Differentiation by substitution Questions in English

(D) Given $y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$.
Using properties of logarithms,$\log(e/x^2) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.
Let $u = 2\log x$. Then the first term is $\tan^{-1}\left(\frac{1-u}{1+u}\right) = \tan^{-1}(1) - \tan^{-1}(u) = \frac{\pi}{4} - \tan^{-1}(2\log x)$.
For the second term,$\tan^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right] = \tan^{-1}(3) + \tan^{-1}(2\log x)$.
Adding these,$y = \frac{\pi}{4} - \tan^{-1}(2\log x) + \tan^{-1}(3) + \tan^{-1}(2\log x) = \frac{\pi}{4} + \tan^{-1}(3)$.
Since $y$ is a constant,$\frac{dy}{dx} = 0$ and $\frac{d^2 y}{dx^2} = 0$.

(C) Given $y = \tan^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right]$.
Let $x = \cos \theta$,then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ into the expression:
$y = \tan^{-1} \left[ \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right]$
Divide numerator and denominator by $\cos \theta$:
$y = \tan^{-1} \left[ \frac{1 - \tan \theta}{1 + \tan \theta} \right]$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$y = \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - \theta \right) \right] = \frac{\pi}{4} - \theta$
Substitute back $\theta = \cos^{-1} x$:
$y = \frac{\pi}{4} - \cos^{-1} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \cos^{-1} x \right) = 0 - \left( -\frac{1}{\sqrt{1 - x^2}} \right) = \frac{1}{\sqrt{1 - x^2}}$.

(A) Given $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$.
Method $1$: Using the chain rule:
$f^{\prime}(x) = \frac{d}{dx}\left[\sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)\right] = \frac{1}{\sqrt{1-\left(\sqrt{\frac{1-x}{2}}\right)^{2}}} \cdot \frac{d}{dx}\left(\sqrt{\frac{1-x}{2}}\right)$
$= \frac{1}{\sqrt{1-\frac{1-x}{2}}} \cdot \frac{1}{2\sqrt{\frac{1-x}{2}}} \cdot \left(-\frac{1}{2}\right)$
$= \frac{1}{\sqrt{\frac{1+x}{2}}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}\right)$
$= \frac{\sqrt{2}}{\sqrt{1+x}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}\right) = \frac{-1}{2\sqrt{1-x^{2}}}$.
Method $2$: Using substitution:
Let $x = \cos \theta$,then $\sqrt{\frac{1-x}{2}} = \sqrt{\frac{1-\cos \theta}{2}} = \sqrt{\frac{2\sin^{2}(\theta/2)}{2}} = \sin(\theta/2)$.
So,$f(x) = \sin^{-1}(\sin(\theta/2)) = \theta/2 = \frac{1}{2}\cos^{-1}(x)$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{-1}{2\sqrt{1-x^{2}}}$.
Thus,the correct option is $A$.

(A) Let $x = \cos(2\theta)$,so $2\theta = \cos^{-1}(x)$ and $\theta = \frac{1}{2} \cos^{-1}(x)$.
Then $\sqrt{1+x} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2} \cos(\theta)$ and $\sqrt{1-x} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2} \sin(\theta)$.
Substituting these into the expression for $y$:
$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{2} \right] = \sin^{-1} \left[ \frac{1}{\sqrt{2}} \cos(\theta) + \frac{1}{\sqrt{2}} \sin(\theta) \right]$.
Using the identity $\sin(\frac{\pi}{4} + \theta) = \sin(\frac{\pi}{4}) \cos(\theta) + \cos(\frac{\pi}{4}) \sin(\theta) = \frac{1}{\sqrt{2}} \cos(\theta) + \frac{1}{\sqrt{2}} \sin(\theta)$.
So,$y = \sin^{-1} [\sin(\frac{\pi}{4} + \theta)] = \frac{\pi}{4} + \theta = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{1}{2\sqrt{1-x^2}}$.

(A) Given $f(x) = \cos^{-1} \left( \frac{1 - (\log x)^2}{1 + (\log x)^2} \right)$.
Let $u = \log x$. Then the expression becomes $f(x) = \cos^{-1} \left( \frac{1 - u^2}{1 + u^2} \right)$.
Using the trigonometric substitution $u = \tan \theta$,we know that $\cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) = \cos^{-1} (\cos 2\theta) = 2\theta = 2 \tan^{-1} u$.
Thus,$f(x) = 2 \tan^{-1} (\log x)$.
Differentiating with respect to $x$ using the chain rule:
$f'(x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x) = \frac{2}{1 + (\log x)^2} \cdot \frac{1}{x}$.
Now,evaluating at $x = e$:
$f'(e) = \frac{2}{1 + (\log e)^2} \cdot \frac{1}{e} = \frac{2}{1 + 1^2} \cdot \frac{1}{e} = \frac{2}{2} \cdot \frac{1}{e} = \frac{1}{e}$.

(B) Let $y = \tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$.
Substitute $x = \cos 2\theta$,which implies $\theta = \frac{1}{2} \cos ^{-1} x$.
Then $y = \tan ^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$.
Using trigonometric identities $1+\cos 2\theta = 2\cos^2\theta$ and $1-\cos 2\theta = 2\sin^2\theta$,we get:
$y = \tan ^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)$.
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1-\tan\theta}{1+\tan\theta}$,we have:
$y = \tan ^{-1}\tan(\frac{\pi}{4} - \theta) = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2} \times \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2 \sqrt{1-x^2}}$.

(B) $h(x) = f(g(x))$
$h(x) = f(\sin^{-1} x) = e^{\sin^{-1} x}$
Differentiating $h(x)$ with respect to $x$ using the chain rule:
$h^{\prime}(x) = \frac{d}{dx}(e^{\sin^{-1} x}) = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x)$
$h^{\prime}(x) = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$
Now,calculating the ratio $\frac{h^{\prime}(x)}{h(x)}$:
$\frac{h^{\prime}(x)}{h(x)} = \frac{e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin^{-1} x}}$
$\frac{h^{\prime}(x)}{h(x)} = \frac{1}{\sqrt{1-x^2}}$

(B) Let $u = \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$. Put $x = \tan \theta$,then $u = \tan ^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2} \tan ^{-1} x$.
Thus,$\frac{du}{dx} = \frac{1}{2(1+x^2)}$.
Let $v = \tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$. Put $x = \sin \phi$,then $v = \tan ^{-1}\left(\frac{2 \sin \phi \cos \phi}{1-2 \sin^2 \phi}\right) = \tan ^{-1}(\tan 2\phi) = 2\phi = 2 \sin ^{-1} x$.
Thus,$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{2(1+x^2)} \times \frac{\sqrt{1-x^2}}{2} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$.
At $x=0$,$\frac{du}{dv} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

(D) Given $y = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$.
Using the identities $1+\sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$ and $1-\sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$,we have:
$y = \tan^{-1}\left(\sqrt{\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))^2}}\right) = \tan^{-1}\left(\frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}\right)$.
Dividing numerator and denominator by $\cos(x/2)$,we get:
$y = \tan^{-1}\left(\frac{1 + \tan(x/2)}{1 - \tan(x/2)}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$.
Since $0 \leqslant x < \frac{\pi}{2}$,we have $\frac{\pi}{4} \leqslant \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$,so $y = \frac{\pi}{4} + \frac{x}{2}$.
Differentiating with respect to $x$,we get $y' = \frac{1}{2}$.
Thus,$y'\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

(A) Given $y = \tan^{-1}(\sec x + \tan x)$.
We can rewrite the expression inside the inverse tangent function as:
$y = \tan^{-1}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right)$.
Using the half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$,we get:
$y = \tan^{-1}\left[\frac{\cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}\right]$
$y = \tan^{-1}\left[\frac{(\cos\frac{x}{2} + \sin\frac{x}{2})^2}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})}\right]$
$y = \tan^{-1}\left[\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right]$
Dividing numerator and denominator by $\cos\frac{x}{2}$:
$y = \tan^{-1}\left(\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right)$
$y = \frac{\pi}{4} + \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} = \frac{1}{2}$.

(A) Given $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$.
Let $x=\tan \theta$,then $\theta=\tan ^{-1} x$.
$u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\frac{2 \sin ^{2}(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \frac{\theta}{2} = \frac{1}{2} \tan ^{-1} x$.
Thus,$\frac{d u}{d x} = \frac{1}{2(1+x^{2})}$.
Given $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$.
Let $x=\sin \theta$,then $\theta=\sin ^{-1} x$.
$v=\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin ^{2} \theta}\right) = \tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right) = \tan ^{-1}(\tan 2 \theta) = 2 \theta = 2 \sin ^{-1} x$.
Thus,$\frac{d v}{d x} = \frac{2}{\sqrt{1-x^{2}}}$.
Then $\frac{d u}{d v} = \frac{d u/d x}{d v/d x} = \frac{1}{2(1+x^{2})} \times \frac{\sqrt{1-x^{2}}}{2} = \frac{\sqrt{1-x^{2}}}{4(1+x^{2})}$.
At $x=0$,$\frac{d u}{d v} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

(C) Given,$y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$.
Using trigonometric identities $1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right)$ and $\sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$,we have:
$y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{3x}{2} \right)}{2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{3x}{2} \right)} \right)$
$y = \tan^{-1} \left( \frac{\sin \left( \frac{3x}{2} \right)}{\cos \left( \frac{3x}{2} \right)} \right)$
$y = \tan^{-1} \left( \tan \left( \frac{3x}{2} \right) \right)$
$y = \frac{3x}{2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{3x}{2} \right) = \frac{3}{2}$.

(A) Given $y=\tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right)$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$,we get:
$y = \tan^{-1}\left(\frac{2 \sin x \cos x}{2 \cos^2 x}\right)$
$y = \tan^{-1}(\tan x)$
$y = x$
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(A) Let $y = \tan^{-1} \left( \frac{1-x}{1+x} \right)$.
We know that $\tan^{-1} \left( \frac{a-b}{1+ab} \right) = \tan^{-1} a - \tan^{-1} b$.
Here,$a = 1$ and $b = x$,so $y = \tan^{-1}(1) - \tan^{-1}(x)$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$,we have $y = \frac{\pi}{4} - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} \right) - \frac{d}{dx} \left( \tan^{-1} x \right)$.
$\frac{dy}{dx} = 0 - \frac{1}{1+x^2} = \frac{-1}{1+x^2}$.
Thus,the correct option is $A$.

(C) Let $u = \tan^{-1} x$ and $v = \cot^{-1} x$.
We know that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x \in R$.
Therefore,$u + v = \frac{\pi}{2}$,which implies $u = \frac{\pi}{2} - v$.
Now,differentiate $u$ with respect to $v$:
$\frac{du}{dv} = \frac{d}{dv} (\frac{\pi}{2} - v) = 0 - 1 = -1$.
Thus,the derivative of $\tan^{-1} x$ with respect to $\cot^{-1} x$ is $-1$.

(C) Let $y = \tan^{-1} \left( \frac{x}{1+6x^2} \right)$.
We can rewrite the argument of $\tan^{-1}$ as $\frac{3x - 2x}{1 + (3x)(2x)}$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$,we get:
$y = \tan^{-1}(3x) - \tan^{-1}(2x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1}(3x)) - \frac{d}{dx} (\tan^{-1}(2x))$.
Using the chain rule $\frac{d}{dx} (\tan^{-1}(u)) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = \frac{1}{1+(3x)^2} \cdot 3 - \frac{1}{1+(2x)^2} \cdot 2$.
$\frac{dy}{dx} = \frac{3}{1+9x^2} - \frac{2}{1+4x^2}$.
Thus,the correct option is $C$.

(D) Given that,$y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$.
Dividing the numerator and denominator by $\cos x$,we get:
$y = \tan^{-1}\left(\frac{\tan x + 1}{1 - \tan x}\right)$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,where $A = \pi/4$ and $B = x$:
$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + x\right)\right) = \frac{\pi}{4} + x$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + x\right) = 0 + 1 = 1$.

(B) Given,$f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}$.
Let $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = \frac{3}{\sqrt{13}}$.
Then,$f(x)=\cos ^{-1}(\cos \alpha \cos x - \sin \alpha \sin x)$.
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$f(x) = \cos ^{-1}(\cos(x+\alpha)) = x+\alpha$.
Now,differentiating with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x+\alpha) = 1 + 0 = 1$.
Therefore,$f^{\prime}(0.5) = 1$.

(A) Let $y = \cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)$.
Put $x = 2 \cos \theta$,then $\cos \theta = \frac{x}{2}$.
Then,$\sqrt{\frac{2+x}{2-x}} = \sqrt{\frac{2+2\cos \theta}{2-2\cos \theta}} = \sqrt{\frac{2(1+\cos \theta)}{2(1-\cos \theta)}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot(\theta/2)$.
Substituting this into the expression,we get $y = \cos^2(\cot^{-1}(\cot(\theta/2))) = \cos^2(\theta/2)$.
Using the identity $\cos^2(\theta/2) = \frac{1+\cos \theta}{2}$,we have $y = \frac{1}{2} + \frac{1}{2}\cos \theta$.
Since $\cos \theta = \frac{x}{2}$,we have $y = \frac{1}{2} + \frac{x}{4}$.
Now,differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} + \frac{x}{4}) = 0 + \frac{1}{4} = \frac{1}{4}$.

(C) Given $u = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$.
Using the substitution $x = \tan \theta$,we have $u = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x$.
Therefore,$\frac{du}{dx} = \frac{2}{1+x^2}$.
Given $v = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$.
Using the substitution $x = \tan \theta$,we have $v = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}x$.
Therefore,$\frac{dv}{dx} = \frac{2}{1+x^2}$.
Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} = 1$.

(B) Let $u = \tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ and $v = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$.
For $u$:
$u = \tan ^{-1}\left(\frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}\right) = \tan ^{-1}(\tan(x/2)) = x/2$.
Thus,$\frac{du}{dx} = \frac{1}{2}$.
For $v$:
$v = \tan ^{-1}\left(\frac{\cos^2(x/2) - \sin^2(x/2)}{(\cos(x/2) + \sin(x/2))^2}\right) = \tan ^{-1}\left(\frac{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))}{(\cos(x/2) + \sin(x/2))^2}\right)$.
$v = \tan ^{-1}\left(\frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}\right) = \tan ^{-1}\left(\frac{1 - \tan(x/2)}{1 + \tan(x/2)}\right) = \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = \frac{\pi}{4} - \frac{x}{2}$.
Thus,$\frac{dv}{dx} = -\frac{1}{2}$.
Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/2}{-1/2} = -1$.

(A) Given $f(x) = \sin^{-1}\left[\frac{2 \cdot 2^x}{1 + (2^x)^2}\right]$.
Let $2^x = \tan \theta$,then $\theta = \tan^{-1}(2^x)$.
Substituting this into the function:
$f(x) = \sin^{-1}\left[\frac{2 \tan \theta}{1 + \tan^2 \theta}\right]$
Using the identity $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get:
$f(x) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(2^x)$.
Now,differentiating with respect to $x$:
$f'(x) = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x)$
$f'(x) = \frac{2}{1 + 4^x} \cdot 2^x \log_e 2$.
Evaluating at $x = 0$:
$f'(0) = \frac{2}{1 + 4^0} \cdot 2^0 \log_e 2 = \frac{2}{1 + 1} \cdot 1 \cdot \log_e 2 = \frac{2}{2} \log_e 2 = \log_e 2$.

(B) We have,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$.
Let $x=\tan \theta$,then $\theta=\tan ^{-1} x$.
Substituting this into the function,we get $f(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$.
Since $\sin 2\theta = \frac{2 \tan \theta}{1+\tan ^{2} \theta}$,we have $f(x)=\sin ^{-1}(\sin 2 \theta) = 2 \theta = 2 \tan ^{-1} x$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = \frac{2}{1+x^{2}}$.
Substituting $x=\sqrt{3}$,we get $f^{\prime}(\sqrt{3}) = \frac{2}{1+(\sqrt{3})^{2}} = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2}$.

(D) Given $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$.
We know the trigonometric identity $2\operatorname{Tan}^{-1}(x) = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.
Substituting this into the equation,we get $y = 2\operatorname{Tan}^{-1}(x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = 2 \times \frac{1}{1+x^2} = \frac{2}{1+x^2}$.
Substitute $x = \frac{1}{2}$ into the derivative:
$\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = \frac{2}{1 + (\frac{1}{2})^2} = \frac{2}{1 + \frac{1}{4}} = \frac{2}{\frac{5}{4}} = \frac{8}{5}$.
Thus,the correct option is $D$.

(B) Given $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$.
Let $x = \cos \theta$,then $\theta = \cos^{-1} x$.
Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} = \tan(\theta/2)$.
So,$y = \operatorname{Tanh}^{-1} (\tan(\theta/2))$.
Using the identity $\operatorname{Tanh}^{-1} z = \frac{1}{2} \ln \left( \frac{1+z}{1-z} \right)$:
$y = \frac{1}{2} \ln \left( \frac{1+\tan(\theta/2)}{1-\tan(\theta/2)} \right) = \frac{1}{2} \ln \left( \tan \left( \frac{\pi}{4} + \frac{\theta}{2} \right) \right) = \frac{1}{2} \left( \ln \tan \left( \frac{\pi}{4} + \frac{\cos^{-1} x}{2} \right) \right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\tan(\frac{\pi}{4} + \frac{\theta}{2})} \cdot \sec^2(\frac{\pi}{4} + \frac{\theta}{2}) \cdot \frac{1}{2} \cdot \frac{d}{dx}(\cos^{-1} x)$.
$\frac{dy}{dx} = \frac{1}{4} \cdot \frac{1}{\sin(\frac{\pi}{4} + \frac{\theta}{2}) \cos(\frac{\pi}{4} + \frac{\theta}{2})} \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right)$.
$\frac{dy}{dx} = \frac{1}{2 \sin(\frac{\pi}{2} + \theta)} \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right) = \frac{1}{2 \cos \theta} \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{1}{2x \sqrt{1-x^2}}$.

(B) We know that $\operatorname{Tan}^{-1}(a) - \operatorname{Tan}^{-1}(b) = \operatorname{Tan}^{-1}\left(\frac{a-b}{1+ab}\right)$.
We can rewrite the terms as:
$\operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{2x-x}{1+(2x)(x)}\right) = \operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)$.
$\operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)$.
Substituting these into the expression for $y$:
$y = (\operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)) + (\operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\operatorname{Tan}^{-1}(3x)) - \frac{d}{dx}(\operatorname{Tan}^{-1}(x)) = \frac{3}{1+(3x)^2} - \frac{1}{1+x^2} = \frac{3}{9x^2+1} - \frac{1}{x^2+1}$.

(B) Let $u = \sqrt{x^2-1}$. Then $y = \operatorname{Tan}^{-1}(u) + \operatorname{Sinh}^{-1}(u)$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
First,$\frac{dy}{du} = \frac{d}{du}(\operatorname{Tan}^{-1} u) + \frac{d}{du}(\operatorname{Sinh}^{-1} u) = \frac{1}{1+u^2} + \frac{1}{\sqrt{1+u^2}}$.
Substitute $u^2 = x^2-1$: $\frac{dy}{du} = \frac{1}{1+(x^2-1)} + \frac{1}{\sqrt{1+(x^2-1)}} = \frac{1}{x^2} + \frac{1}{x} = \frac{1+x}{x^2}$.
Next,$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x^2-1}) = \frac{1}{2\sqrt{x^2-1}} \cdot 2x = \frac{x}{\sqrt{x^2-1}}$.
Therefore,$\frac{dy}{dx} = \left(\frac{1+x}{x^2}\right) \cdot \left(\frac{x}{\sqrt{x^2-1}}\right) = \frac{1+x}{x \sqrt{x^2-1}}$.
Thus,the correct option is $B$.

(B) Given,$y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$.
Using the chain rule,$\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x}\right)^2} \times \frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$.
Simplify the denominator: $1+\left(\frac{2-3 \sin x}{3-2 \sin x}\right)^2 = \frac{(3-2 \sin x)^2 + (2-3 \sin x)^2}{(3-2 \sin x)^2} = \frac{9+4 \sin^2 x - 12 \sin x + 4 + 9 \sin^2 x - 12 \sin x}{(3-2 \sin x)^2} = \frac{13 \sin^2 x - 24 \sin x + 13}{(3-2 \sin x)^2}$.
Now,differentiate the inner function using the quotient rule: $\frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}\right) = \frac{(3-2 \sin x)(-3 \cos x) - (2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}$.
$= \frac{-9 \cos x + 6 \sin x \cos x + 4 \cos x - 6 \sin x \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{(3-2 \sin x)^2}$.
Combining these,$\frac{d y}{d x} = \frac{(3-2 \sin x)^2}{13 \sin^2 x - 24 \sin x + 13} \times \frac{-5 \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{13 \sin^2 x - 24 \sin x + 13}$.

(C) Given that $y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right)$.
Let $x^2 = \cos 2\theta$,then $\theta = \frac{1}{2} \cos^{-1} (x^2)$.
Substituting $x^2 = \cos 2\theta$ into the expression:
$y = \tan^{-1} \left( \frac{\sqrt{1 + \cos 2\theta} + \sqrt{1 - \cos 2\theta}}{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}} \right)$.
Using the identities $1 + \cos 2\theta = 2 \cos^2 \theta$ and $1 - \cos 2\theta = 2 \sin^2 \theta$:
$y = \tan^{-1} \left( \frac{\sqrt{2} \cos \theta + \sqrt{2} \sin \theta}{\sqrt{2} \cos \theta - \sqrt{2} \sin \theta} \right) = \tan^{-1} \left( \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \right)$.
Dividing numerator and denominator by $\cos \theta$:
$y = \tan^{-1} \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right) = \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \theta \right) \right) = \frac{\pi}{4} + \theta$.
Substituting $\theta = \frac{1}{2} \cos^{-1} (x^2)$:
$y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1} (x^2)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} \left( \frac{-1}{\sqrt{1 - (x^2)^2}} \right) \cdot \frac{d}{dx}(x^2) = \frac{1}{2} \left( \frac{-1}{\sqrt{1 - x^4}} \right) \cdot 2x = \frac{-x}{\sqrt{1 - x^4}}$.

(B) Let $y = \sin^2 \left( \cot^{-1} \sqrt{\frac{1 + x}{1 - x}} \right)$.
Substitute $x = \cos 2\theta$,so $\theta = \frac{1}{2} \cos^{-1} x$.
Then $\sqrt{\frac{1 + x}{1 - x}} = \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}} = \sqrt{\frac{2 \cos^2 \theta}{2 \sin^2 \theta}} = \cot \theta$.
Thus,$y = \sin^2 \left( \cot^{-1} (\cot \theta) \right) = \sin^2 \theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $y = \frac{1 - x}{2} = \frac{1}{2} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} - \frac{x}{2} \right) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(C) Given $y = \tan^{-1}\left(\frac{a \cos x - b \sin x}{b \cos x + a \sin x}\right)$.
Divide the numerator and denominator by $b \cos x$:
$y = \tan^{-1}\left(\frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x}\right)$.
Let $\frac{a}{b} = \tan \theta$,where $\theta = \tan^{-1}(\frac{a}{b})$.
Then $y = \tan^{-1}\left(\frac{\tan \theta - \tan x}{1 + \tan \theta \tan x}\right)$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$y = \tan^{-1}(\tan(\theta - x)) = \theta - x$.
Since $\theta = \tan^{-1}(\frac{a}{b})$ is a constant,its derivative is $0$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\theta - x) = 0 - 1 = -1$.

(A) Given $y = \sin^{-1}(x \sqrt{1-x^2} - \sqrt{x} \sqrt{1-x})$.
Let $x = \sin \alpha$ and $\sqrt{x} = \sin \beta$. Then $\sqrt{1-x^2} = \cos \alpha$ and $\sqrt{1-x} = \cos \beta$.
Substituting these into the expression,we get $y = \sin^{-1}(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.
Using the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$,we have $y = \sin^{-1}(\sin(\alpha - \beta)) = \alpha - \beta$.
Thus,$y = \sin^{-1} x - \sin^{-1} \sqrt{x}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) - \frac{d}{dx}(\sin^{-1} \sqrt{x})$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x(1-x)}} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x-x^2}}$.

(C) Let $y = \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ and $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$.
Substitute $x = \sin \theta$,then $\sqrt{1-x^2} = \cos \theta$.
$y = \tan ^{-1}\left(\frac{\sin \theta}{1+\cos \theta}\right) = \tan ^{-1}\left(\frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \frac{\theta}{2}$.
Now,$u = \sec ^{-1}\left(\frac{1}{2 \sin^2 \theta - 1}\right) = \sec ^{-1}\left(\frac{1}{-(1-2 \sin^2 \theta)}\right) = \sec ^{-1}\left(\frac{1}{-\cos 2 \theta}\right) = \sec ^{-1}(-\sec 2 \theta) = \sec ^{-1}(\sec(\pi - 2 \theta)) = \pi - 2 \theta$.
Since $x = \sin \theta$,$\theta = \sin^{-1} x$.
Thus,$y = \frac{1}{2} \sin^{-1} x$ and $u = \pi - 2 \sin^{-1} x$.
Then $\frac{dy}{dx} = \frac{1}{2 \sqrt{1-x^2}}$ and $\frac{du}{dx} = \frac{-2}{\sqrt{1-x^2}}$.
Therefore,$\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{1/(2 \sqrt{1-x^2})}{-2/\sqrt{1-x^2}} = -\frac{1}{4}$.

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-x\right)$.
We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}\right)^3 - 3\left(\frac{x}{3}\right)\right)$.
Let $\frac{x}{3} = \cos A$,then $A = \cos ^{-1}\left(\frac{x}{3}\right)$.
Substituting this into the expression,we get $y = \cos ^{-1}(4 \cos ^3 A - 3 \cos A)$.
Using the trigonometric identity $\cos(3A) = 4 \cos ^3 A - 3 \cos A$,we have $y = \cos ^{-1}(\cos 3A) = 3A$.
Thus,$y = 3 \cos ^{-1}\left(\frac{x}{3}\right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \times \left(-\frac{1}{\sqrt{1 - (x/3)^2}}\right) \times \frac{d}{dx}\left(\frac{x}{3}\right)$.
$\frac{dy}{dx} = 3 \times \left(-\frac{1}{\sqrt{(9-x^2)/9}}\right) \times \frac{1}{3} = 3 \times \left(-\frac{3}{\sqrt{9-x^2}}\right) \times \frac{1}{3} = -\frac{3}{\sqrt{9-x^2}}$.

(D) Given $y = \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$.
We know that $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Thus,$\sqrt{1 + \sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$ and $\sqrt{1 - \sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$.
Substituting these into the expression for $y$:
$y = \tan^{-1} \left[ \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})} \right]$ (assuming $0 < x < \frac{\pi}{2}$).
$y = \tan^{-1} \left( \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} \right) = \tan^{-1} (\cot \frac{x}{2}) = \tan^{-1} \left( \tan (\frac{\pi}{2} - \frac{x}{2}) \right) = \frac{\pi}{2} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{2} - \frac{x}{2}) = -\frac{1}{2}$.
Depending on the interval of $x$,the derivative can be $\pm \frac{1}{2}$.
Therefore,the correct option is $(D)$.

(B) Given,$y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$.
Let $x=\tan \theta$,then $\theta = \tan ^{-1} x$.
Substituting $x$ in the expression:
$y = \tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)$
$y = \tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$
$y = \tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$
Using trigonometric identities $1-\cos \theta = 2\sin ^2(\theta/2)$ and $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$:
$y = \tan ^{-1}\left(\frac{2\sin ^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \theta/2$.
Since $\theta = \tan ^{-1} x$,we have $y = \frac{1}{2} \tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.
Hence,option $B$ is correct.

(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$
Let $x = \cos 2\theta$,then $2\theta = \cos^{-1} x$ or $\theta = \frac{1}{2} \cos^{-1} x$.
Substituting $x$ in the expression:
$y = \sin^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}{2}\right)$
Using trigonometric identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$:
$y = \sin^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{2}\right) = \sin^{-1}\left(\frac{\cos \theta - \sin \theta}{\sqrt{2}}\right)$
$y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\cos \theta - \frac{1}{\sqrt{2}}\sin \theta\right)$
$y = \sin^{-1}\left(\sin \frac{\pi}{4} \cos \theta - \cos \frac{\pi}{4} \sin \theta\right)$
$y = \sin^{-1}\left(\sin\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$
Substituting $\theta = \frac{1}{2} \cos^{-1} x$:
$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2\sqrt{1-x^2}}$
Thus,option $D$ is correct.

(B) Given $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$.
Using the formula $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right)$,the first term is $\tan^{-1}(5x) - \tan^{-1}(x)$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right)$,the second term is $\tan^{-1}(2/3) + \tan^{-1}(x)$.
Thus,$y = \tan^{-1}(5x) - \tan^{-1}(x) + \tan^{-1}(2/3) + \tan^{-1}(x) = \tan^{-1}(5x) + \tan^{-1}(2/3)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1}(5x)) + \frac{d}{dx} (\tan^{-1}(2/3))$.
Since $\tan^{-1}(2/3)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = \frac{1}{1 + (5x)^2} \times \frac{d}{dx}(5x) + 0 = \frac{5}{1 + 25x^2}$.

(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} \right)$.
Let $\cos \alpha = \frac{3}{5}$,then $\sin \alpha = \frac{4}{5}$.
Also,let $\sin \theta = \frac{x}{\sqrt{1+x^2}}$,then $\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
Substituting these,we get $y = \sin^{-1} (\cos \alpha \cos \theta + \sin \alpha \sin \theta) = \sin^{-1} (\cos(\alpha - \theta)) = \sin^{-1} (\sin(\frac{\pi}{2} - (\alpha - \theta))) = \frac{\pi}{2} - \alpha + \theta$.
Since $\theta = \tan^{-1} x$,we have $y = \frac{\pi}{2} - \alpha + \tan^{-1} x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 0 - 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.

(D) Given,$f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$.
Let $y = \cot ^{-1}\left(\frac{x^{2x}-1}{2x^x}\right)$.
Put $x^x = \tan \theta$,then $y = \cot ^{-1}\left(\frac{\tan^2 \theta - 1}{2 \tan \theta}\right)$.
Since $\cot 2\theta = \frac{1-\tan^2 \theta}{2 \tan \theta}$,we have $y = \cot ^{-1}(-\cot 2\theta)$.
Using the property $\cot ^{-1}(-z) = \pi - \cot ^{-1}(z)$,we get $y = \pi - \cot ^{-1}(\cot 2\theta) = \pi - 2\theta$.
Substituting back $\theta = \tan ^{-1}(x^x)$,we have $y = \pi - 2 \tan ^{-1}(x^x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -2 \cdot \frac{1}{1+(x^x)^2} \cdot \frac{d}{dx}(x^x)$.
Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we have $\frac{dy}{dx} = -\frac{2x^x(1 + \ln x)}{1 + x^{2x}}$.
At $x=1$,$\frac{dy}{dx} = -\frac{2(1)^1(1 + \ln 1)}{1 + (1)^2} = -\frac{2(1)(1+0)}{2} = -1$.

(A) Given the function $f(x) = \cos^2 \left( \tan^{-1} \sqrt{\frac{1-x}{1+x}} \right)$ for $-1 < x < 1$.
Let $x = \cos(2\theta)$,where $0 < 2\theta < \pi$,so $0 < \theta < \frac{\pi}{2}$.
Then,$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \tan\theta$.
Substituting this into the function,we get $f(x) = \cos^2(\tan^{-1}(\tan\theta)) = \cos^2\theta$.
Using the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$,we have $f(x) = \frac{1+x}{2}$.
Differentiating with respect to $x$,we get $f'(x) = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} \right) = \frac{1}{2}$.

(B) Let $y = \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right)$.
Using the substitution $\cos \theta = \frac{b - a \cos x}{a - b \cos x}$,we know that $\cos y = \frac{b - a \cos x}{a - b \cos x}$.
By applying the component-dividendo rule or trigonometric identities for $\tan(y/2)$,we find that $\tan \left( \frac{y}{2} \right) = \sqrt{\frac{1 - \cos y}{1 + \cos y}}$.
Substituting $\cos y = \frac{b - a \cos x}{a - b \cos x}$:
$1 - \cos y = 1 - \frac{b - a \cos x}{a - b \cos x} = \frac{a - b \cos x - b + a \cos x}{a - b \cos x} = \frac{(a - b)(1 + \cos x)}{a - b \cos x}$.
$1 + \cos y = 1 + \frac{b - a \cos x}{a - b \cos x} = \frac{a - b \cos x + b - a \cos x}{a - b \cos x} = \frac{(a + b)(1 - \cos x)}{a - b \cos x}$.
Thus,$\tan^2 \left( \frac{y}{2} \right) = \frac{(a - b)(1 + \cos x)}{(a + b)(1 - \cos x)} = \frac{a - b}{a + b} \cot^2 \left( \frac{x}{2} \right)$.
Taking the square root,$y = 2 \tan^{-1} \left( \sqrt{\frac{a - b}{a + b}} \cot \frac{x}{2} \right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + \frac{a - b}{a + b} \cot^2 \frac{x}{2}} \cdot \sqrt{\frac{a - b}{a + b}} \cdot \left( -\frac{1}{2} \csc^2 \frac{x}{2} \right)$.
Simplifying this expression yields $\frac{dy}{dx} = \frac{-\sqrt{a^2 - b^2}}{a - b \cos x}$.

(B) Let $x^2 = \cos(2\theta)$,where $2\theta \in (0, \pi)$,so $\theta = \frac{1}{2} \cos^{-1}(x^2)$.
Then $\sqrt{1+x^2} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2\theta} = \sqrt{2}\cos\theta$ and $\sqrt{1-x^2} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta$.
Substituting these into $f(x)$:
$f(x) = \operatorname{Tan}^{-1} \left[ \frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta} \right] = \operatorname{Tan}^{-1} \left[ \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \right] = \operatorname{Tan}^{-1} \left[ \frac{1 + \tan\theta}{1 - \tan\theta} \right] = \operatorname{Tan}^{-1} [\tan(\frac{\pi}{4} + \theta)] = \frac{\pi}{4} + \theta$.
Substituting $\theta$ back: $f(x) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.
Differentiating with respect to $x$:
$f'(x) = 0 + \frac{1}{2} \cdot \frac{-1}{\sqrt{1-(x^2)^2}} \cdot \frac{d}{dx}(x^2) = \frac{-1}{2\sqrt{1-x^4}} \cdot 2x = \frac{-x}{\sqrt{1-x^4}}$.

(D) Let $x^2 = \cos(2\theta)$,so $2\theta = \cos^{-1}(x^2)$,which implies $\theta = \frac{1}{2} \cos^{-1}(x^2)$.
Then $\sqrt{1+x^2} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2}\cos(\theta)$ and $\sqrt{1-x^2} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2}\sin(\theta)$.
Substituting these into the expression for $y$:
$y = \operatorname{Tan}^{-1}\left(\frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{\sqrt{2}\cos(\theta) - \sqrt{2}\sin(\theta)}\right) = \operatorname{Tan}^{-1}\left(\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)}\right)$.
Dividing numerator and denominator by $\cos(\theta)$:
$y = \operatorname{Tan}^{-1}\left(\frac{1 + \tan(\theta)}{1 - \tan(\theta)}\right) = \operatorname{Tan}^{-1}(\tan(\frac{\pi}{4} + \theta)) = \frac{\pi}{4} + \theta$.
Substituting $\theta = \frac{1}{2} \cos^{-1}(x^2)$:
$y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} \left(-\frac{1}{\sqrt{1-(x^2)^2}}\right) \cdot (2x) = -\frac{x}{\sqrt{1-x^4}}$.
Thus,the correct option is $D$.

(A) Given that,$y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\}$.
Let $x = \cos 2\theta$,then $\theta = \frac{1}{2} \cos^{-1} x$.
Substituting $x = \cos 2\theta$ into the expression:
$y = \tan^{-1} \left\{ \frac{\cos 2\theta}{1 + \sin 2\theta} \right\} + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} \right\}$
$y = \tan^{-1} \left\{ \frac{\cos^2 \theta - \sin^2 \theta}{(\cos \theta + \sin \theta)^2} \right\} + \sin \left\{ 2 \tan^{-1} (\tan \theta) \right\}$
$y = \tan^{-1} \left\{ \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right\} + \sin 2\theta$
$y = \tan^{-1} \left\{ \tan \left( \frac{\pi}{4} - \theta \right) \right\} + \sin 2\theta$
$y = \frac{\pi}{4} - \theta + \sin 2\theta$
Substituting $\theta = \frac{1}{2} \cos^{-1} x$ and $\sin 2\theta = \sqrt{1 - x^2}$:
$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x + \sqrt{1 - x^2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - x^2}} \right) + \frac{1}{2\sqrt{1 - x^2}} (-2x)$
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \frac{1 - 2x}{2\sqrt{1 - x^2}}$.

(D) Let $u = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $v = \sqrt{1-x^2}$.
Using the identity $\operatorname{Sec}^{-1}(\frac{1}{z}) = \cos^{-1}(z)$,we have $u = \cos^{-1}(2x^2-1)$.
Let $x = \cos \theta$,then $u = \cos^{-1}(2\cos^2 \theta - 1) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\cos^{-1}x$.
Thus,$\frac{du}{dx} = 2 \times (-\frac{1}{\sqrt{1-x^2}}) = -\frac{2}{\sqrt{1-x^2}}$.
Now,$v = \sqrt{1-x^2}$,so $\frac{dv}{dx} = \frac{1}{2\sqrt{1-x^2}} \times (-2x) = -\frac{x}{\sqrt{1-x^2}}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}} = \frac{2}{x}$.
At $x = \frac{1}{2}$,$\frac{du}{dv} = \frac{2}{1/2} = 4$.

(D) Given that $px+my+n=0$ is the tangent to the curve $y=f(x)$ at $x=\alpha$.
The slope of the tangent line $px+my+n=0$ is $m_t = -\frac{p}{m}$.
Therefore,$f'(\alpha) = -\frac{p}{m}$.
Now,we need to find $\frac{d}{dx} [f(\alpha e^{2x})]$ at $x=0$.
Using the chain rule:
$\frac{d}{dx} [f(\alpha e^{2x})] = f'(\alpha e^{2x}) \cdot \frac{d}{dx}(\alpha e^{2x}) = f'(\alpha e^{2x}) \cdot (2\alpha e^{2x})$.
At $x=0$,the expression becomes:
$f'(\alpha e^0) \cdot (2\alpha e^0) = f'(\alpha) \cdot (2\alpha)$.
Substituting $f'(\alpha) = -\frac{p}{m}$:
$(-\frac{p}{m}) \cdot (2\alpha) = -\frac{2p\alpha}{m}$.

(C) Let $\theta = \cos^{-1} \sqrt{\frac{1+x^2}{2}}$. Then $\cos \theta = \sqrt{\frac{1+x^2}{2}}$.
Squaring both sides,$\cos^2 \theta = \frac{1+x^2}{2}$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1 = \frac{1}{\cos^2 \theta} - 1$,we get:
$y = \frac{1}{\cos^2 \theta} - 1 = \frac{1}{\frac{1+x^2}{2}} - 1 = \frac{2}{1+x^2} - 1 = \frac{2 - (1+x^2)}{1+x^2} = \frac{1-x^2}{1+x^2}$.
Now,differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2}$.
Thus,the correct option is $C$.

(D) Given,$y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$.
Substitute $ax = \tan \theta$,so $\theta = \tan ^{-1}(ax)$.
Then $y = \tan ^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2}$.
Thus,$y = \frac{1}{2} \tan ^{-1}(ax)$.
Differentiating with respect to $x$,we get $y^{\prime} = \frac{1}{2} \cdot \frac{a}{1+a^2 x^2}$.
This implies $(1+a^2 x^2) y^{\prime} = \frac{a}{2}$.
Differentiating again with respect to $x$,we get $(1+a^2 x^2) y^{\prime \prime} + (2a^2 x) y^{\prime} = 0$.
Comparing this with $(1+a^2 x^2) y^{\prime \prime} + g(x) y^{\prime} = 0$,we find $g(x) = 2a^2 x$.
The equation $1+a^2 x^2 + g(x) = 0$ becomes $a^2 x^2 + 2a^2 x + 1 = 0$.
The sum of the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is $-\frac{B}{A}$.
Here,the sum of the roots is $-\frac{2a^2}{a^2} = -2$.