\displaystyle \textbf{Question 1. }\text{Find the derivative of }y=\log x+\frac{1}{x}\text{ with respect to }x.\text{ \hspace{0.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=\log x+\frac{1}{x}
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\log x)+\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{x^{2}}
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\displaystyle \textbf{Question 2. }\text{If }y=3\cos(\log x)+4\sin(\log x),\text{ show that} \ \
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0.\text{ \hspace{0.2cm} ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=3\cos(\log x)+4\sin(\log x) \qquad \cdots(i)
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=-\frac{3}{x}\sin(\log x)+\frac{4}{x}\cos(\log x) \qquad \cdots(ii)
\displaystyle \text{On multiplying by } x \text{ both sides, we get}
\displaystyle x\frac{dy}{dx}=-3\sin(\log x)+4\cos(\log x) \qquad \cdots(iii)
\displaystyle \text{Again, on differentiating Eq. (ii) w.r.t. } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{3}{x^{2}}\sin(\log x)-\frac{3}{x^{2}}\cos(\log x)-\frac{4}{x^{2}}\cos(\log x)-\frac{4}{x^{2}}\sin(\log x)
\displaystyle \text{On multiplying by } x^{2} \text{ both sides, we get}
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}=3\sin(\log x)-3\cos(\log x)-4\cos(\log x)-4\sin(\log x)
\displaystyle =-\left[-3\sin(\log x)+4\cos(\log x)\right]-\left[3\cos(\log x)+4\sin(\log x)\right]
\displaystyle \Rightarrow x^{2}\frac{d^{2}y}{dx^{2}}=-x\frac{dy}{dx}-y \quad \text{[using Eqs. (i) and (iii)]}
\displaystyle \Rightarrow x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0
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\displaystyle \textbf{Question 3. }\text{The derivative of }\log x\text{ with respect to }\frac{1}{x}\text{ is} \ \
\displaystyle \text{(a) }\frac{1}{x}\quad \text{(b) }-\frac{1}{x^{3}}\quad \text{(c) }-\frac{1}{x}\quad \text{(d) }-x\text{ \hspace{0.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(d) Let } u=\log x \text{ and } v=\frac{1}{x}
\displaystyle \therefore \frac{du}{dx}=\frac{1}{x} \text{ and } \frac{dv}{dx}=-\frac{1}{x^{2}}
\displaystyle \Rightarrow \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{1}{x}}{-\frac{1}{x^{2}}}=-x
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\displaystyle \textbf{Question 4. }\text{If }y=e^{ax}\cos bx,\text{ then prove that} \ \
\displaystyle \frac{d^{2}y}{dx^{2}}-2a\frac{dy}{dx}+(a^{2}+b^{2})y=0.\text{ \hspace{0.2cm} ISC 2023} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=e^{ax}\cos bx \qquad \cdots(i)
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=be^{ax}(-\sin bx)+ae^{ax}\cos bx
\displaystyle \Rightarrow \frac{dy}{dx}=-be^{ax}\sin bx+ay \quad \text{[from Eq. (i)]} \qquad \cdots(ii)
\displaystyle \text{Again, on differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=-b^{2}e^{ax}\cos bx-abe^{ax}\sin bx+a\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=-b^{2}y+a\left(\frac{dy}{dx}-ay\right)+a\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=-b^{2}y+a\frac{dy}{dx}-a^{2}y+a\frac{dy}{dx}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}-2a\frac{dy}{dx}+(a^{2}+b^{2})y=0
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\displaystyle \textbf{Question 5. }\text{Find }\frac{dy}{dx}\text{ if }x^{3}+y^{3}=3axy.\text{ \hspace{0.2cm} ISC 2020} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } x^{3}+y^{3}=3axy
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle 3x^{2}+3y^{2}\frac{dy}{dx}=3a\left(x\frac{dy}{dx}+y\right)
\displaystyle \Rightarrow 3x^{2}+3y^{2}\frac{dy}{dx}=3ax\frac{dy}{dx}+3ay
\displaystyle \Rightarrow 3y^{2}\frac{dy}{dx}-3ax\frac{dy}{dx}=3ay-3x^{2}
\displaystyle \Rightarrow \frac{dy}{dx}(3y^{2}-3ax)=3ay-3x^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3(ay-x^{2})}{3(y^{2}-ax)}
\displaystyle \therefore \frac{dy}{dx}=\frac{ay-x^{2}}{y^{2}-ax}
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\displaystyle \textbf{Question 6. }\text{If }y=e^{m\sin^{-1}x},\text{ prove that }(1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=m^{2}y.\text{ \hspace{0.2cm} ISC 2020} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=e^{m\sin^{-1}x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \log y=m\sin^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{m}{\sqrt{1-x^{2}}}
\displaystyle \Rightarrow \sqrt{1-x^{2}}\frac{dy}{dx}=my
\displaystyle \text{On squaring both sides, we get}
\displaystyle (1-x^{2})\left(\frac{dy}{dx}\right)^{2}=m^{2}y^{2}
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle (1-x^{2})\left(2\frac{dy}{dx}\frac{d^{2}y}{dx^{2}}\right)-2x\left(\frac{dy}{dx}\right)^{2}=2m^{2}y\frac{dy}{dx}
\displaystyle \text{On dividing both sides by } 2\frac{dy}{dx}, \text{ we get}
\displaystyle (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=m^{2}y \quad \text{Hence proved.}
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\displaystyle \textbf{Question 7. }\text{If }y=e^{\sin^{-1}x}\text{ and }z=e^{\cos^{-1}x},\text{ prove that } \frac{dy}{dz}=e^{\pi/2}.\text{ \hspace{0.2cm} ISC 2019} \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=e^{\sin^{-1}x} \text{ and } z=e^{\cos^{-1}x}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=e^{\sin^{-1}x}\cdot \frac{1}{\sqrt{1-x^{2}}}
\displaystyle \text{and } \frac{dz}{dx}=e^{\cos^{-1}x}\cdot \left(-\frac{1}{\sqrt{1-x^{2}}}\right)
\displaystyle \therefore \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{e^{\sin^{-1}x}\cdot \frac{1}{\sqrt{1-x^{2}}}}{-e^{\cos^{-1}x}\cdot \frac{1}{\sqrt{1-x^{2}}}}
\displaystyle =-e^{\sin^{-1}x-\cos^{-1}x}
\displaystyle =-e^{\frac{\pi}{2}-2\cos^{-1}x} \quad \left[\because \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\right]
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\displaystyle \textbf{Question 8. }\text{If }y=\cos(\sin x),\text{ show that }  \frac{d^{2}y}{dx^{2}}+\tan x\frac{dy}{dx}+y\cos^{2}x=0.\text{ \hspace{0.2cm} ISC 2017} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\cos(\sin x)
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=-\sin(\sin x)\cdot \cos x \qquad \cdots(i)
\displaystyle \text{On differentiating again w.r.t. } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=-\cos(\sin x)\cdot (\cos x)^{2}+\sin(\sin x)\cdot \sin x \qquad \cdots(ii)
\displaystyle \text{Now, LHS }=\frac{d^{2}y}{dx^{2}}+\tan x\frac{dy}{dx}+y\cos^{2}x
\displaystyle =\left[-\cos(\sin x)\cos^{2}x+\sin(\sin x)\sin x\right]+\tan x[-\sin(\sin x)\cos x]+\cos(\sin x)\cos^{2}x
\displaystyle =\sin x\sin(\sin x)-\sin x\sin(\sin x)=0=\text{RHS}
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 9. }\text{If }\log y=\tan^{-1}x,\text{ then prove that} \ \
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-1)\frac{dy}{dx}=0.\text{ \hspace{0.2cm} ISC 2016} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \log y=\tan^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x^{2}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{1+x^{2}} \Rightarrow (1+x^{2})\frac{dy}{dx}=y
\displaystyle \text{Again, on differentiating w.r.t. } x, \text{ we get}
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}=\frac{dy}{dx}
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-1)\frac{dy}{dx}=0 \quad \text{Hence proved.}
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\displaystyle \textbf{Question 10. }\text{If }y=e^{m\cos^{-1}x},\text{ then prove that }  (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=m^{2}y.\text{ \hspace{0.2cm} ISC 2015, 13} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=e^{m\cos^{-1}x} \qquad \cdots(i)
\displaystyle \text{Taking log on both sides, we get } \log y=m\cos^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=m\left(-\frac{1}{\sqrt{1-x^{2}}}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{my}{\sqrt{1-x^{2}}} \qquad \cdots(ii)
\displaystyle \text{Again, on differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=-m\left[\frac{\sqrt{1-x^{2}}\frac{dy}{dx}-y\cdot \frac{-x}{\sqrt{1-x^{2}}}}{1-x^{2}}\right]
\displaystyle =-m\left[\frac{\sqrt{1-x^{2}}\frac{dy}{dx}+\frac{xy}{\sqrt{1-x^{2}}}}{1-x^{2}}\right]
\displaystyle \text{Substituting } \frac{dy}{dx}=-\frac{my}{\sqrt{1-x^{2}}} \text{ from (ii), we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=-m\left[\frac{-my+\frac{xy}{\sqrt{1-x^{2}}}}{1-x^{2}}\right]
\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}}=m^{2}y-\frac{mxy}{\sqrt{1-x^{2}}}
\displaystyle \text{Using } \frac{dy}{dx}=-\frac{my}{\sqrt{1-x^{2}}}, \text{ we get}
\displaystyle (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=m^{2}y \quad \text{Hence proved.}
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\displaystyle \textbf{Question 11. }\text{If }y=\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}},\text{ then prove that }  (1-x^{2})\frac{dy}{dx}=x+\frac{y}{x}.\text{ \hspace{0.2cm} ISC 2014} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}
\displaystyle \text{Using quotient rule, } \frac{dy}{dx}=\frac{\sqrt{1-x^{2}}\cdot \frac{d}{dx}(x\sin^{-1}x)-x\sin^{-1}x\cdot \frac{d}{dx}(\sqrt{1-x^{2}})}{( \sqrt{1-x^{2}} )^{2}}
\displaystyle \text{Now, } \frac{d}{dx}(x\sin^{-1}x)=\sin^{-1}x+\frac{x}{\sqrt{1-x^{2}}}
\displaystyle \text{and } \frac{d}{dx}(\sqrt{1-x^{2}})=\frac{-x}{\sqrt{1-x^{2}}}
\displaystyle \therefore \frac{dy}{dx}=\frac{\sqrt{1-x^{2}}\left(\sin^{-1}x+\frac{x}{\sqrt{1-x^{2}}}\right)+x\sin^{-1}x\cdot \frac{x}{\sqrt{1-x^{2}}}}{1-x^{2}}
\displaystyle =\frac{\sqrt{1-x^{2}}\sin^{-1}x+x+x^{2}\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}}{1-x^{2}}
\displaystyle \Rightarrow (1-x^{2})\frac{dy}{dx}=x+\frac{(1-x^{2})\sin^{-1}x+x^{2}\sin^{-1}x}{\sqrt{1-x^{2}}}
\displaystyle =(1-x^{2})\frac{dy}{dx}=x+\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}
\displaystyle \therefore \frac{dy}{dx}=\frac{x}{1-x^{2}}+\frac{\sin^{-1}x}{(1-x^{2})^{3/2}}
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\displaystyle \textbf{Question 12. }\text{If }y=(\cot^{-1}x)^{2},\text{ then show that} \ \
\displaystyle (1+x^{2})^{2}\frac{d^{2}y}{dx^{2}}+2x(1+x^{2})\frac{dy}{dx}=2.\text{ \hspace{0.2cm} ISC 2013} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=(\cot^{-1}x)^{2}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=2(\cot^{-1}x)\cdot \frac{d}{dx}(\cot^{-1}x)
\displaystyle =2(\cot^{-1}x)\left(-\frac{1}{1+x^{2}}\right)
\displaystyle \Rightarrow (1+x^{2})\frac{dy}{dx}=-2\cot^{-1}x \qquad \cdots(i)
\displaystyle \text{Again, on differentiating w.r.t. } x, \text{ we get}
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}=-2\left(-\frac{1}{1+x^{2}}\right)
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}=\frac{2}{1+x^{2}}
\displaystyle \text{Multiplying both sides by } (1+x^{2}), \text{ we get}
\displaystyle (1+x^{2})^{2}\frac{d^{2}y}{dx^{2}}+2x(1+x^{2})\frac{dy}{dx}=2 \quad \text{Hence proved.}
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\displaystyle \textbf{Question 13. }\text{Using a suitable substitution, find the derivative of} \ \
\displaystyle \tan^{-1}\sqrt{\frac{a-x}{a+x}}\text{ with respect to }x.\text{ \hspace{0.2cm} ISC 2010} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\tan^{-1}\sqrt{\frac{a-x}{a+x}}
\displaystyle \text{Put } x=a\cos 2\theta
\displaystyle \Rightarrow y=\tan^{-1}\sqrt{\frac{a-a\cos 2\theta}{a+a\cos 2\theta}}
\displaystyle =\tan^{-1}\sqrt{\frac{a(1-\cos 2\theta)}{a(1+\cos 2\theta)}}
\displaystyle =\tan^{-1}\sqrt{\frac{2\sin^{2}\theta}{2\cos^{2}\theta}}
\displaystyle =\tan^{-1}(\tan\theta)=\theta
\displaystyle \Rightarrow \theta=\tan^{-1}\sqrt{\frac{a-x}{a+x}}
\displaystyle \text{But } x=a\cos 2\theta \Rightarrow \theta=\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right)
\displaystyle \therefore y=\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right)
\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-\frac{x^{2}}{a^{2}}}}\cdot \frac{1}{a}
\displaystyle =-\frac{1}{2\sqrt{a^{2}-x^{2}}}
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\displaystyle \textbf{Question 14. }\text{Find the derivative of }\sin x^{2}\text{ with respect to }x^{3}.\text{ \hspace{0.2cm} ISC 2009} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } u=\sin x^{2} \text{ and } v=x^{3}
\displaystyle \text{To find } \frac{du}{dv}
\displaystyle \text{Now, } \frac{du}{dx}=\cos(x^{2})\cdot \frac{d}{dx}(x^{2})=2x\cos(x^{2})
\displaystyle \text{and } \frac{dv}{dx}=3x^{2}
\displaystyle \therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{2x\cos(x^{2})}{3x^{2}}=\frac{2\cos(x^{2})}{3x}
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\displaystyle \textbf{Question 15. }\text{If }y=\sqrt{\frac{1-\cos x}{1+\cos x}},\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2008} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\sqrt{\frac{1-\cos x}{1+\cos x}}
\displaystyle \text{On multiplying and dividing RHS by } (1-\cos x), \text{ we get}
\displaystyle y=\sqrt{\frac{(1-\cos x)^{2}}{(1+\cos x)(1-\cos x)}}
\displaystyle =\sqrt{\frac{(1-\cos x)^{2}}{1-\cos^{2}x}}=\sqrt{\frac{(1-\cos x)^{2}}{\sin^{2}x}}
\displaystyle =\frac{1-\cos x}{\sin x}
\displaystyle =\frac{1}{\sin x}-\frac{\cos x}{\sin x}
\displaystyle \Rightarrow y=\mathrm{cosec}\,x-\cot x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=-\mathrm{cosec}\,x\cot x+\mathrm{cosec}^{2}x
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\displaystyle \textbf{Question 16. }\text{If }e^{x+y}=xy,\text{ then show that }\frac{dy}{dx}=\frac{y(1-x)}{x(y-1)}.\text{ \hspace{0.2cm} ISC 2007} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } e^{x+y}=xy
\displaystyle \text{Taking log on both sides, we get}
\displaystyle x+y=\log x+\log y
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle 1+\frac{dy}{dx}=\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}
\displaystyle \Rightarrow \left(1-\frac{1}{y}\right)\frac{dy}{dx}=\frac{1}{x}-1
\displaystyle \therefore \frac{dy}{dx}=\frac{y(1-x)}{x(y-1)} \quad \text{Hence proved.}
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\displaystyle \textbf{Question 17. }\text{If }y=e^{\sin x^{2}},\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2006} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=e^{\sin x^{2}}
\displaystyle \text{Let } y=e^{z},\ z=\sin t \text{ and } t=x^{2}
\displaystyle \text{On differentiating both sides of } t \text{ w.r.t. } x, \text{ we get } \frac{dt}{dx}=2x \qquad (i)
\displaystyle \text{On differentiating both sides of } z \text{ w.r.t. } t, \text{ we get } \frac{dz}{dt}=\cos t \qquad (ii)
\displaystyle \text{On differentiating both sides of } y \text{ w.r.t. } z, \text{ we get } \frac{dy}{dz}=e^{z} \qquad (iii)
\displaystyle \text{Using chain rule, } \frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dt}\cdot \frac{dt}{dx}
\displaystyle =e^{z}\cdot \cos t \cdot 2x
\displaystyle =2x\, e^{\sin x^{2}} \cos x^{2}
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\displaystyle \textbf{Question 18. }\text{If }\sin xy+\cos xy=1\text{ and }\tan xy\ne 1,\text{ then show that} \ \
\displaystyle \frac{dy}{dx}=-\frac{y}{x}.\text{ \hspace{0.2cm} ISC 2005} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \sin xy+\cos xy=1
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{d}{dx}(\sin xy)+\frac{d}{dx}(\cos xy)=0
\displaystyle \Rightarrow \cos(xy)\cdot \frac{d(xy)}{dx}-\sin(xy)\cdot \frac{d(xy)}{dx}=0
\displaystyle \Rightarrow (\cos xy-\sin xy)\left(x\frac{dy}{dx}+y\right)=0
\displaystyle \Rightarrow \left(1-\tan xy\right)\left(x\frac{dy}{dx}+y\right)=0 \quad \left[\text{divide by } \cos xy\right]
\displaystyle \Rightarrow x\frac{dy}{dx}+y=0 \quad \left[\because \tan xy\neq 1\right]
\displaystyle \therefore \frac{dy}{dx}=-\frac{y}{x}
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\displaystyle \textbf{Question 19. }\text{If }x^{y}y^{x}=5,\text{ then show that } \frac{dy}{dx}=-\frac{\log y+\frac{y}{x}}{\log x+\frac{x}{y}}.\text{ \hspace{0.2cm} ISC 2004} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } x^{y}y^{x}=5
\displaystyle \text{Taking log on both sides, we get}
\displaystyle y\log x+x\log y=\log 5
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle y\cdot \frac{1}{x}+\log x\frac{dy}{dx}+\log y+x\cdot \frac{1}{y}\frac{dy}{dx}=0
\displaystyle \Rightarrow \left(\log x+\frac{x}{y}\right)\frac{dy}{dx}=-\left(\frac{y}{x}+\log y\right)
\displaystyle \therefore \frac{dy}{dx}=-\frac{\frac{y}{x}+\log y}{\frac{x}{y}+\log x}
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\displaystyle \textbf{Question 20. }\text{If }x=a\sin^{3}t\text{ and }y=a\cos^{3}t,\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2004} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } x=a\sin^{3}t \text{ and } y=a\cos^{3}t
\displaystyle \text{Now, differentiating both sides of } x \text{ w.r.t. } t, \text{ we get}
\displaystyle \frac{dx}{dt}=a\cdot 3\sin^{2}t\cos t=3a\sin^{2}t\cos t \qquad (i)
\displaystyle \text{Now, differentiating both sides of } y \text{ w.r.t. } t, \text{ we get}
\displaystyle \frac{dy}{dt}=a\cdot 3\cos^{2}t(-\sin t)=-3a\sin t\cos^{2}t \qquad (ii)
\displaystyle \text{Now, } \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-3a\sin t\cos^{2}t}{3a\sin^{2}t\cos t}
\displaystyle =-\frac{\cos t}{\sin t}=-\cot t
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\displaystyle \textbf{Question 21. }\text{If }y=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right),\text{ then prove that } \frac{dy}{dx}=\frac{2}{1+x^{2}}.\text{ \hspace{0.2cm} ISC 2003} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right)
\displaystyle \text{Put } x=\tan\theta \Rightarrow y=\tan^{-1}\left(\frac{2\tan\theta}{1-\tan^{2}\theta}\right)
\displaystyle =\tan^{-1}(\tan 2\theta)=2\theta
\displaystyle \Rightarrow y=2\tan^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=2\cdot \frac{1}{1+x^{2}}
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\displaystyle \textbf{Question 22. }\text{If }y=e^{x}\log\tan 2x,\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2002} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=e^{x}\log(\tan 2x)
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=e^{x}\frac{d}{dx}[\log(\tan 2x)]+\log(\tan 2x)\frac{d}{dx}(e^{x})
\displaystyle =e^{x}\left(\frac{1}{\tan 2x}\cdot \frac{d}{dx}(\tan 2x)\right)+e^{x}\log(\tan 2x)
\displaystyle =e^{x}\left(\frac{1}{\tan 2x}\cdot \sec^{2}2x \cdot 2\right)+e^{x}\log(\tan 2x)
\displaystyle =\frac{2e^{x}\sec^{2}2x}{\tan 2x}+e^{x}\log(\tan 2x)
\displaystyle \frac{dy}{dx}=\frac{2e^{x}\sec^{2}2x}{\tan 2x}+e^{x}\log(\tan 2x)
\displaystyle =\frac{2e^{x}}{\sin 2x\cos 2x}+e^{x}\log(\tan 2x)
\displaystyle =\frac{4e^{x}}{\sin 4x}+e^{x}\log(\tan 2x)
\displaystyle =e^{x}\left(4\,\mathrm{cosec}\,4x+\log(\tan 2x)\right)
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\displaystyle \textbf{Question 23. }\text{If }y=(\cos x)^{\cos x},\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2001} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=(\cos x)^{\cos x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \log y=\cos x\log(\cos x)
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=\cos x\cdot \frac{-\sin x}{\cos x}+ \log(\cos x)\cdot (-\sin x)
\displaystyle =-\sin x-\sin x\log(\cos x)
\displaystyle \therefore \frac{dy}{dx}=y[-\sin x(1+\log(\cos x))]
\displaystyle =(\cos x)^{\cos x}\left[-\sin x(1+\log(\cos x))\right]
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\displaystyle \textbf{Question 24. }\text{If }y=\log\sqrt{\frac{1-\cos x}{1+\cos x}},\text{ then find }\frac{dy}{dx}.\text{ \hspace{0.2cm} ISC 2000} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\log\sqrt{\frac{1-\cos x}{1+\cos x}}
\displaystyle \Rightarrow y=\frac{1}{2}\log\left(\frac{1-\cos x}{1+\cos x}\right)
\displaystyle =\frac{1}{2}\left[\log(1-\cos x)-\log(1+\cos x)\right]
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{1-\cos x}\cdot \sin x-\frac{1}{1+\cos x}\cdot (-\sin x)\right]
\displaystyle =\frac{1}{2}\left[\frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}\right]
\displaystyle =\frac{1}{2}\left[\frac{\sin x(1+\cos x+1-\cos x)}{1-\cos^{2}x}\right]
\displaystyle =\frac{1}{2}\cdot \frac{2\sin x}{\sin^{2}x}=\frac{1}{\sin x}
\displaystyle \therefore \frac{dy}{dx}=\mathrm{cosec}\,x
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