Class 12: Higher Order Derivative

$\displaystyle \textbf{Question 1: }~\text{Find the second order derivatives of each of the following functions:}$

$\displaystyle \textbf{(i) }~x^3+\tan x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = x^{3} + \tan x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 3x^{2} + \sec^{2}x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 6x + 2\sec^{2}x\tan x$

$\displaystyle \textbf{(ii) }~\sin(\log x)$
$\displaystyle \text{Answer:}$

$\displaystyle y = \sin(\log x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \cos(\log x)\cdot\frac{1}{x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\sin(\log x)\cdot\frac{1}{x}\cdot\frac{1}{x} + \cos(\log x)\cdot\left(-\frac{1}{x^{2}}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\sin(\log x)+\cos(\log x)}{x^{2}}$

$\displaystyle \textbf{(iii) }~\log(\sin x)$
$\displaystyle \text{Answer:}$

$\displaystyle y = \log(\sin x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sin x}\cdot\cos x$

$\displaystyle \Rightarrow \frac{dy}{dx} = \cot x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\mathrm{cosec}^{2}x$

$\displaystyle \textbf{(iv) }~e^x\sin 5x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = e^{x}\sin(5x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{x}\sin(5x) + e^{x}\cdot 5\cos(5x)$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = e^{x}\sin(5x) + 5e^{x}\cos(5x) + 5e^{x}\cos(5x) - 25e^{x}\sin(5x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -24e^{x}\sin(5x) + 10e^{x}\cos(5x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2e^{x}\big(5\cos(5x) - 12\sin(5x)\big)$

$\displaystyle \textbf{(v) }~e^{6x}\cos 3x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = e^{6x}\cos(3x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{6x}\cdot 6\cos(3x) + e^{6x}\cdot(-3\sin(3x))$

$\displaystyle \Rightarrow \frac{dy}{dx} = 6e^{6x}\cos(3x) - 3e^{6x}\sin(3x)$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 6e^{6x}\cdot 6\cos(3x) - 6e^{6x}\cdot 3\sin(3x) - 3e^{6x}\cdot 6\sin(3x) - 3e^{6x}\cdot 3\cos(3x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 27e^{6x}\cos(3x) - 36e^{6x}\sin(3x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 9e^{6x}\big(3\cos(3x) - 4\sin(3x)\big)$

$\displaystyle \textbf{(vi) }~x^3\log x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = x^{3}\log x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 3x^{2}\log x + x^{3}\cdot\frac{1}{x}$

$\displaystyle \Rightarrow \frac{dy}{dx} = 3x^{2}\log x + x^{2}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 6x\log x + 3x^{2}\cdot\frac{1}{x} + 2x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 6x\log x + 5x$

$\displaystyle \textbf{(vii) }~\tan^{-1}x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = \tan^{-1}x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{1+x^{2}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{2x}{(1+x^{2})^{2}}$

$\displaystyle \textbf{(viii) }~x\cos x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = x\cos x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \cos x - x\sin x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\sin x - \sin x - x\cos x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -(2\sin x + x\cos x)$

$\displaystyle \textbf{(ix) }~\log(\log x)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle y = \log(\log x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{\log x}\cdot\frac{1}{x}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x\log x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\big((x\log x)^{-1}\big)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{(x\log x)' }{(x\log x)^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\log x + 1}{(x\log x)^{2}}$

$\displaystyle \textbf{Question 2: }~\text{If }y=e^{-x}\cos x,\text{ show that }\frac{d^2y}{dx^2}=2e^{-x}\sin x.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = e^{-x}\cos x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = -e^{-x}\sin x - e^{-x}\cos x$

$\displaystyle \Rightarrow \frac{dy}{dx} = -(e^{-x}\sin x + e^{-x}\cos x)$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\big(-e^{-x}\cos x - e^{-x}\sin x - e^{-x}\sin x + e^{-x}\cos x\big)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2e^{-x}\sin x$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 3: }~\text{If }y=x+\tan x,\text{ show that }\cos^2x\frac{d^2y}{dx^2}-2y+2x=0.\;[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = x + \tan x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 1 + \sec^{2}x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 2\sec^{2}x\tan x$

$\displaystyle \text{Dividing both sides by } \sec^{2}x,$

$\displaystyle \cos^{2}x\,\frac{d^{2}y}{dx^{2}} = 2\tan x$

$\displaystyle \Rightarrow \cos^{2}x\,\frac{d^{2}y}{dx^{2}} = 2(y-x)\quad[\because y=x+\tan x \Rightarrow \tan x = y-x]$

$\displaystyle \Rightarrow \cos^{2}x\,\frac{d^{2}y}{dx^{2}} - 2y + 2x = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 4: }~\text{If }y=x^3\log x,\text{ prove that }\frac{d^4y}{dx^4}=\frac{6}{x}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = x^{3}\log x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 3x^{2}\log x + x^{3}\cdot\frac{1}{x}$

$\displaystyle \Rightarrow \frac{dy}{dx} = 3x^{2}\log x + x^{2}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 6x\log x + 3x^{2}\cdot\frac{1}{x} + 2x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 6x\log x + 5x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{3}y}{dx^{3}} = 6\log x + 6x\cdot\frac{1}{x} + 5$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} = 6\log x + 11$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{4}y}{dx^{4}} = \frac{6}{x}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 5: }~\text{If }y=\log(\sin x),\text{ prove that }\frac{d^3y}{dx^3}=2\cos x\,\mathrm{cosec}^3x.$
$\displaystyle \text{Answer:}$

$\displaystyle y = \log(\sin x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sin x}\cdot\cos x$

$\displaystyle \Rightarrow \frac{dy}{dx} = \cot x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\mathrm{cosec}^{2}x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{3}y}{dx^{3}} = -2\mathrm{cosec}x\cdot\frac{d}{dx}(\mathrm{cosec}x)$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} = -2\mathrm{cosec}x\cdot(-\mathrm{cosec}x\cot x)$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} = 2\mathrm{cosec}^{2}x\cot x$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} = 2\cos x\,\mathrm{cosec}^{3}x$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 6: }~\text{If }y=2\sin x+3\cos x,\text{ show that }\frac{d^2y}{dx^2}+y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = 2\sin x + 3\cos x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2\cos x - 3\sin x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -2\sin x - 3\cos x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -(2\sin x + 3\cos x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -y$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + y = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 7: }~\text{If }y=\frac{\log x}{x},\text{ show that }\frac{d^2y}{dx^2}=\frac{2\log x-3}{x^3}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \frac{\log x}{x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1-\log x}{x^{2}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left((1-\log x)x^{-2}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = (-1)\cdot x^{-3} + (1-\log x)(-2)x^{-3}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-x - 2x + 2x\log x}{x^{4}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{2\log x - 3}{x^{3}}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 8: }~\text{If }x=a\sec\theta,\;y=b\tan\theta,\text{ prove that }\frac{d^2y}{dx^2}=-\frac{b^4}{a^2y^3}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a\sec\theta \text{ and } y = b\tan\theta$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = a\sec\theta\tan\theta \quad \text{and} \quad \frac{dy}{d\theta} = b\sec^{2}\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy}{d\theta}\cdot\frac{d\theta}{dx} = \frac{b\sec^{2}\theta}{a\sec\theta\tan\theta}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{b}{a}\,\mathrm{cosec}\,\theta$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{b}{a}\cdot\frac{d}{d\theta}(\mathrm{cosec}\,\theta)\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{b}{a}(-\mathrm{cosec}\,\theta\cot\theta)\cdot\frac{1}{a\sec\theta\tan\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b}{a^{2}}\cot^{3}\theta$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b}{a^{2}}\left(\frac{b}{y}\right)^{3} \quad [\because y = b\tan\theta]$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b^{4}}{a^{2}y^{3}}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 9: }~\text{If }x=a(\cos\theta+\theta\sin\theta),\;y=a(\sin\theta-\theta\cos\theta),\text{ prove that }\frac{d^2x}{d\theta^2}=a(\cos\theta-\theta\sin\theta),\;\frac{d^2y}{d\theta^2}=a(\sin\theta+\theta\cos\theta)\text{ and }\frac{d^2y}{dx^2}=\frac{\sec^3\theta}{a\theta}. \;[\text{CBSE 2012, 2017}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{It is given that } x = a(\cos t + t\sin t) \text{ and } y = a(\sin t - t\cos t)$

$\displaystyle \text{Differentiating } x \text{ with respect to } t,$

$\displaystyle \frac{dx}{dt} = a\frac{d}{dt}(\cos t + t\sin t)$

$\displaystyle \Rightarrow \frac{dx}{dt} = a\left[-\sin t + \sin t\cdot\frac{d}{dt}(t) + t\cdot\frac{d}{dt}(\sin t)\right]$

$\displaystyle \Rightarrow \frac{dx}{dt} = a[-\sin t + \sin t + t\cos t]$

$\displaystyle \Rightarrow \frac{dx}{dt} = at\cos t$

$\displaystyle \text{Differentiating } y \text{ with respect to } t,$

$\displaystyle \frac{dy}{dt} = a\frac{d}{dt}(\sin t - t\cos t)$

$\displaystyle \Rightarrow \frac{dy}{dt} = a\left[\cos t - \left(\cos t\cdot\frac{d}{dt}(t) + t\cdot\frac{d}{dt}(\cos t)\right)\right]$

$\displaystyle \Rightarrow \frac{dy}{dt} = a[\cos t - (\cos t - t\sin t)]$

$\displaystyle \Rightarrow \frac{dy}{dt} = at\sin t$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t$

$\displaystyle \text{Then, } \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\tan t)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \sec^{2}t\cdot\frac{dt}{dx}$

$\displaystyle \text{Since } \frac{dx}{dt} = at\cos t \Rightarrow \frac{dt}{dx} = \frac{1}{at\cos t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{\sec^{2}t}{at\cos t} = \frac{\sec^{3}t}{at}, \quad 0<t<\frac{\pi}{2}$

$\displaystyle \textbf{Question 10: }~\text{If }y=e^x\cos x,\text{ prove that }\frac{d^2y}{dx^2}=2e^x\cos\!\left(x+\frac{\pi}{2}\right).\;[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$

$\displaystyle y = e^{x}\cos x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{x}\cos x - e^{x}\sin x = e^{x}(\cos x - \sin x)$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = e^{x}(\cos x - \sin x) + e^{x}(-\sin x - \cos x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = e^{x}\cos x - e^{x}\sin x - e^{x}\sin x - e^{x}\cos x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -2e^{x}\sin x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2e^{x}\cos\!\left(x+\frac{\pi}{2}\right)$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 11: }~\text{If }x=a\cos\theta,\;y=b\sin\theta,\text{ show that }\frac{d^2y}{dx^2}=-\frac{b^4}{a^2y^3}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a\cos\theta \text{ and } y = b\sin\theta$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = -a\sin\theta \quad \text{and} \quad \frac{dy}{d\theta} = b\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\cos\theta}{-a\sin\theta} = -\frac{b}{a}\cot\theta$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{b}{a}\cdot\frac{d}{d\theta}(\cot\theta)\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b}{a}(-\mathrm{cosec}^{2}\theta)\cdot\frac{1}{-a\sin\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b}{a^{2}}\cdot\frac{1}{\sin^{3}\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b}{a^{2}}\left(\frac{b}{y}\right)^{3} \quad [\because y = b\sin\theta]$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{b^{4}}{a^{2}y^{3}}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 12: }~\text{If }x=a(1-\cos^3\theta),\;y=a\sin^3\theta,\text{ prove that }\frac{d^2y}{dx^2}=\frac{32}{27a}\text{ at }\theta=\frac{\pi}{6}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a(1-\cos^{3}\theta), \quad y = a\sin^{3}\theta$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = a\cdot 3\cos^{2}\theta\sin\theta = 3a\cos^{2}\theta\sin\theta$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{d\theta} = a\cdot 3\sin^{2}\theta\cos\theta = 3a\sin^{2}\theta\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\sin^{2}\theta\cos\theta}{3a\cos^{2}\theta\sin\theta} = \tan\theta$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\tan\theta) = \sec^{2}\theta\cdot\frac{d\theta}{dx}$

$\displaystyle \text{Since } \frac{dx}{d\theta} = 3a\cos^{2}\theta\sin\theta, \quad \frac{d\theta}{dx} = \frac{1}{3a\cos^{2}\theta\sin\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{\sec^{2}\theta}{3a\cos^{2}\theta\sin\theta} = \frac{\sec^{4}\theta}{3a\sin\theta}$

$\displaystyle \text{At } \theta=\frac{\pi}{6},$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{\sec^{4}\!\left(\frac{\pi}{6}\right)} {3a\sin\!\left(\frac{\pi}{6}\right)} = \frac{32}{27a}$

$\displaystyle \textbf{Question 13: }~\text{If }x=a(\theta+\sin\theta),\;y=a(1+\cos\theta),\text{ prove that }\frac{d^2y}{dx^2}=-\frac{a}{y^2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a(\theta + \sin\theta) \text{ and } y = a(1+\cos\theta)$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = a + a\cos\theta \quad \text{and} \quad \frac{dy}{d\theta} = -a\sin\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-a\sin\theta}{a(1+\cos\theta)} = -\frac{\sin\theta}{1+\cos\theta}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(-\frac{\sin\theta}{1+\cos\theta}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{(1+\cos\theta)\cos\theta - \sin\theta(-\sin\theta)} {(1+\cos\theta)^{2}}\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\cos\theta + \cos^{2}\theta + \sin^{2}\theta} {(1+\cos\theta)^{2}}\cdot\frac{1}{a(1+\cos\theta)}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{1+\cos\theta}{a(1+\cos\theta)^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{1}{a(1+\cos\theta)^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{a}{y^{2}} \quad [\because y = a(1+\cos\theta)]$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 14: }~\text{If }x=a(\theta-\sin\theta),\;y=a(1+\cos\theta)\text{ find }\frac{d^2y}{dx^2}. \;[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a(\theta-\sin\theta) \text{ and } y = a(1+\cos\theta)$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = a-a\cos\theta \quad \text{and} \quad \frac{dy}{d\theta} = -a\sin\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-a\sin\theta}{a(1-\cos\theta)} = -\frac{\sin\theta}{1-\cos\theta}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(-\frac{\sin\theta}{1-\cos\theta}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{(1-\cos\theta)\cos\theta-\sin\theta(\sin\theta)} {(1-\cos\theta)^{2}}\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{-\cos\theta+\cos^{2}\theta+\sin^{2}\theta} {(1-\cos\theta)^{2}}\cdot\frac{1}{a(1-\cos\theta)}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{1-\cos\theta}{a(1-\cos\theta)^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{1}{a(1-\cos\theta)^{2}}$

$\displaystyle \textbf{Question 15: }~\text{If }x=a(1-\cos\theta),\;y=a(\theta+\sin\theta),\text{ prove that }\frac{d^2y}{dx^2}=-\frac{1}{a}\text{ at }\theta=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a(1-\cos\theta) \text{ and } y = a(\theta+\sin\theta)$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = a\sin\theta \quad \text{and} \quad \frac{dy}{d\theta} = a + a\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1+\cos\theta)}{a\sin\theta} = \frac{1+\cos\theta}{\sin\theta}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{1+\cos\theta}{\sin\theta}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{\sin\theta(-\sin\theta)-(1+\cos\theta)\cos\theta} {\sin^{2}\theta}\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-\sin^{2}\theta-(1+\cos\theta)\cos\theta} {\sin^{2}\theta}\cdot\frac{1}{a\sin\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-1-\cos\theta}{a\sin^{3}\theta}$

$\displaystyle \text{At } \theta=\frac{\pi}{2},$

$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{\theta=\pi/2} = \frac{-1-0}{a\cdot 1} = -\frac{1}{a}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 16: }~\text{If }x=a(1+\cos\theta),\;y=a(\theta+\sin\theta),\text{ prove that }\frac{d^2y}{dx^2}=-\frac{1}{a}\text{ at }\theta=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = a(1+\cos\theta) \text{ and } y = a(\theta+\sin\theta)$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = -a\sin\theta \quad \text{and} \quad \frac{dy}{d\theta} = a + a\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1+\cos\theta)}{-a\sin\theta} = -\frac{1+\cos\theta}{\sin\theta}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(-\frac{1+\cos\theta}{\sin\theta}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\sin\theta(-\sin\theta)-(1+\cos\theta)\cos\theta} {\sin^{2}\theta}\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\sin^{2}\theta-(1+\cos\theta)\cos\theta} {\sin^{2}\theta}\cdot\frac{1}{-a\sin\theta}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{1+\cos\theta}{a\sin^{3}\theta}$

$\displaystyle \text{At } \theta=\frac{\pi}{2},$

$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{\theta=\pi/2} = -\frac{1+\cos\frac{\pi}{2}}{a(\sin\frac{\pi}{2})^{3}} = -\frac{1}{a}$

$\displaystyle \textbf{Question 17: }~\text{If }x=\cos\theta,\;y=\sin^3\theta,\text{ prove that }y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=3\sin^2\theta(5\cos^2\theta-1).\;[\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = \cos\theta \text{ and } y = \sin^{3}\theta$

$\displaystyle \text{Differentiating with respect to } \theta,$

$\displaystyle \frac{dx}{d\theta} = -\sin\theta \quad \text{and} \quad \frac{dy}{d\theta} = 3\sin^{2}\theta\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\sin^{2}\theta\cos\theta}{-\sin\theta} = -3\sin\theta\cos\theta$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{d\theta}\!\left(-3\sin\theta\cos\theta\right)\cdot\frac{d\theta}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = (-3\cos^{2}\theta + 3\sin^{2}\theta)\cdot\frac{1}{-\sin\theta}$

$\displaystyle \text{Now,}$

$\displaystyle \text{LHS } = y\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2}$

$\displaystyle = \sin^{3}\theta\cdot \frac{-3\cos^{2}\theta + 3\sin^{2}\theta}{-\sin\theta} + ( -3\sin\theta\cos\theta )^{2}$

$\displaystyle = 3\sin^{2}\theta\cos^{2}\theta - 3\sin^{4}\theta + 9\sin^{2}\theta\cos^{2}\theta$

$\displaystyle = 12\sin^{2}\theta\cos^{2}\theta - 3\sin^{4}\theta$

$\displaystyle = 3\sin^{2}\theta(4\cos^{2}\theta - \sin^{2}\theta)$

$\displaystyle = 3\sin^{2}\theta(5\cos^{2}\theta - 1) \quad [\because \sin^{2}\theta + \cos^{2}\theta = 1]$

$\displaystyle = \text{RHS}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 18: }~\text{If }y=\sin(\sin x),\text{ prove that }\frac{d^2y}{dx^2}+\tan x\frac{dy}{dx}+y\cos^2x=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \sin(\sin x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \cos(\sin x)\cos x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\sin(\sin x)\cos x\cdot\cos x - \cos(\sin x)\sin x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\sin(\sin x)\cos^{2}x - \cos(\sin x)\sin x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -y\cos^{2}x - \sin x\,\frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + \tan x\,\frac{dy}{dx} + y\cos^{2}x = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 19: }~\text{If }x=\sin t,\;y=\sin pt,\text{ prove that }(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = \sin t \text{ and } y = \sin pt$

$\displaystyle \text{Differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = \cos t \quad \text{and} \quad \frac{dy}{dt} = p\cos pt$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p\cos pt}{\cos t}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\!\left(\frac{p\cos pt}{\cos t}\right)\cdot\frac{dt}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-p^{2}\sin pt\cos t + p\cos pt\sin t}{\cos^{2}t}\cdot\frac{1}{\cos t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-p^{2}\sin pt\cos t + p\cos pt\sin t}{\cos^{3}t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{p^{2}\sin pt}{\cos^{2}t} + \frac{p\cos pt\sin t}{\cos^{3}t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{p^{2}y}{\cos^{2}t} + \frac{x}{\cos^{2}t}\frac{dy}{dx}$

$\displaystyle \Rightarrow \cos^{2}t\,\frac{d^{2}y}{dx^{2}} = -p^{2}y + x\frac{dy}{dx}$

$\displaystyle \Rightarrow (1-\sin^{2}t)\frac{d^{2}y}{dx^{2}} = -p^{2}y + x\frac{dy}{dx}$

$\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} + p^{2}y = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 20: }~\text{If }y=(\sin^{-1}x)^2,\text{ prove that }(1-x^2)y_2-xy_1-2=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = (\sin^{-1}x)^{2}$

$\displaystyle \text{Let } y_{1} = \frac{dy}{dx} \text{ and } y_{2} = \frac{d^{2}y}{dx^{2}}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle y_{1} = 2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle y_{2} = 2\cdot\frac{1}{1-x^{2}} + 2\sin^{-1}x\cdot\frac{x}{(1-x^{2})^{3/2}}$

$\displaystyle \Rightarrow y_{2} = \frac{2}{1-x^{2}} + \frac{2x\sin^{-1}x}{(1-x^{2})\sqrt{1-x^{2}}}$

$\displaystyle \Rightarrow y_{2} = \frac{2}{1-x^{2}} + \frac{x\,y_{1}}{1-x^{2}}$

$\displaystyle \Rightarrow y_{2}(1-x^{2}) = 2 + xy_{1}$

$\displaystyle \Rightarrow (1-x^{2})y_{2} - xy_{1} - 2 = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 21: }~\text{If }y=e^{\tan^{-1}x},\text{ prove that }(1+x^2)y_2+(2x-1)y_1=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = e^{\tan^{-1}x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{\tan^{-1}x}\cdot\frac{1}{1+x^{2}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = e^{\tan^{-1}x}\cdot\frac{1}{(1+x^{2})^{2}} + e^{\tan^{-1}x}\cdot\frac{-2x}{(1+x^{2})^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{e^{\tan^{-1}x}}{(1+x^{2})^{2}} - \frac{2x e^{\tan^{-1}x}}{(1+x^{2})^{2}}$

$\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} = \frac{e^{\tan^{-1}x}}{1+x^{2}} - \frac{2x e^{\tan^{-1}x}}{1+x^{2}}$

$\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} - 2x\frac{dy}{dx}$

$\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} + (2x-1)\frac{dy}{dx} = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 22: }~\text{If }y=3\cos(\log x)+4\sin(\log x),\text{ prove that }x^2y_2+xy_1+y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = 3\cos(\log x) + 4\sin(\log x)$

$\displaystyle \text{Let } y_1 = \frac{dy}{dx} \text{ and } y_2 = \frac{d^2y}{dx^2}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle y_1 = -3\sin(\log x)\cdot\frac{1}{x} + 4\cos(\log x)\cdot\frac{1}{x}$

$\displaystyle \Rightarrow y_1 = \frac{-3\sin(\log x)+4\cos(\log x)}{x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle y_2 = \frac{d}{dx}\left(\frac{-3\sin(\log x)+4\cos(\log x)}{x}\right)$

$\displaystyle \Rightarrow y_2 = \frac{\left(-3\cos(\log x)-4\sin(\log x)\right)x -\left(-3\sin(\log x)+4\cos(\log x)\right)}{x^2}$

$\displaystyle \Rightarrow y_2 = \frac{-3\cos(\log x)-4\sin(\log x)}{x^2} - \frac{-3\sin(\log x)+4\cos(\log x)}{x^2}$

$\displaystyle \Rightarrow y_2 = -\frac{3\cos(\log x)+4\sin(\log x)}{x^2} - \frac{-3\sin(\log x)+4\cos(\log x)}{x^2}$

$\displaystyle \Rightarrow y_2 = -\frac{y}{x^2} - \frac{y_1}{x}$

$\displaystyle \Rightarrow x^2y_2 = -y - xy_1$

$\displaystyle \Rightarrow x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0$

$\displaystyle \textbf{Question 23: }~\text{If }y=e^{2x}(ax+b),\text{ show that }y_2-4y_1+4y=0.\;[\text{ CBSE 2009, 2012, 2016}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } y = e^{2x}(ax+b)\text{. To prove: } y_2 - 4y_1 + 4y = 0$

$\displaystyle \text{Proof:}$

$\displaystyle \text{We have,}$

$\displaystyle y = e^{2x}(ax+b) \qquad \text{(i)}$

$\displaystyle y_1 = \frac{dy}{dx} = ae^{2x} + 2e^{2x}(ax+b) \qquad \text{(ii)}$

$\displaystyle y_2 = \frac{d^2y}{dx^2} = 2ae^{2x} + 4e^{2x}(ax+b) + 2ae^{2x}$

$\displaystyle \Rightarrow y_2 = 4ae^{2x} + 4e^{2x}(ax+b) \qquad \text{(iii)}$

$\displaystyle \text{LHS } = y_2 - 4y_1 + 4y$

$\displaystyle = \big[4ae^{2x} + 4e^{2x}(ax+b)\big] - \big[4ae^{2x} + 8e^{2x}(ax+b)\big] + 4e^{2x}(ax+b)$

$\displaystyle = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 24: }~\text{If }x=\sin\!\left(\frac{1}{a}\log y\right),\text{ show that }(1-x^2)y_2-xy_1-a^2y=0.\;[\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given,}$

$\displaystyle x = \sin\!\left(\frac{1}{a}\log y\right)$

$\displaystyle \Rightarrow \log y = a\sin^{-1}x$

$\displaystyle \Rightarrow y = e^{a\sin^{-1}x} \qquad \text{(i)}$

$\displaystyle \text{To prove: } (1-x^{2})y_2 - xy_1 - a^{2}y = 0$

$\displaystyle \text{Let } y_1 = \frac{dy}{dx} \text{ and } y_2 = \frac{d^{2}y}{dx^{2}}$

$\displaystyle \text{Differentiating (i) with respect to } x,$

$\displaystyle y_1 = e^{a\sin^{-1}x}\cdot\frac{a}{\sqrt{1-x^{2}}} \qquad \text{(ii)}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle y_2 = \frac{d}{dx}\!\left( \frac{ae^{a\sin^{-1}x}}{\sqrt{1-x^{2}}} \right)$

$\displaystyle \Rightarrow y_2 = ae^{a\sin^{-1}x}\frac{d}{dx}\!\left((1-x^{2})^{-1/2}\right) + \frac{a}{\sqrt{1-x^{2}}}\frac{d}{dx}\!\left(e^{a\sin^{-1}x}\right)$

$\displaystyle \Rightarrow y_2 = ae^{a\sin^{-1}x}\cdot\frac{x}{(1-x^{2})^{3/2}} + \frac{a^{2}e^{a\sin^{-1}x}}{1-x^{2}}$

$\displaystyle \Rightarrow (1-x^{2})y_2 = \frac{axe^{a\sin^{-1}x}}{\sqrt{1-x^{2}}} + a^{2}e^{a\sin^{-1}x}$

$\displaystyle \text{Using (i) and (ii),}$

$\displaystyle (1-x^{2})y_2 = x y_1 + a^{2}y$

$\displaystyle \Rightarrow (1-x^{2})y_2 - xy_1 - a^{2}y = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 25: }~\text{If }\log y=\tan^{-1}x,\text{ show that }(1+x^2)y_2+(2x-1)y_1=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle \log y = \tan^{-1}x$

$\displaystyle \text{Let } y_1 = \frac{dy}{dx} \text{ and } y_2 = \frac{d^{2}y}{dx^{2}}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{1}{y}\,y_1 = \frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow (1+x^{2})y_1 = y \qquad \text{(i)}$

$\displaystyle \text{Differentiating (i) with respect to } x,$

$\displaystyle (1+x^{2})y_2 + 2x y_1 = y_1$

$\displaystyle \Rightarrow (1+x^{2})y_2 + (2x-1)y_1 = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 26: }~\text{If }y=\tan^{-1}x,\text{ show that }(1+x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \tan^{-1}x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{1+x^{2}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{2x}{(1+x^{2})^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{2x}{1+x^{2}}\cdot\frac{1}{1+x^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{2x}{1+x^{2}}\frac{dy}{dx}$

$\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} = -2x\frac{dy}{dx}$

$\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} + 2x\frac{dy}{dx} = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 27: }~\text{If }y=\{\log(x+\sqrt{x^2+1})\}^2,\text{ show that }(1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=2.\;[\text{CBSE 2008}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y=\left[\log\!\left(x+\sqrt{x^{2}+1}\right)\right]^{2}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2\log\!\left(x+\sqrt{x^{2}+1}\right)\cdot \frac{1}{x+\sqrt{x^{2}+1}} \left(1+\frac{x}{\sqrt{x^{2}+1}}\right)$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2\log\!\left(x+\sqrt{x^{2}+1}\right)}{x+\sqrt{x^{2}+1}} \cdot\frac{x+\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2\log\!\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left( \frac{2\log\!\left(x+\sqrt{x^{2}+1}\right)}{\sqrt{x^{2}+1}} \right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{2}{x^{2}+1} - \frac{2x\log\!\left(x+\sqrt{x^{2}+1}\right)}{(x^{2}+1)^{3/2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{2 - x\frac{dy}{dx}}{x^{2}+1}$

$\displaystyle \Rightarrow (x^{2}+1)\frac{d^{2}y}{dx^{2}} = 2 - x\frac{dy}{dx}$

$\displaystyle \Rightarrow (x^{2}+1)\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} = 2$

$\displaystyle \textbf{Question 28: }~\text{If }y=(\tan^{-1}x)^2,\text{ then prove that }(1+x^2)^2y_2+2x(1+x^2)y_1=2.\;[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = (\tan^{-1}x)^{2}$

$\displaystyle \text{Let } y_1 = \frac{dy}{dx} \text{ and } y_2 = \frac{d^{2}y}{dx^{2}}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle y_1 = 2\tan^{-1}x\cdot\frac{1}{1+x^{2}} = \frac{2\tan^{-1}x}{1+x^{2}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle y_2 = \frac{d}{dx}\!\left(\frac{2\tan^{-1}x}{1+x^{2}}\right)$

$\displaystyle \Rightarrow y_2 = \frac{2}{(1+x^{2})^{2}} - \frac{4x\tan^{-1}x}{(1+x^{2})^{2}}$

$\displaystyle \Rightarrow y_2 = \frac{2}{(1+x^{2})^{2}} - \frac{2x}{1+x^{2}}\,y_1$

$\displaystyle \Rightarrow (1+x^{2})^{2}y_2 = 2 - 2x(1+x^{2})y_1$

$\displaystyle \Rightarrow (1+x^{2})^{2}y_2 + 2x(1+x^{2})y_1 = 2$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 29: }~\text{If }y=\cot x,\text{ show that }\frac{d^2y}{dx^2}+2y\frac{dy}{dx}=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \cot x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = -\mathrm{cosec}^{2}x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -2\,\mathrm{cosec}x \cdot (-\mathrm{cosec}x\cot x)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2\,\mathrm{cosec}^{2}x\cot x$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -2y\frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + 2y\frac{dy}{dx} = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 30: }~\text{Find }\frac{d^2y}{dx^2},\text{ where }y=\log\!\left(\frac{x^2}{e^2}\right).\;[\text{CBSE 2000}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \log\!\left(\frac{x^{2}}{e^{2}}\right)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{1}{\frac{x^{2}}{e^{2}}}\cdot\frac{2x}{e^{2}} = \frac{2}{x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{2}{x^{2}}$

$\displaystyle \textbf{Question 31: }~\text{If }y=ae^{2x}+be^{-x},\text{ show that }\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0.\;[\text{CBSE 2000C}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = ae^{2x} + be^{-x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2ae^{2x} - be^{-x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 4ae^{2x} + be^{-x}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = (2ae^{2x} - be^{-x}) + 2(ae^{2x} + be^{-x})$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + 2y$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} - 2y = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 32: }~\text{If }y=e^x(\sin x+\cos x),\text{ prove that }\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0.\;[\text{CBSE 2002, 2009}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = e^{x}(\sin x + \cos x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{x}(\sin x + \cos x) + e^{x}(\cos x - \sin x)$

$\displaystyle \Rightarrow \frac{dy}{dx} = 2e^{x}\cos x$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 2e^{x}\cos x - 2e^{x}\sin x$

$\displaystyle \text{Now,}$

$\displaystyle \text{LHS } = \frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y$

$\displaystyle = (2e^{x}\cos x - 2e^{x}\sin x) - 4e^{x}\cos x + 2e^{x}(\sin x + \cos x)$

$\displaystyle = 0 = \text{RHS}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 33: }~\text{If }y=\cos^{-1}x,\text{ find }\frac{d^2y}{dx^2}\text{ in terms of }y\text{ alone. }\;$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \cos^{-1}x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^{2}}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{2x}{2(1-x^{2})^{3/2}} = -\frac{x}{(1-x^{2})^{3/2}}$

$\displaystyle \text{Now,}$

$\displaystyle y = \cos^{-1}x$

$\displaystyle \Rightarrow x = \cos y$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\cos y}{(1-\cos^{2}y)^{3/2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\cos y}{(\sin^{2}y)^{3/2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\cot y\,\mathrm{cosec}^{2}y$

$\displaystyle \textbf{Question 34: }~\text{If }y=e^{a\cos^{-1}x},\text{ prove that }(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0.\;[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = e^{a\cos^{-1}x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = e^{a\cos^{-1}x}\cdot\frac{d}{dx}(a\cos^{-1}x) = -\frac{a\,e^{a\cos^{-1}x}}{\sqrt{1-x^{2}}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(-\frac{a\,e^{a\cos^{-1}x}}{\sqrt{1-x^{2}}}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = e^{a\cos^{-1}x}\cdot\frac{a^{2}}{1-x^{2}} + \frac{xa\,e^{a\cos^{-1}x}}{(1-x^{2})^{3/2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = y\cdot\frac{a^{2}}{1-x^{2}} - \frac{x}{1-x^{2}}\frac{dy}{dx}$

$\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} = a^{2}y - x\frac{dy}{dx}$

$\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} - a^{2}y = 0$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 35: }~\text{If }y=500e^{7x}+600e^{-7x},\text{ show that }\frac{d^2y}{dx^2}=49y.\;$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = 500e^{7x} + 600e^{-7x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 3500e^{7x} - 4200e^{-7x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 24500e^{7x} + 29400e^{-7x}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 49(500e^{7x} + 600e^{-7x})$

Recognize the original expression:

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 49y$

$\displaystyle \textbf{Question 36: }~\text{If }x=2\cos t-\cos2t,\;y=2\sin t-\sin2t,\text{ find }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = 2\cos t - \cos 2t \text{ and } y = 2\sin t - \sin 2t$

$\displaystyle \text{Differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = -2\sin t + 2\sin 2t$

$\displaystyle \frac{dy}{dt} = 2\cos t - 2\cos 2t$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t - 2\cos 2t}{-2\sin t + 2\sin 2t} = \frac{\cos t - \cos 2t}{-\sin t + \sin 2t}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\!\left(\frac{\cos t - \cos 2t}{-\sin t + \sin 2t}\right)\cdot\frac{dt}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{(-\sin t + 2\sin 2t)(-\sin t + \sin 2t) - (\cos t - \cos 2t)(-\cos t + 2\cos 2t)} {(-\sin t + \sin 2t)^{2}}\cdot\frac{1}{-2\sin t + 2\sin 2t}$

$\displaystyle \text{At } t=\frac{\pi}{2},$

$\displaystyle \sin t = 1,\quad \cos t = 0,\quad \sin 2t = 0,\quad \cos 2t = -1$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{(-1+0)(-1+0)-(0+1)(0-2)}{(-1+0)^{2}(-2+0)} = \frac{1+2}{-2} = -\frac{3}{2}$

$\displaystyle \textbf{Question 37: }~\text{If }x=4z^2+5,\;y=6z^2+7z+3,\text{ find }\frac{d^2y}{dx^2}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x = 4z^{2} + 5 \text{ and } y = 6z^{2} + 7z + 3$

$\displaystyle \text{Differentiating with respect to } z,$

$\displaystyle \frac{dx}{dz} = 8z \quad \text{and} \quad \frac{dy}{dz} = 12z + 7$

$\displaystyle \therefore \frac{dy}{dx} = \frac{dy/dz}{dx/dz} = \frac{12z + 7}{8z}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dz}\!\left(\frac{12z + 7}{8z}\right)\cdot\frac{dz}{dx}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{8z(12) - 8(12z + 7)}{64z^{2}}\cdot\frac{1}{8z}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{96z - 96z - 56}{512z^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{7}{64z^{3}}$

$\displaystyle \textbf{Question 38: }~\text{If }y=\log(1+\cos x),\text{ prove that }\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}\frac{dy}{dx}=0.\;[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \log(1+\cos x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \frac{-\sin x}{1+\cos x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{-(\cos x+\cos^{2}x+\sin^{2}x)}{(1+\cos x)^{2}} = -\frac{1}{1+\cos x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{3}y}{dx^{3}} = -\frac{\sin x}{(1+\cos x)^{2}}$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} + \frac{\sin x}{(1+\cos x)^{2}} = 0$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} + \left(-\frac{1}{1+\cos x}\right) \left(-\frac{\sin x}{1+\cos x}\right) = 0$

$\displaystyle \Rightarrow \frac{d^{3}y}{dx^{3}} + \frac{d^{2}y}{dx^{2}}\cdot\frac{dy}{dx} = 0$

$\displaystyle \textbf{Question 39: }~\text{If }y=\sin(\log x),\text{ prove that }x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0.\;[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \sin(\log x)$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = \cos(\log x)\cdot\frac{1}{x} = \frac{\cos(\log x)}{x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{\cos(\log x)}{x}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{-\sin(\log x)\cdot\frac{1}{x}\cdot x - \cos(\log x)} {x^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{\sin(\log x)}{x^{2}} - \frac{\cos(\log x)}{x^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{y}{x^{2}} - \frac{1}{x}\frac{dy}{dx}$

$\displaystyle \Rightarrow x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + y = 0$

$\displaystyle \textbf{Question 40: }~\text{If }y=3e^{2x}+2e^{3x},\text{ prove that }\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0.\;[\text{CBSE 2007, 2009}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = 3e^{2x} + 2e^{3x}$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 6e^{2x} + 6e^{3x}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = 12e^{2x} + 18e^{3x}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 5(6e^{2x} + 6e^{3x}) - 6(3e^{2x} + 2e^{3x})$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 5\frac{dy}{dx} - 6y$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} - 5\frac{dy}{dx} + 6y = 0$

$\displaystyle \textbf{Question 41: }~\text{If }y=(\cot^{-1}x)^2,\text{ prove that }y_2(x^2+1)^2+2x(x^2+1)y_1=2.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \big(\cot^{-1} x\big)^2$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle y_1 = 2\cot^{-1}x \cdot \left(-\frac{1}{1+x^2}\right) = -\frac{2\cot^{-1}x}{1+x^2}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle y_2 = \frac{2 + 4x\cot^{-1}x}{(1+x^2)^2}$

$\displaystyle \Rightarrow y_2 = \frac{2}{(1+x^2)^2} + \frac{2x\cdot 2\cot^{-1}x}{(1+x^2)(1+x^2)}$

$\displaystyle \Rightarrow y_2 = \frac{2}{(1+x^2)^2} - \frac{2x y_1}{1+x^2}$

$\displaystyle \Rightarrow (1+x^2)^2 y_2 = 2 - 2x y_1(1+x^2)$

$\displaystyle \Rightarrow (1+x^2)^2 y_2 + 2x y_1(1+x^2) = 2$

$\displaystyle \textbf{Question 42: }~\text{If }y=\mathrm{cosec}^{-1}x,\;x>1,\text{ then show that }x(x^2-1)\frac{d^2y}{dx^2}+(2x^2-1)\frac{dy}{dx}=0.\;[\text{CBSE 2010}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle y = \mathrm{cosec}^{-1} x$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = -\frac{1}{x\sqrt{x^{2}-1}}$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(-\frac{1}{x\sqrt{x^{2}-1}}\right)$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{\sqrt{x^{2}-1} + \dfrac{x^{2}}{\sqrt{x^{2}-1}}} {x^{2}(x^{2}-1)}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{x^{2}-1 + x^{2}} {x^{2}(x^{2}-1)\sqrt{x^{2}-1}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{2x^{2}-1} {x^{2}(x^{2}-1)\sqrt{x^{2}-1}}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{2}{(x^{2}-1)\sqrt{x^{2}-1}} - \frac{1}{x^{2}\sqrt{x^{2}-1}}$

$\displaystyle \Rightarrow (x^{2}-1)\frac{d^{2}y}{dx^{2}} = \frac{2}{\sqrt{x^{2}-1}} - \frac{1}{x^{2}\sqrt{x^{2}-1}}$

$\displaystyle \Rightarrow (x^{2}-1)\frac{d^{2}y}{dx^{2}} = -2x\frac{dy}{dx} + \frac{1}{x}\frac{dy}{dx}$

$\displaystyle \Rightarrow x(x^{2}-1)\frac{d^{2}y}{dx^{2}} = -(2x^{2}-1)\frac{dy}{dx}$

$\displaystyle \Rightarrow x(x^{2}-1)\frac{d^{2}y}{dx^{2}} + (2x^{2}-1)\frac{dy}{dx} = 0$

$\displaystyle \text{Hence Proved.}$

$\displaystyle \textbf{Question 43: }~\text{If }x=\cos t+\log\tan\frac{t}{2},\;y=\sin t,\text{ then find the value of }\frac{d^2y}{dt^2}\text{ and }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{4}.\;[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle x = \cos t + \log \tan\frac{t}{2} \quad \text{and} \quad y = \sin t$

$\displaystyle \text{Differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}\!\left(\cos t + \log \tan\frac{t}{2}\right)$

$\displaystyle \frac{dx}{dt} = -\sin t + \frac{1}{\tan\frac{t}{2}}\cdot \sec^{2}\frac{t}{2}\cdot\frac{1}{2}$

$\displaystyle \frac{dx}{dt} = -\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}$

$\displaystyle \frac{dx}{dt} = -\sin t + \frac{1}{\sin t} = \frac{-\sin^{2}t + 1}{\sin t} = \frac{\cos^{2}t}{\sin t}$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}(\sin t) = \cos t$

$\displaystyle \text{Now, } \frac{d^{2}y}{dt^{2}} = \frac{d}{dt}(\cos t) = -\sin t$

$\displaystyle \left(\frac{d^{2}y}{dt^{2}}\right)_{t=\pi/4} = -\sin\frac{\pi}{4} = -\frac{1}{\sqrt{2}} \quad (1)$

$\displaystyle \text{Also, } \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{\dfrac{\cos^{2}t}{\sin t}} = \tan t$

$\displaystyle \text{Now, } \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dt}(\tan t)\cdot\frac{dt}{dx}$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \sec^{2}t \cdot \frac{\sin t}{\cos^{2}t} = \frac{\sin t}{\cos^{4}t}$

$\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)_{t=\pi/4} = \frac{\sin(\pi/4)}{\cos^{4}(\pi/4)} = 2\sqrt{2} \quad (2)$

$\displaystyle \text{Hence, at } t=\frac{\pi}{4},\; \frac{d^{2}y}{dt^{2}}=-\frac{1}{\sqrt{2}} \text{ and } \frac{d^{2}y}{dx^{2}}=2\sqrt{2}$

$\displaystyle \textbf{Question 44: }~\text{If }x=a\sin t\text{ and }y=a\!\left(\cos t+\log\tan\frac{t}{2}\right),\text{ find }\frac{d^2y}{dx^2}.\;[\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle x = a\sin t \quad \text{and} \quad y = a\!\left(\cos t + \log \tan\frac{t}{2}\right)$

$\displaystyle \text{Differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}(a\sin t) = a\cos t$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}\!\left[a\!\left(\cos t + \log \tan\frac{t}{2}\right)\right]$

$\displaystyle \frac{dy}{dt} = a\!\left(-\sin t + \frac{1}{\tan\frac{t}{2}}\cdot\sec^{2}\frac{t}{2}\cdot\frac{1}{2}\right)$

$\displaystyle \frac{dy}{dt} = a\!\left(-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right)$

$\displaystyle \frac{dy}{dt} = a\!\left(-\sin t + \frac{1}{\sin t}\right) = a\!\left(\frac{1-\sin^{2}t}{\sin t}\right)$

$\displaystyle \frac{dy}{dt} = a\!\left(\frac{\cos^{2}t}{\sin t}\right)$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\left(\dfrac{\cos^{2}t}{\sin t}\right)}{a\cos t} = \cot t$

$\displaystyle \text{Therefore, } \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\cot t)$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\cot t)\cdot\frac{dt}{dx} = (-\mathrm{cosec}^{2}t)\cdot\frac{1}{a\cos t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{1}{a\sin^{2}t\cos t}$

$\displaystyle \textbf{Question 45: }~\text{If }x=a(\cos t+t\sin t)\text{ and }y=a(\sin t-t\cos t),\text{ then find the value of }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{4}.\;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle x = a(\cos t + t\sin t) \quad \text{and} \quad y = a(\sin t - t\cos t)$

$\displaystyle \text{Differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}\big[a(\cos t + t\sin t)\big]$

$\displaystyle \frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}\big[a(\sin t - t\cos t)\big]$

$\displaystyle \frac{dy}{dt} = a(\cos t - \cos t + t\sin t) = at\sin t$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t$

$\displaystyle \text{Differentiating again with respect to } x,$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\tan t)$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\tan t)\cdot\frac{dt}{dx} = \sec^{2}t\cdot\frac{1}{at\cos t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{1}{at\cos^{3}t}$

$\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)_{t=\pi/4} = \frac{1}{a(\pi/4)\cos^{3}(\pi/4)} = \frac{8\sqrt{2}}{a\pi}$

$\displaystyle \text{Hence, at } t=\frac{\pi}{4},\; \frac{d^{2}y}{dx^{2}} = \frac{8\sqrt{2}}{a\pi}$

$\displaystyle \textbf{Question 46: }~\text{If }x=a\!\left(\cos t+\log\tan\frac{t}{2}\right),\;y=a\sin t,\text{ evaluate }\frac{d^2y}{dx^2}\text{ at }t=\frac{\pi}{3}.\;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have,}$

$\displaystyle x = a\!\left(\cos t + \log \tan\frac{t}{2}\right) \quad \text{and} \quad y = a\sin t$

$\displaystyle \text{On differentiating with respect to } t,$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}\!\left[a\!\left(\cos t + \log \tan\frac{t}{2}\right)\right]$

$\displaystyle \frac{dx}{dt} = a\!\left(-\sin t + \frac{1}{\tan\frac{t}{2}}\cdot \sec^{2}\frac{t}{2}\cdot \frac{1}{2}\right)$

$\displaystyle \frac{dx}{dt} = a\!\left(-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right)$

$\displaystyle \frac{dx}{dt} = a\!\left(-\sin t + \frac{1}{\sin t}\right) = a\!\left(\frac{1-\sin^{2}t}{\sin t}\right)$

$\displaystyle \frac{dx}{dt} = a\!\left(\frac{\cos^{2}t}{\sin t}\right)$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}(a\sin t) = a\cos t$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\cos t}{a\left(\dfrac{\cos^{2}t}{\sin t}\right)} = \tan t$

$\displaystyle \text{Therefore, } \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\tan t)$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}(\tan t)\cdot\frac{dt}{dx} = \sec^{2}t\cdot\frac{\sin t}{a\cos^{2}t}$

$\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{\sin t}{a\cos^{4}t}$

$\displaystyle \left(\frac{d^{2}y}{dx^{2}}\right)_{t=\pi/3} = \frac{\sin(\pi/3)}{a\cos^{4}(\pi/3)} = \frac{\dfrac{\sqrt{3}}{2}}{a\left(\dfrac{1}{16}\right)} = \frac{8\sqrt{3}}{a}$

$\displaystyle \text{Hence, at } t=\frac{\pi}{3},\; \frac{d^{2}y}{dx^{2}} = \frac{8\sqrt{3}}{a}$

$\displaystyle \textbf{Question 47: }~\text{If }x=a(\cos2t+2t\sin2t)\text{ and }y=a(\sin2t-2t\cos2t),\text{ then find }\frac{d^2y}{dx^2}.\;[\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } x = a(\cos 2t + 2t \sin 2t) \text{ and } y = a(\sin 2t - 2t \cos 2t)$

$\displaystyle \text{On differentiating with respect to } t, \text{ we get}$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}\left[a(\cos 2t + 2t \sin 2t)\right]$

$\displaystyle = a(-2\sin 2t + 2\sin 2t + 4t \cos 2t)$

$\displaystyle = a(4t \cos 2t)$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}\left[a(\sin 2t - 2t \cos 2t)\right]$

$\displaystyle = a(2\cos 2t - 2\cos 2t + 4t \sin 2t)$

$\displaystyle = a(4t \sin 2t)$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$\displaystyle = \frac{a(4t \sin 2t)}{a(4t \cos 2t)} = \tan 2t$

$\displaystyle \text{Therefore, } \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\tan 2t)$

$\displaystyle = \frac{d}{dt}(\tan 2t)\times\frac{dt}{dx}$

$\displaystyle = 2\sec^2 2t \times \frac{1}{a(4t \cos 2t)}$

$\displaystyle = \frac{1}{2at \cos^3 2t}$

$\displaystyle = \frac{1}{2at}\sec^3 2t$

$\displaystyle \textbf{Question 48: }~\text{If }x=3\cot t-2\cos^3t,\;y=3\sin t-2\sin^3t,\text{ find }\frac{d^2y}{dx^2}.\;[\text{CBSE 2017}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } x = 3\cos t - 2\cos^3 t$

$\displaystyle \Rightarrow \frac{dx}{dt} = 3(-\sin t) - 6\cos^2 t(-\sin t)$

$\displaystyle = -3\sin t + 6\sin t \cos^2 t$

$\displaystyle \text{Also, } y = 3\sin t - 2\sin^3 t$

$\displaystyle \Rightarrow \frac{dy}{dt} = 3\cos t - 6\sin^2 t \cos t$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$\displaystyle = \frac{3\cos t - 6\sin^2 t \cos t}{-3\sin t + 6\sin t \cos^2 t}$

$\displaystyle = \frac{3\cos t(1 - 2\sin^2 t)}{3\sin t(-1 + 2\cos^2 t)}$

$\displaystyle = \frac{\cos t(\cos 2t)}{\sin t(\cos 2t)}$

$\displaystyle = \cot t$

$\displaystyle \text{So, } \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\displaystyle = \frac{d}{dx}(\cot t)$

$\displaystyle = -\mathrm{cosec}^2 t \cdot \frac{dt}{dx}$

$\displaystyle = -\frac{\mathrm{cosec}^2 t}{\dfrac{dx}{dt}}$

$\displaystyle = -\frac{\mathrm{cosec}^2 t}{-3\sin t + 6\sin t \cos^2 t}$

$\displaystyle = -\frac{\mathrm{cosec}^2 t}{3\sin t(-1 + 2\cos^2 t)}$

$\displaystyle = \frac{\mathrm{cosec}^3 t}{3\cos 2t}$

$\displaystyle \textbf{Question 49: }~\text{If }x=a\sin t-b\cos t,\;y=a\cos t+b\sin t,\text{ prove that }\frac{d^2y}{dx^2}=-\frac{x^2+y^2}{y^3}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } x = a\sin t - b\cos t,\; y = a\cos t + b\sin t$

$\displaystyle \text{On differentiating with respect to } t, \text{ we get}$

$\displaystyle \frac{dx}{dt} = \frac{d}{dt}(a\sin t - b\cos t) = a\cos t + b\sin t$

$\displaystyle \text{and}$

$\displaystyle \frac{dy}{dt} = \frac{d}{dt}(a\cos t + b\sin t) = -a\sin t + b\cos t$

$\displaystyle \text{Now, } \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \frac{-a\sin t + b\cos t}{a\cos t + b\sin t}$

$\displaystyle \text{Therefore, } \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{-a\sin t + b\cos t}{a\cos t + b\sin t}\right)$

$\displaystyle = \frac{d}{dt}\left(\frac{-a\sin t + b\cos t}{a\cos t + b\sin t}\right)\cdot\frac{dt}{dx}$

$\displaystyle = \frac{(a\cos t + b\sin t)\dfrac{d}{dt}(-a\sin t + b\cos t) - (-a\sin t + b\cos t)\dfrac{d}{dt}(a\cos t + b\sin t)}{(a\cos t + b\sin t)^2}\cdot\frac{1}{a\cos t + b\sin t}$

$\displaystyle = \frac{(a\cos t + b\sin t)(-a\cos t - b\sin t) - (-a\sin t + b\cos t)(-a\sin t + b\cos t)}{(a\cos t + b\sin t)^3}$

$\displaystyle = \frac{-(a\cos t + b\sin t)^2 - (-a\sin t + b\cos t)^2}{(a\cos t + b\sin t)^3}$

$\displaystyle = \frac{-(a\cos t + b\sin t)^2 - (a\sin t - b\cos t)^2}{(a\cos t + b\sin t)^3}$

$\displaystyle = -\frac{x^2 + y^2}{y^3}$

$\displaystyle \text{Hence, } \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3}$

$\displaystyle \textbf{Question 50: }~\text{Find }A\text{ and }B\text{ so that }y=A\sin3x+B\cos3x\text{ satisfies }\frac{d^2y}{dx^2}+4\frac{dy}{dx}+3y=10\cos3x.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } y = A\sin 3x + B\cos 3x$

$\displaystyle \frac{dy}{dx} = 3A\cos 3x - 3B\sin 3x$

$\displaystyle \frac{d^2y}{dx^2} = -9A\sin 3x - 9B\cos 3x$

$\displaystyle \text{Therefore,}$

$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 3y$

$\displaystyle = (-9A\sin 3x - 9B\cos 3x) + 4(3A\cos 3x - 3B\sin 3x) + 3(A\sin 3x + B\cos 3x)$

$\displaystyle = (-6A - 12B)\sin 3x + (-6B + 12A)\cos 3x$

$\displaystyle \text{It is given that,}$

$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 3y = 10\cos 3x$

$\displaystyle \text{Comparing the coefficients of } \sin 3x \text{ and } \cos 3x, \text{ we get}$

$\displaystyle -6A - 12B = 0 \quad \text{and} \quad -6B + 12A = 10$

$\displaystyle \text{Solving, } A = \frac{2}{3} \text{ and } B = -\frac{1}{3}$

$\displaystyle \text{Hence, } A = \frac{2}{3},\; B = -\frac{1}{3}$

$\displaystyle \textbf{Question 51: }~\text{If }y=Ae^{-kt}\cos(pt+c),\text{ prove that }\frac{d^2y}{dt^2}+2k\frac{dy}{dt}+n^2y=0,\text{ where }n^2=p^2+k^2.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We know, }$

$\displaystyle y = Ae^{-kt}\cos(pt + c) \quad \text{...(1)}$

$\displaystyle \text{Differentiating } y \text{ with respect to } t, \text{ we get}$

$\displaystyle \frac{dy}{dt} = -kAe^{-kt}\cos(pt + c) - pAe^{-kt}\sin(pt + c)$

$\displaystyle = -ky - pAe^{-kt}\sin(pt + c) \quad [\text{From (1)}]$

$\displaystyle \Rightarrow pAe^{-kt}\sin(pt + c) = -ky - \frac{dy}{dt} \quad \text{...(2)}$

$\displaystyle \text{Differentiating } \frac{dy}{dt} \text{ with respect to } t, \text{ we get}$

$\displaystyle \frac{d^2y}{dt^2} = -k\frac{dy}{dt} + pkAe^{-kt}\sin(pt + c) - p^2Ae^{-kt}\cos(pt + c)$

$\displaystyle = -k\frac{dy}{dt} + k\left(-ky - \frac{dy}{dt}\right) - p^2y \quad [\text{From (1) and (2)}]$

$\displaystyle = -2k\frac{dy}{dt} - (k^2 + p^2)y$

$\displaystyle \Rightarrow \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + (k^2 + p^2)y = 0$

$\displaystyle \Rightarrow \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + n^2y = 0,\; \text{where } n^2 = p^2 + k^2$

$\displaystyle \text{Hence, } \frac{d^2y}{dt^2} + 2k\frac{dy}{dt} + n^2y = 0,\; \text{where } n^2 = p^2 + k^2$

$\displaystyle \textbf{Question 52: }~\text{If }y=x^n\{a\cos(\log x)+b\sin(\log x)\},\text{ prove that }x^2\frac{d^2y}{dx^2}+(1-2n)x\frac{dy}{dx}+(1+n^2)y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } y = x^n\{a\cos(\log x) + b\sin(\log x)\} \quad \text{...(1)}$

$\displaystyle \text{Differentiating } y \text{ with respect to } x, \text{ we get}$

$\displaystyle \frac{dy}{dx} = nx^{\,n-1}\{a\cos(\log x) + b\sin(\log x)\} + x^n\left\{-a\sin(\log x)\frac{1}{x} + b\cos(\log x)\frac{1}{x}\right\}$

$\displaystyle = \frac{n}{x}x^n\{a\cos(\log x) + b\sin(\log x)\} + x^{\,n-1}\{-a\sin(\log x) + b\cos(\log x)\}$

$\displaystyle = \frac{n}{x}y + x^{\,n-1}\{-a\sin(\log x) + b\cos(\log x)\} \quad [\text{From (1)}]$

$\displaystyle \Rightarrow x^{\,n-1}\{-a\sin(\log x) + b\cos(\log x)\} = \frac{dy}{dx} - \frac{n}{x}y \quad \text{...(2)}$

$\displaystyle \text{Differentiating } \frac{dy}{dx} \text{ with respect to } x, \text{ we get}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{n}{x}\frac{dy}{dx} - \frac{ny}{x^2} + (n-1)x^{\,n-2}\{-a\sin(\log x) + b\cos(\log x)\}$

$\displaystyle \quad + x^{\,n-1}\left\{-a\cos(\log x)\frac{1}{x} - b\sin(\log x)\frac{1}{x}\right\}$

$\displaystyle = \frac{n}{x}\frac{dy}{dx} - \frac{ny}{x^2} + \frac{n-1}{x}\left(\frac{dy}{dx} - \frac{n}{x}y\right) - \frac{y}{x^2} \quad [\text{From (1) and (2)}]$

$\displaystyle = \frac{n}{x}\frac{dy}{dx} - \frac{ny}{x^2} + \frac{n-1}{x}\frac{dy}{dx} - \frac{n(n-1)y}{x^2} - \frac{y}{x^2}$

$\displaystyle = \left(\frac{2n-1}{x}\right)\frac{dy}{dx} - \frac{n^2+1}{x^2}y$

$\displaystyle \Rightarrow x^2\frac{d^2y}{dx^2} - x(2n-1)\frac{dy}{dx} + (n^2+1)y = 0$

$\displaystyle \text{Hence, } x^2\frac{d^2y}{dx^2} + (1-2n)x\frac{dy}{dx} + (1+n^2)y = 0$

$\displaystyle \textbf{Question 53: }~\text{If }y=a\{x+\sqrt{x^2+1}\}^n+b\{x-\sqrt{x^2+1}\}^{-n},\text{ prove that }(x^2+1)\frac{d^2y}{dx^2}+x\frac{dy}{dx}-n^2y=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We have, } y = a\{x+\sqrt{x^2+1}\}^n + b\{x-\sqrt{x^2+1}\}^{-n} \quad \text{...(1)}$

$\displaystyle \text{Differentiating } y \text{ with respect to } x, \text{ we get}$

$\displaystyle \frac{dy}{dx} = an\{x+\sqrt{x^2+1}\}^{\,n-1}\left(1+\frac{1}{2\sqrt{x^2+1}}\cdot 2x\right) - bn\{x-\sqrt{x^2+1}\}^{-n-1}\left(1-\frac{1}{2\sqrt{x^2+1}}\cdot 2x\right)$

$\displaystyle = an\{x+\sqrt{x^2+1}\}^{\,n-1}\left(1+\frac{x}{\sqrt{x^2+1}}\right) - bn\{x-\sqrt{x^2+1}\}^{-n-1}\left(1-\frac{x}{\sqrt{x^2+1}}\right)$

$\displaystyle = an\{x+\sqrt{x^2+1}\}^{\,n-1}\left(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\right) + bn\{x-\sqrt{x^2+1}\}^{-n-1}\left(\frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}}\right)$