =\frac{1}{x+\sqrt{x^2+1}}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}
\displaystyle =\frac{1}{\sqrt{x^2+1}}
\displaystyle \Rightarrow \sqrt{x^2+1}\frac{dy}{dx}=1
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \sqrt{x^2+1}\frac{d^2y}{dx^2}+\frac{x}{\sqrt{x^2+1}}\cdot\frac{dy}{dx}=0
\displaystyle \text{Multiplying both sides by }\sqrt{x^2+1}\text{:}
\displaystyle (x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 122. }\text{If }\log\left(\sqrt{1+x^2}-x\right)=y\sqrt{1+x^2}\text{, then show that}
\displaystyle (1+x^2)\dfrac{dy}{dx}+xy+1=0.    \hspace{2.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\log(\sqrt{1+x^2}-x)=y\sqrt{1+x^2}\qquad\ldots\text{(i)} 
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{1}{\sqrt{1+x^2}-x}\cdot\frac{d}{dx}[\sqrt{1+x^2}-x]
\displaystyle =y\frac{d}{dx}\sqrt{1+x^2}+\sqrt{1+x^2}\frac{dy}{dx}
\displaystyle \text{[by using chain rule and product rule of derivative]}
\displaystyle \Rightarrow \frac{1}{\sqrt{1+x^2}-x}\left[\frac{1}{2\sqrt{1+x^2}}\cdot\frac{d}{dx}(1+x^2)-1\right]
\displaystyle =\frac{y}{2\sqrt{1+x^2}}\cdot\frac{d}{dx}(1+x^2)+\sqrt{1+x^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{1}{\sqrt{1+x^2}-x}\left[\frac{2x}{2\sqrt{1+x^2}}-1\right]=y\times\frac{2x}{2\sqrt{1+x^2}}+\sqrt{1+x^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{1}{\sqrt{1+x^2}-x}\left[\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right]=\frac{xy}{\sqrt{1+x^2}}+\sqrt{1+x^2}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{-1}{\sqrt{1+x^2}}=\frac{xy+(1+x^2)\frac{dy}{dx}}{\sqrt{1+x^2}}
\displaystyle \Rightarrow -1=xy+(1+x^2)\frac{dy}{dx}
\displaystyle \therefore (1+x^2)\frac{dy}{dx}+xy+1=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 123. }\text{If }x=a(\theta+\sin\theta)\text{ and }y=a(1-\cos\theta)\text{, then find }\dfrac{d^2y}{dx^2}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{If }x=a(\theta+\sin\theta)\text{ and }y=a(1-\cos\theta)\text{, then find }\frac{d^2y}{dx^2} 
\displaystyle \frac{dx}{d\theta}=a(1+\cos\theta)\text{ and }\frac{dy}{d\theta}=a\sin\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\sin\theta}{a(1+\cos\theta)}
\displaystyle =\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}=\tan\frac{\theta}{2}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\tan\frac{\theta}{2}\right)=\frac{d}{d\theta}\left(\tan\frac{\theta}{2}\right)\cdot\frac{d\theta}{dx}
\displaystyle =\frac{1}{2}\sec^2\frac{\theta}{2}\cdot\frac{1}{a(1+\cos\theta)}
\displaystyle =\frac{1}{2}\sec^2\frac{\theta}{2}\cdot\frac{1}{a\cdot2\cos^2\frac{\theta}{2}}
\displaystyle =\frac{\sec^4\frac{\theta}{2}}{4a}
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\displaystyle \textbf{Question 124. }\text{If }y=a\sin x+b\cos x\text{, then prove that }   y^2+\left(\dfrac{dy}{dx}\right)^2=a^2+b^2.    \hspace{0.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove }y^2+\left(\frac{dy}{dx}\right)^2=a^2+b^2\qquad\ldots\text{(i)} 
\displaystyle \text{Given, }y=a\sin x+b\cos x\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eq. (ii) w.r.t. }x\text{, we get}
\displaystyle \frac{dy}{dx}=a\cos x-b\sin x
\displaystyle \text{Now, let us take LHS of Eq. (i)}
\displaystyle \text{Here, LHS}=y^2+\left(\frac{dy}{dx}\right)^2
\displaystyle \text{On putting the value of }y\text{ and }dy/dx\text{, we get}
\displaystyle \text{LHS}=(a\sin x+b\cos x)^2+(a\cos x-b\sin x)^2
\displaystyle =a^2\sin^2x+b^2\cos^2x+2ab\sin x\cos x
\displaystyle +a^2\cos^2x+b^2\sin^2x-2ab\sin x\cos x
\displaystyle =a^2\sin^2x+b^2\cos^2x+a^2\cos^2x+b^2\sin^2x
\displaystyle =a^2(\sin^2x+\cos^2x)+b^2(\sin^2x+\cos^2x)
\displaystyle =a^2+b^2\qquad[\because\sin^2x+\cos^2x=1]
\displaystyle =\text{RHS}\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 125. }\text{If }x=a(\cos\theta+\theta\sin\theta)\text{ and }y=a(\sin\theta-\theta\cos\theta)\text{, then find }\dfrac{d^2y}{dx^2}.
\displaystyle  \hspace{2.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{If }x=a(\cos\theta+\theta\sin\theta)\text{ and }y=a(\sin\theta-\theta\cos\theta)\text{, then find }\frac{d^2y}{dx^2} 
\displaystyle \frac{dx}{d\theta}=a(-\sin\theta+\sin\theta+\theta\cos\theta)=a\theta\cos\theta
\displaystyle \frac{dy}{d\theta}=a(\cos\theta-\cos\theta+\theta\sin\theta)=a\theta\sin\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}(\tan\theta)=\frac{d}{d\theta}(\tan\theta)\cdot\frac{d\theta}{dx}
\displaystyle =\sec^2\theta\cdot\frac{1}{a\theta\cos\theta}
\displaystyle =\frac{\sec^3\theta}{a\theta}
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\displaystyle \textbf{Question 126. }\text{If }x=a(\theta-\sin\theta)\text{ and }y=a(1+\cos\theta)
\displaystyle  \text{, then find }\dfrac{dy}{dx}\text{ at }\theta=\dfrac{\pi}{3}.  \hspace{2.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=a(\theta-\sin\theta)\text{ and }y=a(1+\cos\theta) 
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dx}{d\theta}=a(1-\cos\theta)
\displaystyle \text{and }\frac{dy}{d\theta}=-a\sin\theta
\displaystyle \therefore \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-a\sin\theta}{a(1-\cos\theta)}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{a\times2\sin^2\frac{\theta}{2}}\qquad\left[\because\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\text{ and }1-\cos x=2\sin^2\frac{x}{2}\right]
\displaystyle \Rightarrow \frac{dy}{dx}=-\cot\frac{\theta}{2}
\displaystyle \text{On putting }\theta=\frac{\pi}{3}\text{, we get}
\displaystyle \left[\frac{dy}{dx}\right]_{\theta=\frac{\pi}{3}}=-\cot\frac{\pi}{6}=-\sqrt{3}\qquad\left[\because\cot\frac{\pi}{6}=\sqrt{3}\right]
\displaystyle \text{Hence, }\frac{dy}{dx}\text{ at }\theta=\frac{\pi}{3}\text{ is }-\sqrt{3}
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\displaystyle \textbf{Question 127. }\text{If }x=\tan\left(\dfrac{1}{a}\log y\right)\text{, then show that}
\displaystyle (1+x^2)\dfrac{d^2y}{dx^2}+(2x-a)\dfrac{dy}{dx}=0.   \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x=\tan\left(\frac{1}{a}\log y\right) 
\displaystyle \Rightarrow \tan^{-1}x=\frac{1}{a}\log y\qquad[\because\tan\theta=a\Rightarrow\theta=\tan^{-1}a]
\displaystyle \Rightarrow a\tan^{-1}x=\log y
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle a\times\frac{1}{1+x^2}=\frac{1}{y}\frac{dy}{dx}\qquad\left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\right]
\displaystyle \Rightarrow (1+x^2)\frac{dy}{dx}=ay
\displaystyle \text{Again, differentiating both sides w.r.t. }x\text{, we get}
\displaystyle (1+x^2)\cdot\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\cdot\frac{d}{dx}(1+x^2)=\frac{d}{dx}(ay)
\displaystyle \text{[by using product rule of derivative]}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+\frac{dy}{dx}\cdot(2x)=a\cdot\frac{dy}{dx}
\displaystyle \Rightarrow (1+x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}-a\frac{dy}{dx}=0
\displaystyle \therefore (1+x^2)\frac{d^2y}{dx^2}+(2x-a)\frac{dy}{dx}=0\qquad \text{Hence proved.}
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\displaystyle \textbf{Question 128. }\text{If }f(x)=\begin{cases}ax+b,\ 0<x\leq1\\2x^2-x,\ 1<x<2\end{cases}\text{ is a differentiable function}
\displaystyle \text{in }(0,2),\text{ then find the values of }a\text{ and }b.  \hspace{2.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\begin{cases}ax+b, & 0<x\leq 1\\2x^2-x, & 1<x<2\end{cases}\text{ is differentiable function in }(0,2) 
\displaystyle \therefore f(x)\text{ is also differentiable at }x=1\in(0,2)\text{ and }f(x)\text{ is continuous at }x=1
\displaystyle \lim_{x\to 1^-}f(x)=\lim_{h\to 0}f(1-h)
\displaystyle =\lim_{h\to 0}(a(1-h)+b)
\displaystyle =\lim_{h\to 0}(a-ah+b)
\displaystyle =a+b
\displaystyle \lim_{x\to 1^+}f(x)=\lim_{h\to 0}f(1+h)
\displaystyle =\lim_{h\to 0}\{2(1+h)^2-(1+h)\}
\displaystyle =\lim_{h\to 0}\{2(1+h^2+2h)-(1+h)\}
\displaystyle =\lim_{h\to 0}(2+2h^2+4h-1-h)
\displaystyle =\lim_{h\to 0}(2h^2+3h+1)
\displaystyle =2\times0+3\times0+1=1
\displaystyle \text{and }f(1)=a+b
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=1
\displaystyle \therefore \lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)=f(1)
\displaystyle \Rightarrow a+b=1=a+b
\displaystyle \Rightarrow a+b=1\qquad\ldots\text{(i)}
\displaystyle \text{Now, }f(x)\text{ is differentiable at }x=1\text{, we get}
\displaystyle Lf'(1)=Rf'(1)
\displaystyle \Rightarrow \lim_{h\to 0}\frac{f(1-h)-f(1)}{-h}=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}
\displaystyle \Rightarrow \lim_{h\to 0}\frac{\{a(1-h)+b\}-(a+b)}{-h}
\displaystyle =\lim_{h\to 0}\frac{\{2(1+h)^2-(1+h)\}-(a+b)}{h}
\displaystyle \Rightarrow \lim_{h\to 0}\frac{(a-ah+b-a-b)}{-h}
\displaystyle =\lim_{h\to 0}\frac{(2+2h^2+4h-1-h-a-b)}{h}
\displaystyle \Rightarrow \lim_{h\to 0}\left(\frac{-ah}{-h}\right)=\lim_{h\to 0}\frac{2h^2+3h+1-a-b}{h}
\displaystyle \Rightarrow a=\lim_{h\to 0}\frac{2h^2+3h+1-1}{h}\qquad\text{[using (i)]}
\displaystyle \Rightarrow a=\lim_{h\to 0}\frac{h(2h+3)}{h}\Rightarrow a=\lim_{h\to 0}(2h+3)
\displaystyle \Rightarrow a=2\times0+3\Rightarrow a=3
\displaystyle \text{From Eq. (i), }3+b=1\Rightarrow b=-2
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