CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 06

Question. If \( (\cos x)^y = (\cos y)^x \), find \( \frac{dy}{dx} \).
Answer: We have, \( (\cos x)^y = (\cos y)^x \)
Taking log on both sides, we get
\( y \log (\cos x) = x \log (\cos y) \) ...(i)
Differentiating w.r.t. \( x \), we get
\( y \cdot \frac{1}{\cos x}(-\sin x) + \log(\cos x)\frac{dy}{dx} = x \cdot \frac{1}{\cos y}\left(-\sin y \frac{dy}{dx}\right) + \log(\cos y) \cdot 1 \)
\( \Rightarrow -y \tan x + \log(\cos x)\frac{dy}{dx} = -x \tan y \frac{dy}{dx} + \log(\cos y) \)
\( \Rightarrow \frac{dy}{dx} \left[ x \tan y + \log(\cos x) \right] = y \tan x + \log(\cos y) \)
\( \Rightarrow \frac{dy}{dx} = \frac{y\tan x + \log(\cos y)}{x\tan y + \log(\cos x)} \)

Question. If \( y = x^{\sin x - \cos x} + \frac{x^2 - 1}{x^2 + 1} \), find \( \frac{dy}{dx} \).
Answer: Here \( y = x^{\sin x - \cos x} + \frac{x^2 - 1}{x^2 + 1} \)
Let \( u = x^{\sin x - \cos x} \) and \( v = \frac{x^2 - 1}{x^2 + 1} \)
\( \therefore y = u + v \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)
Now, \( u = x^{\sin x - \cos x} \)
\( \Rightarrow \log u = (\sin x - \cos x)\log x \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = (\sin x - \cos x) \cdot \frac{1}{x} + (\cos x + \sin x) \cdot \log x \)
\( \Rightarrow \frac{du}{dx} = x^{\sin x - \cos x} \left[ \frac{\sin x - \cos x}{x} + (\cos x + \sin x) \log x \right] \) ...(2)
Now, \( v = \frac{x^2-1}{x^2+1} = 1 - \frac{2}{x^2+1} \)
\( \Rightarrow \frac{dv}{dx} = 0 - 2 \cdot (-1)(x^2+1)^{-2} \cdot 2x = \frac{4x}{(x^2+1)^2} \) ...(3)
From (1), (2) and (3), we get
\( \frac{dy}{dx} = x^{\sin x - \cos x} \left[ \frac{\sin x - \cos x}{x} + (\cos x + \sin x) \log x \right] + \frac{4x}{(x^2+1)^2} \)

Question. Find \( \frac{dy}{dx} \) when \( y = x^{\cot x} + \frac{2x^2 - 3}{x^2 + x + 2} \).
Answer: Let \( u = x^{\cot x} \), \( v = \frac{2x^2 - 3}{x^2 + x + 2} \)
\( \Rightarrow y = u + v \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)
Now, \( u = x^{\cot x} \Rightarrow \log u = \cot x \cdot \log x \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = \cot x \cdot \frac{1}{x} - \csc^2 x \cdot \log x \)
\( \Rightarrow \frac{du}{dx} = x^{\cot x} \left[ \frac{\cot x}{x} - \csc^2 x \log x \right] \) ...(2)
Also, \( v = \frac{2x^2 - 3}{x^2 + x + 2} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dv}{dx} = \frac{(x^2 + x + 2) \cdot 4x - (2x^2 - 3) \cdot (2x + 1)}{(x^2 + x + 2)^2} \)
\( = \frac{(4x^3 + 4x^2 + 8x) - (4x^3 + 2x^2 - 6x - 3)}{(x^2 + x + 2)^2} \)
\( = \frac{2x^2 + 14x + 3}{(x^2 + x + 2)^2} \) ...(3)
From (1), (2) and (3), we get
\( \frac{dy}{dx} = x^{\cot x} \left[ \frac{\cot x}{x} - \csc^2 x \log x \right] + \frac{2x^2 + 14x + 3}{(x^2 + x + 2)^2} \)

Question. Differentiate \( x^{\circ \cos x} + \frac{x^2 + 1}{x^2 - 1} \) with respect to \( x \).
Answer: Let \( y = x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1} \)
Let \( u = x^{x \cos x} \) and \( v = \frac{x^2 + 1}{x^2 - 1} \)
\( \therefore y = u + v \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(1)
Now, \( u = x^{x \cos x} \)
Taking log on both sides, we get
\( \log u = x \cos x \cdot \log x \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = x \cos x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(x \cos x) \)
\( = x \cos x \cdot \frac{1}{x} + \log x [ \cos x - x \sin x ] \)
\( = \cos x - x \sin x \log x + \cos x \log x \)
\( \therefore \frac{du}{dx} = x^{x \cos x} [ \cos x - x \sin x \log x + \cos x \log x ] \) ...(2)
Also, \( v = \frac{x^2 + 1}{x^2 - 1} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dv}{dx} = \frac{(x^2 - 1)(2x) - (x^2 + 1)(2x)}{(x^2 - 1)^2} \)
\( = \frac{2x [x^2 - 1 - x^2 - 1]}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2} \) ...(3)
From (1), (2) and (3), we get
\( \frac{dy}{dx} = x^{x \cos x} [ \cos x - x \sin x \log x + \cos x \log x ] - \frac{4x}{(x^2 - 1)^2} \)
\( = x^{x \cos x} [ \cos x(1 + \log x) - x \sin x \log x ] - \frac{4x}{(x^2 - 1)^2} \)

Question. If \( x^y = e^{x-y} \), show that \( \frac{dy}{dx} = \frac{\log x}{\{\log(xe)\}^2} \).
Answer: We have, \( x^y = e^{x-y} \)
Taking log on both sides, we get
\( y \log x = (x - y)\log e = x - y \)
\( \Rightarrow y(1 + \log x) = x \Rightarrow y = \frac{x}{1 + \log x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \left(\frac{1}{x}\right)}{(1 + \log x)^2} = \frac{\log x}{(1 + \log x)^2} \)
Since \( 1 = \log e \), we can write:
\( \frac{dy}{dx} = \frac{\log x}{(\log e + \log x)^2} = \frac{\log x}{\{\log(xe)\}^2} \)

Question. Find \( \frac{dy}{dx} \), if \( y = (\cos x)^x + (\sin x)^{1/x} \).
Answer: We have, \( y = (\cos x)^x + (\sin x)^{1/x} \)
\( \Rightarrow y = e^{x \log \cos x} + e^{\frac{\log \sin x}{x}} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{x \log \cos x} \left[ x \cdot \frac{1}{\cos x}(-\sin x) + \log \cos x \right] + e^{\frac{\log \sin x}{x}} \left[ \frac{1}{x} \cdot \frac{1}{\sin x}(\cos x) + \log \sin x \left(-\frac{1}{x^2}\right) \right] \)
\( = (\cos x)^x [\log \cos x - x \tan x] + (\sin x)^{1/x} \left[ \frac{\cot x}{x} - \frac{\log \sin x}{x^2} \right] \)

Question. If \( y = (\sin x - \cos x)^{\sin x - \cos x} \), \( \frac{pi}{4} < x < \frac{3\pi}{4} \), then find \( \frac{dy}{dx} \).
Answer: We have, \( y = (\sin x - \cos x)^{\sin x - \cos x} \)
Taking log on both sides, we get
\( \log y = (\sin x - \cos x) \log (\sin x - \cos x) \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{y} \frac{dy}{dx} = (\sin x - \cos x) \cdot \frac{\cos x + \sin x}{\sin x - \cos x} + \log(\sin x - \cos x) \cdot (\cos x + \sin x) \)
\( \Rightarrow \frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x)[1 + \log(\sin x - \cos x)] \)
\( \dots \frac{dy}{dx} = (\sin x - \cos x)^{\sin x - \cos x} \times (\cos x + \sin x)[1 + \log(\sin x - \cos x)] \)

Question. Differentiate the following with respect to \( x \): \( (x)^{\cos x} + (\sin x)^{\tan x} \).
Answer: Let \( y = (x)^{\cos x} + (\sin x)^{\tan x} \)
\( \Rightarrow y = e^{\cos x \log x} + e^{\tan x \log \sin x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{\cos x \log x} \left[ \frac{\cos x}{x} - \sin x \log x \right] + e^{\tan x \log \sin x} \left[ \frac{\tan x \cos x}{\sin x} + \sec^2 x \log \sin x \right] \)
\( = x^{\cos x} \left[ \frac{\cos x}{x} - \sin x \log x \right] + (\sin x)^{\tan x} \left[ 1 + \sec^2 x \log \sin x \right] \)

Question. If \( y = (\log x)^x + (x)^{\cos x} \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = (\log x)^x + (x)^{\cos x} \)
\( \Rightarrow y = e^{x \log(\log x)} + e^{\cos x \cdot \log x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{x \log(\log x)} \left[ \frac{x}{x \log x} + \log(\log x) \right] + e^{\cos x \cdot \log x} \left[ \frac{\cos x}{x} + \log x (-\sin x) \right] \)
\( = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] + x^{\cos x} \left[ \frac{\cos x}{x} - \sin x \cdot \log x \right] \)

Question. If \( y = (x)^{\sin x} + (\log x)^x \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = (x)^{\sin x} + (\log x)^x \)
\( \Rightarrow y = e^{\sin x \log x} + e^{x \log(\log x)} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{\sin x \log x} \left[ \sin x \cdot \frac{1}{x} + \cos x \log x \right] + e^{x \log(\log x)} \left[ x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 \right] \)
\( = x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right] + (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] \)

Question. If \( y = x^x - (\sin x)^x \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = x^x - (\sin x)^x \)
\( \Rightarrow y = e^{x \log x} - e^{x \log \sin x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{x \log x} \left[ \frac{x}{x} + \log x \right] - e^{x \log \sin x} \left[ x \cdot \frac{\cos x}{\sin x} + \log \sin x \cdot 1 \right] \)
\( = (x)^x(1 + \log x) - (\sin x)^x(x \cot x + \log \sin x) \)

Question. If \( y = (\log x)^{\cos x} + \frac{x^2 + 1}{x^2 - 1} \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = (\log x)^{\cos x} + \frac{x^2 + 1}{x^2 - 1} \)
\( \Rightarrow y = e^{\cos x \log(\log x)} + \frac{x^2 + 1}{x^2 - 1} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{\log x \cdot x} + \log(\log x)(-\sin x) \right] + \left[ \frac{(x^2 - 1)2x - (x^2 + 1)(2x)}{(x^2 - 1)^2} \right] \)
\( = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right] - \left[ \frac{4x}{(x^2 - 1)^2} \right] \)

Question. Differentiate \( (\sin x)^{\tan x} + (\cos x)^{\sec x} \) w.r.t. \( x \).
Answer: Let \( y = (\sin x)^{\tan x} + (\cos x)^{\sec x} \)
\( \Rightarrow y = e^{\tan x \cdot \log \sin x} + e^{\sec x \cdot \log \cos x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = (\sin x)^{\tan x} \left\{ \sec^2 x \cdot \log \sin x + \tan x \cdot \frac{1}{\sin x} \cos x \right\} + (\cos x)^{\sec x} \left\{ \sec x \tan x \cdot \log \cos x + \sec x \cdot \frac{1}{\cos x} (-\sin x) \right\} \)
\( \Rightarrow \frac{dy}{dx} = (\sin x)^{\tan x} \left\{ \sec^2 x \log \sin x + 1 \right\} + (\cos x)^{\sec x} \{ \sec x \tan x \cdot \log \cos x - \sec x \tan x \} \)

Question. If \( x = a \sin 2t(1 + \cos 2t) \) and \( y = b \cos 2t(1 - \cos 2t) \), find the values of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \) and \( t = \frac{\pi}{3} \).
Answer: \( x = a \sin 2t (1 + \cos 2t) \), \( y = b \cos 2t(1 - \cos 2t) \)
Now, \( \frac{dx}{dt} = 2a \cos 2t(1 + \cos 2t) + a \sin 2t(-2 \sin 2t) \)
\( = 2a \cos 2t + 2a[\cos^2 2t - \sin^2 2t] \)
\( = 2a \cos 2t + 2a \cos 4t \)
Also, \( \frac{dy}{dt} = -2b \sin 2t(1 - \cos 2t) + b \cos 2t(2 \sin 2t) \)
\( = -2b \sin 2t + 4b (\sin 2t \cos 2t) \)
\( = -2b \sin 2t + 2b \sin 4t \)
So, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2b(\sin 4t - \sin 2t)}{2a(\cos 4t + \cos 2t)} \)
\( \therefore \left. \frac{dy}{dx} \right|_{t=\pi/4} = \frac{b}{a} \left[ \frac{\sin \pi - \sin(\pi/2)}{\cos \pi + \cos(\pi/2)} \right] = \frac{b}{a} \left[ \frac{0 - 1}{-1 + 0} \right] = \frac{b}{a} \br />\( \left. \frac{dy}{dx} \right|_{t=\pi/3} = \frac{b}{a} \left[ \frac{\sin(4\pi/3) - \sin(2\pi/3)}{\cos(4\pi/3) + \cos(2\pi/3)} \right] = \frac{b}{a} \left[ \frac{-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}}{-\frac{1}{2} - \frac{1}{2}} \right] = \frac{\sqrt{3}b}{a} \)

Question. Differentiate \( \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) w.r.t. \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \), if \( x \in (-1, 1) \).
Answer: Let \( u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \)
Put \( x = \tan \theta \Rightarrow \theta = \tan^{-1} x \)
\( \therefore u = \tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan \theta}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\frac{\sec\theta-1}{\tan \theta}\right) \Rightarrow u = \tan^{-1}\left(\frac{1-\cos\theta}{\sin \theta}\right) \)
\( \Rightarrow u = \tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) \Rightarrow u = \tan^{-1}\left(\tan\frac{\theta}{2}\right) \)
\( \dots u = \frac{\theta}{2} \Rightarrow u = \frac{1}{2}\tan^{-1} x \)
Differentiating w.r.t. \( x \), we get
\( \frac{du}{dx} = \frac{1}{2(1+x^2)} \)
Also, let \( v = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \Rightarrow v = 2\tan^{-1} x \)
Differentiating w.r.t. \( x \), we get
\( \frac{dv}{dx} = \frac{2}{1+x^2} \)
\( \therefore \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2(1+x^2)}}{\frac{2}{1+x^2}} \Rightarrow \frac{du}{dv} = \frac{1}{4} \)

Question. If \( x = ae^t(\sin t + \cos t) \) and \( y = ae^t(\sin t - \cos t) \), prove that \( \frac{dy}{dx} = \frac{x+y}{x-y} \).
Answer: We have \( x = ae^t(\sin t + \cos t) \)
\( \Rightarrow \frac{dx}{dt} = ae^t(\sin t + \cos t) + ae^t(\cos t - \sin t) = 2ae^t \cos t \)
and \( y = ae^t(\sin t - \cos t) \)
\( \Rightarrow \frac{dy}{dt} = ae^t(\sin t - \cos t) + ae^t(\cos t + \sin t) = 2ae^t \sin t \)
\( \therefore \text{L.H.S.} = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2ae^t \sin t}{2ae^t \cos t} = \tan t \)
Also, \( \text{R.H.S.} = \frac{x+y}{x-y} = \frac{ae^t(\sin t + \cos t) + ae^t(\sin t - \cos t)}{ae^t(\sin t + \cos t) - ae^t(\sin t - \cos t)} = \frac{2ae^t \sin t}{2ae^t \cos t} = \tan t = \text{L.H.S.} \)

Question. Differentiate \( \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \) with respect to \( \cos^{-1}(2x\sqrt{1-x^2}) \), when \( x \neq 0 \).
Answer: Let \( u = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \)
Put \( x = \cos \theta \)
\( \therefore u = \tan^{-1}\left(\frac{\sqrt{1-\cos^2\theta}}{\cos\theta}\right) = \tan^{-1}\left(\frac{\sin\theta}{\cos\theta}\right) = \tan^{-1}(\tan\theta) = \theta \)
\( \Rightarrow \frac{du}{d\theta} = 1 \)
Also let, \( v = \cos^{-1}(2x\sqrt{1-x^2}) \Rightarrow v = \cos^{-1}(2\cos\theta\sqrt{1-\cos^2\theta}) = \cos^{-1}(2\cos\theta\sin\theta) = \cos^{-1}(\sin 2\theta) = \cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right) = \frac{\pi}{2}-2\theta \)
\( \Rightarrow \frac{dv}{d\theta} = -2 \)
Now, \( \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{1}{-2} = -\frac{1}{2} \)

Question. If \( x = \cos t(3 - 2 \cos^2 t) \) and \( y = \sin t(3 - 2 \sin^2 t) \), find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \).
Answer: Here, \( x = \cos t(3 - 2\cos^2 t) \), \( y = \sin t(3 - 2\sin^2 t) \)
\( \Rightarrow \frac{dx}{dt} = -\sin t(3 - 2\cos^2 t) + \cos t[-2 \cdot 2\cos t(-\sin t)] = -3\sin t + 6\cos^2 t\sin t \)
and \( \frac{dy}{dt} = \cos t(3 - 2\sin^2 t) + \sin t(-2 \cdot 2\sin t\cos t) = -3\cos t + 6\sin^2 t\cos t \)
\( \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t - 6\sin^2 t\cos t}{-3\sin t + 6\cos^2 t\sin t} = \frac{3\cos t\cos 2t}{3\sin t\cos 2t} = \cot t \)
\( \Rightarrow \left. \frac{dy}{dx} \right| _ {t=\pi/4} = \cot\frac{\pi}{4} = 1 \)

Question. If \( x = 2 \cos \theta - \cos 2\theta \) and \( y = 2\sin \theta - \sin 2\theta \), then prove that \( \frac{dy}{dx} = \tan\left(\frac{3\theta}{2}\right) \).
Answer: Here, \( x = 2\cos\theta - \cos 2\theta \), \( y = 2\sin\theta - \sin 2\theta \)
\( \frac{dx}{d\theta} = -2\sin\theta + 2\sin 2\theta \) and \( \frac{dy}{d\theta} = 2\cos\theta - 2\cos 2\theta \)
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(\cos\theta - \cos 2\theta)}{2(\sin 2\theta - \sin\theta)} = \frac{2\sin\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)}{2\cos\left(\frac{3\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)} = \tan\left(\frac{3\theta}{2}\right) \)

Question. If \( x = \sqrt{a^{\sin^{-1} t}} \), \( y = \sqrt{a^{\cos^{-1} t}} \), show that \( \frac{dy}{dx} = -\frac{y}{x} \).
Answer: Here, \( x = \sqrt{a^{\sin^{-1} t}} \Rightarrow \log x = \frac{1}{2}\sin^{-1} t\cdot\log a \)
Differentiating w.r.t. \( t \), we get
\( \frac{1}{x}\frac{dx}{dt} = \frac{1}{2}\frac{1}{\sqrt{1-t^2}}\log a \) ...(1)
Also, \( y = \sqrt{a^{\cos^{-1} t}} \Rightarrow \log y = \frac{1}{2}\cos^{-1} t\cdot\log a \)
Differentiating w.r.t. \( t \), we get
\( \frac{1}{y}\frac{dy}{dt} = \frac{1}{2}\frac{-1}{\sqrt{1-t^2}}\log a \) ...(2)
Now dividing (2) by (1), we get
\( \frac{\frac{1}{y}\frac{dy}{dt}}{\frac{1}{x}\frac{dx}{dt}} = -1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \)

Question. If \( x = a(\theta - \sin \theta) \) and \( y = a(1 + \cos \theta) \), find \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{3} \).
Answer: Here, \( x = a(\theta - \sin\theta) \Rightarrow \frac{dx}{d\theta} = a(1 - \cos\theta) \)
and \( y = a(1 + \cos\theta) \Rightarrow \frac{dy}{d\theta} = a(-\sin\theta) \)
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-a\sin\theta}{a(1 - \cos\theta)} = \frac{-\sin\theta}{1-\cos\theta} \)
\( \therefore \left. \frac{dy}{dx} \right| _ {\theta = \pi/3} = \frac{-\sin(\pi/3)}{1-\cos(\pi/3)} = \frac{-\sqrt{3}/2}{1-(1/2)} = -\sqrt{3} \)

Question. If \( x = a\left(\cos t + \log \tan \frac{t}{2}\right) \) and \( y = a \sin t \), find \( \frac{dy}{dx} \).
Answer: Here, \( x = a\left(\cos t + \log\tan\frac{t}{2}\right) \)
\( \Rightarrow \frac{dx}{dt} = a\left[-\sin t + \frac{1}{\tan\frac{t}{2}}\sec^2\frac{t}{2}\cdot\frac{1}{2}\right] = a\left[-\sin t + \frac{\cos\frac{t}{2}}{\sin\frac{t}{2}}\cdot\frac{1}{\cos^2\frac{t}{2}}\cdot\frac{1}{2}\right] = a\left[-\sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}\right] = a\left[-\sin t + \frac{1}{\sin t}\right] = a\left[\frac{-\sin^2 t + 1}{\sin t}\right] = \frac{a\cos^2 t}{\sin t} \)
Also, \( y = a\sin t \Rightarrow \frac{dy}{dt} = a\cos t \)
\( \dots \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = a\cos t\cdot\frac{\sin t}{a\cos^2 t} = \tan t \)

Question. If \( y = x^x \), prove that \( \frac{d^2y}{dx^2} - \frac{1}{y}\left(\frac{dy}{dx}\right)^2 - \frac{y}{x} = 0 \).
Answer: We have, \( y = x^x \Rightarrow y = e^{x\log x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{x\log x}\left[x\cdot\frac{1}{x} + \log x\right] \Rightarrow \frac{dy}{dx} = x^x(1 + \log x) \Rightarrow \frac{dy}{dx} = y(1 + \log x) \) ...(1)
Again differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = (1 + \log x)\frac{dy}{dx} + y\cdot\frac{1}{x} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{y}\left(\frac{dy}{dx}\right)^2 + \frac{y}{x} \) [From (1)]
\( \Rightarrow \frac{d^2y}{dx^2} - \frac{1}{y}\left(\frac{dy}{dx}\right)^2 - \frac{y}{x} = 0 \)

Question. If \( y = 2\cos(\log x) + 3\sin(\log x) \), prove that \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \).
Answer: We have, \( y = 2\cos(\log x) + 3\sin(\log x) \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = -2\sin(\log x)\cdot\frac{1}{x} + 3\cos(\log x)\cdot\frac{1}{x} \)
\( \Rightarrow x\frac{dy}{dx} = -2\sin(\log x) + 3\cos(\log x) \) ...(1)
Again differentiating w.r.t. \( x \), we get
\( x\frac{d^2y}{dx^2} + \frac{dy}{dx}\cdot 1 = -2\cos(\log x)\cdot\frac{1}{x} - 3\sin(\log x)\cdot\frac{1}{x} \)
\( \Rightarrow x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} = -[2\cos(\log x) + 3\sin(\log x)] \)
\( \Rightarrow x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} = -y \Rightarrow x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0 \)

Question. If \( x = \sin t \) and \( y = \sin pt \). Prove that \( (1-x^2)\frac{d^2y}{dx^2} - x \frac{dy}{dx} + p^2y = 0 \).
Answer: We have, \( x = \sin t \) and \( y = \sin pt \)
\( \frac{dx}{dt} = \cos t \) and \( \frac{dy}{dt} = p\cos pt \)
\( \Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p\cos pt}{\cos t} \)
Differentiating both sides w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = \frac{-p^2\sin pt\cos t + p\cos pt\sin t}{\cos^2 t}\cdot\frac{dt}{dx} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-p^2\sin pt\cos t + p\cos pt\sin t}{\cos^3 t} \) (Since \( \frac{dt}{dx} = \frac{1}{\cos t} \))
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-p^2\sin pt}{\cos^2 t} + \frac{p\cos pt\sin t}{\cos^3 t} \)
\( \Rightarrow \frac{d^2y}{dx^2} = -\frac{p^2y}{\cos^2 t} + \frac{x}{\cos^2 t}\frac{dy}{dx} \)
\( \Rightarrow \cos^2 t\frac{d^2y}{dx^2} = -p^2y + x\frac{dy}{dx} \)
\( \Rightarrow (1 - \sin^2 t)\frac{d^2y}{dx^2} = -p^2y + x\frac{dy}{dx} \)
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2y = 0 \)

Question. If \( x = a \cos \theta + b \sin \theta \), \( y = a \sin \theta - b \cos \theta \), show that \( y^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 \).
Answer: Given, \( x = a\cos\theta + b\sin\theta \), \( y = a\sin\theta - b\cos\theta \)
\( \Rightarrow x^2 = a^2\cos^2\theta + b^2\sin^2\theta + 2ab\cos\theta\sin\theta \) ...(1)
and \( y^2 = a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta \) ...(2)
Adding (1) and (2), we get
\( x^2 + y^2 = a^2 + b^2 \)
Differentiating w.r.t \( x \), we get
\( 2x + 2y\frac{dy}{dx} = 0 \Rightarrow x + y\frac{dy}{dx} = 0 \) ...(1)
Again differentiating w.r.t. \( x \), we get
\( 1 + y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \)
Multiplying by \( y \) on both sides, we get
\( y + y^2\frac{d^2y}{dx^2} + y\left(\frac{dy}{dx}\right)\frac{dy}{dx} = 0 \)
\( \Rightarrow y^2\frac{d^2y}{dx^2} - x\frac{dy}{dx} + y = 0 \) [From (1)]

Question. If \( y = e^{m\sin^{-1} x} \), \( -1 \le x \le 1 \), then show that \( (1-x^2)\frac{d^2y}{dx^2} - x \frac{dy}{dx} - m^2y = 0 \).
Answer: We have, \( y = e^{m\sin^{-1} x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{m\sin^{-1} x}\left(\frac{m}{\sqrt{1-x^2}}\right) = \frac{my}{\sqrt{1-x^2}} \) ...(1)
Again differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = m\left[\frac{\sqrt{1-x^2}\frac{dy}{dx} - y\frac{1}{2\sqrt{1-x^2}}(-2x)}{1-x^2}\right] \)
\( \Rightarrow (1-x^2)\frac{d^2y}{dx^2} = m\left[my + \frac{xy}{\sqrt{1-x^2}}\right] \) [From (1)]
\( \Rightarrow (1-x^2)\frac{d^2y}{dx^2} = m\left[my + x\cdot\left(\frac{1}{m}\frac{dy}{dx}\right)\right] \)
\( \Rightarrow (1-x^2)\frac{d^2y}{dx^2} = m^2y + x\frac{dy}{dx} \)
\( \Rightarrow (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - m^2y = 0 \)

Question. If \( y = (x + \sqrt{1+x^2})^n \), then show that \( (1+x^2)\frac{d^2y}{dx^2} + x \frac{dy}{dx} = n^2y \).
Answer: We have, \( y = (x + \sqrt{1+x^2})^n \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = n(x + \sqrt{1+x^2})^{n-1}\left[1 + \frac{2x}{2\sqrt{1+x^2}}\right] \)
\( = n(x + \sqrt{1+x^2})^{n-1}\left[\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right] = \frac{n(x + \sqrt{1+x^2})^n}{\sqrt{1+x^2}} = \frac{ny}{\sqrt{1+x^2}} \) ...(1)
Again differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = n\left[\frac{\sqrt{1+x^2}\cdot\frac{dy}{dx} - y\frac{2x}{2\sqrt{1+x^2}}}{1+x^2}\right] \)
\( \Rightarrow (1+x^2)\frac{d^2y}{dx^2} = n\left[\sqrt{1+x^2}\frac{dy}{dx} - \frac{xy}{\sqrt{1+x^2}}\right] \)
\( = n\sqrt{1+x^2}\frac{dy}{dx} - \frac{nxy}{\sqrt{1+x^2}} = n^2y - x\frac{dy}{dx} \) [From (1)]
\( \Rightarrow (1+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = n^2y \)