\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}

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