\displaystyle \textbf{Question 1. }\text{If }\sin^{-1}x+\sin^{-1}y=\frac{\pi}{2},\text{ then }\frac{dy}{dx}\text{ is equal to} \ \
\displaystyle \text{(a) }\frac{x}{y}\quad \text{(b) }-\frac{x}{y}\quad \text{(c) }\frac{y}{x}\quad \text{(d) }-\frac{y}{x}  \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } \sin^{-1}x+\sin^{-1}y=\frac{\pi}{2}
\displaystyle \Rightarrow \sin^{-1}x=\frac{\pi}{2}-\sin^{-1}y
\displaystyle \Rightarrow \sin^{-1}x=\cos^{-1}y \quad \left[\because \cos^{-1}\theta=\frac{\pi}{2}-\sin^{-1}\theta\right]
\displaystyle \Rightarrow y=\sqrt{1-x^{2}} \qquad \cdots(i)
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^{2}}}(-2x)=\frac{-x}{\sqrt{1-x^{2}}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-x}{y} \quad \left[\text{from Eq. (i), } y=\sqrt{1-x^{2}}\right]
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\displaystyle \textbf{Question 2. }\text{If }y=t^{2}\text{ and }t=x+3,\text{ then }\frac{dy}{dx}\text{ is equal to} \ \
\displaystyle \text{(a) }(x+3)^{2}\quad \text{(b) }2(x+3)\quad \text{(c) }2t\quad \text{(d) }2(x+3)^{2}  \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) We have, } y=t^{2} \Rightarrow \frac{dy}{dt}=2t
\displaystyle \text{and } t=x+3 \Rightarrow x=t-3
\displaystyle \Rightarrow \frac{dx}{dt}=1
\displaystyle \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\displaystyle =\frac{2t}{1}=2t=2(x+3)
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\displaystyle \textbf{Question 3. }\text{What will be the derivative of }\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\text{ with} \ \
\displaystyle \text{respect to }\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)?  \ \
\displaystyle \text{(a) }-1\quad \text{(b) }1\quad \text{(c) }2\quad \text{(d) }4 \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } u=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)
\displaystyle \text{and } v=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)
\displaystyle \text{Now, let } x=\tan\theta
\displaystyle \text{On evaluating } \frac{du}{d\theta} \text{ and } \frac{dv}{d\theta}, \text{ we get}
\displaystyle \frac{du}{d\theta}=\frac{d}{d\theta}\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)
\displaystyle =\frac{d}{d\theta}\sin^{-1}(\sin 2\theta)
\displaystyle =\frac{d}{d\theta}(2\theta)=2
\displaystyle \frac{dv}{d\theta}=\frac{d}{d\theta}\cos^{-1}\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)
\displaystyle =\frac{d}{d\theta}\cos^{-1}(\cos 2\theta)
\displaystyle =\frac{d}{d\theta}(2\theta)=2
\displaystyle \therefore \frac{du}{dv}=\frac{\frac{du}{d\theta}}{\frac{dv}{d\theta}}=\frac{2}{2}=1
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\displaystyle \textbf{Question 4. }\text{If }y=\sqrt{\sin x+y},\text{ then find }\frac{dy}{dx}.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=(\sin x+y)^{\frac{1}{2}}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{2}(\sin x+y)^{-\frac{1}{2}}\left(\cos x+\frac{dy}{dx}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{2y}\left(\cos x+\frac{dy}{dx}\right) \quad \left[\because (\sin x+y)^{\frac{1}{2}}=y\right]
\displaystyle \Rightarrow \frac{dy}{dx}\left(1-\frac{1}{2y}\right)=\frac{\cos x}{2y}
\displaystyle \therefore \frac{dy}{dx}=\frac{\cos x}{2y}\cdot \frac{2y}{2y-1}=\frac{\cos x}{2y-1}
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\displaystyle \textbf{Question 5. }\text{Differentiate the function with reference to }x: \ \
\displaystyle f(x)=\cos^{-1}\left(\sqrt{\frac{1-\cos x}{2}}\right).  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\cos^{-1}\sqrt{\frac{1-\cos x}{2}}
\displaystyle =\cos^{-1}\sqrt{\frac{2\sin^{2}\frac{x}{2}}{2}}
\displaystyle =\cos^{-1}\sqrt{\sin^{2}\frac{x}{2}}
\displaystyle =\cos^{-1}\left(\sin\frac{x}{2}\right)
\displaystyle =\cos^{-1}\left(\cos\left(\frac{\pi}{2}-\frac{x}{2}\right)\right)
\displaystyle \Rightarrow y=\frac{\pi}{2}-\frac{x}{2}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we obtain}
\displaystyle \frac{dy}{dx}=-\frac{1}{2}
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\displaystyle \textbf{Question 6. }\text{Find }\frac{dy}{dx},\text{ if }x=at^{2}\text{ at }y=2at.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given that, } x=at^{2},\ y=2at
\displaystyle \text{On differentiating both functions w.r.t. } t, \text{ we get}
\displaystyle \frac{dx}{dt}=2at \text{ and } \frac{dy}{dt}=2a
\displaystyle \text{Now, } \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}
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\displaystyle \textbf{Question 7. }\text{The second derivative of }y=x^{3}-5x^{2}+x\text{ is} \ \
\displaystyle \text{(a) }10x-5\quad \text{(b) }6x-10\quad \text{(c) }3x^{2}-10x\quad \text{(d) }3x^{2}-10x+1  \ \
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } y=x^{3}-5x^{2}+x
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(x^{3}-5x^{2}+x)
\displaystyle =3x^{2}-5(2x)+1=3x^{2}-10x+1
\displaystyle \therefore \frac{dy}{dx}=3x^{2}-10x+1
\displaystyle \text{Again, on differentiating w.r.t. } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(\frac{dy}{dx}\right)
\displaystyle =\frac{d}{dx}(3x^{2}-10x+1)
\displaystyle =3(2x)-10(1)+0=6x-10
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\displaystyle \textbf{Question 8. }\text{If }x=a\left(\frac{1-t^{2}}{1+t^{2}}\right)\text{ and }y=\frac{2at}{1+t^{2}},\text{ prove that} \ \
\displaystyle \frac{dy}{dx}=-\frac{x}{y}.\text{ Hence, prove that } \frac{d^{2}y}{dx^{2}}=-\left(\frac{x^{2}+y^{2}}{y^{3}}\right).  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } x=a\left(\frac{1-t^{2}}{1+t^{2}}\right) \text{ and } y=\frac{2at}{1+t^{2}}
\displaystyle \text{Let } t=\tan\theta
\displaystyle \therefore x=a\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)=a\cos 2\theta
\displaystyle \text{and } y=\frac{2a\tan\theta}{1+\tan^{2}\theta}=a\sin 2\theta
\displaystyle \text{Now, } \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{2a\cos 2\theta}{-2a\sin 2\theta}=-\frac{\cos 2\theta}{\sin 2\theta}=-\frac{x}{y}
\displaystyle \text{Now, } \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(-\frac{x}{y}\right)
\displaystyle =-\frac{y-x\frac{dy}{dx}}{y^{2}}
\displaystyle =-\frac{y-x\left(-\frac{x}{y}\right)}{y^{2}}
\displaystyle =-\frac{\frac{y^{2}+x^{2}}{y}}{y^{2}}
\displaystyle \therefore \frac{d^{2}y}{dx^{2}}=-\frac{x^{2}+y^{2}}{y^{3}} \qquad \text{Hence proved.}
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\displaystyle \textbf{Question 9. }\text{If }y=(x+\sqrt{1+x^{2}})^{n},\text{ then prove that} \ \
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=n^{2}y.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\left(x+\sqrt{1+x^{2}}\right)^{n}
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\frac{d}{dx}\left(x+\sqrt{1+x^{2}}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{2x}{2\sqrt{1+x^{2}}}\right)
\displaystyle =n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{x}{\sqrt{1+x^{2}}}\right)
\displaystyle \text{Continuing Q5:}
\displaystyle \frac{dy}{dx}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{x}{\sqrt{1+x^{2}}}\right)
\displaystyle =\frac{n\left(x+\sqrt{1+x^{2}}\right)^{n}}{\sqrt{1+x^{2}}}=\frac{ny}{\sqrt{1+x^{2}}}
\displaystyle \Rightarrow \sqrt{1+x^{2}}\frac{dy}{dx}=ny \qquad \cdots(i)
\displaystyle \text{Again, on differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \sqrt{1+x^{2}}\frac{d^{2}y}{dx^{2}}+\frac{x}{\sqrt{1+x^{2}}}\frac{dy}{dx}=n\frac{dy}{dx}
\displaystyle \text{Substituting } \frac{dy}{dx}=\frac{ny}{\sqrt{1+x^{2}}} \text{ from Eq. (i), we get}
\displaystyle \sqrt{1+x^{2}}\frac{d^{2}y}{dx^{2}}+\frac{x}{\sqrt{1+x^{2}}}\cdot \frac{ny}{\sqrt{1+x^{2}}}=\frac{n^{2}y}{\sqrt{1+x^{2}}}
\displaystyle \text{On multiplying both sides by } \sqrt{1+x^{2}}, \text{ we get}
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=n^{2}y \quad \text{Hence proved.}
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\displaystyle \textbf{Question 10. }\text{If }x=\tan\left(\frac{1}{a}\log y\right),\text{ prove that} \ \
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-a)\frac{dy}{dx}=0.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given that } x=\tan^{-1}\left(\frac{1}{a}\log y\right)
\displaystyle \Rightarrow \tan^{-1}x=\frac{1}{a}\log y
\displaystyle \Rightarrow a\tan^{-1}x=\log y
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we obtain}
\displaystyle \frac{a}{1+x^{2}}=\frac{1}{y}\frac{dy}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{ay}{1+x^{2}}
\displaystyle \Rightarrow (1+x^{2})\frac{dy}{dx}=ay \qquad \cdots(i)
\displaystyle \text{Again, on differentiating both sides w.r.t. } x, \text{ we obtain}
\displaystyle (1+x^{2})\frac{d^{2}y}{dx^{2}}+2x\frac{dy}{dx}=a\frac{dy}{dx}
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-a)\frac{dy}{dx}=0
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\displaystyle \textbf{Question 11. }\text{If }y=e^{a\cos^{-1}x},\text{ where }-1\leq x\leq 1,\text{ show that} \ \
\displaystyle (1-x^{2})y_{2}-xy_{1}-a^{2}y=0.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y=e^{a\cos^{-1}x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \log y=a\cos^{-1}x
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{1}{y}\frac{dy}{dx}=a\left(-\frac{1}{\sqrt{1-x^{2}}}\right)
\displaystyle \therefore \frac{dy}{dx}=-\frac{ay}{\sqrt{1-x^{2}}}
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{a}{\sqrt{1-x^{2}}}
\displaystyle \Rightarrow \sqrt{1-x^{2}}\frac{dy}{dx}=ay
\displaystyle \Rightarrow (1-x^{2})\left(\frac{dy}{dx}\right)^{2}=a^{2}y^{2}
\displaystyle \text{Again, 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}=2a^{2}y\frac{dy}{dx}
\displaystyle \text{On dividing by } 2\frac{dy}{dx}, \text{ we get}
\displaystyle (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=a^{2}y
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