Differentiation by substitution Questions in English

(B) The given equation is $(1+\sin x)y^3 - (\cos x)y^2 + 2(1+\sin x)y - 2\cos x = 0$.
Rearranging the terms,we get $(1+\sin x)y(y^2+2) - \cos x(y^2+2) = 0$.
This simplifies to $(y^2+2)((1+\sin x)y - \cos x) = 0$.
Since $y^2+2 = 0$ has no real solutions for $y$,we must have $(1+\sin x)y - \cos x = 0$,which implies $y = \frac{\cos x}{1+\sin x}$.
Using the half-angle formulas $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$,let $t = \tan(x/2)$.
Then $y = \frac{1-t^2}{1+t^2} / (1 + \frac{2t}{1+t^2}) = \frac{1-t^2}{1+t^2+2t} = \frac{(1-t)(1+t)}{(1+t)^2} = \frac{1-t}{1+t}$.
We want to find $\frac{dy}{dt}$ at $x = \frac{\pi}{2}$,which means $t = \tan(\frac{\pi}{4}) = 1$.
$y = \frac{1-t}{1+t} \Rightarrow \frac{dy}{dt} = \frac{-(1+t) - (1-t)}{(1+t)^2} = \frac{-2}{(1+t)^2}$.
At $t = 1$,$\frac{dy}{dt} = \frac{-2}{(1+1)^2} = \frac{-2}{4} = -\frac{1}{2}$.

(A) Let $P = \sin^{-1}\left(\frac{3y}{2} - \frac{y^3}{2}\right)$.
We can rewrite the expression inside the inverse sine function as $3\left(\frac{y}{2}\right) - 4\left(\frac{y}{2}\right)^3$.
Let $\frac{y}{2} = \sin \theta$,which implies $\theta = \sin^{-1}\left(\frac{y}{2}\right)$.
Substituting this into the expression,we get $P = \sin^{-1}(3\sin \theta - 4\sin^3 \theta)$.
Using the trigonometric identity $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$,we have $P = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting back for $\theta$,we get $P = 3\sin^{-1}\left(\frac{y}{2}\right)$.
Differentiating with respect to $y$,we get $\frac{dP}{dy} = 3 \cdot \frac{1}{\sqrt{1 - (y/2)^2}} \cdot \frac{d}{dy}\left(\frac{y}{2}\right)$.
$\frac{dP}{dy} = 3 \cdot \frac{1}{\sqrt{1 - y^2/4}} \cdot \frac{1}{2} = 3 \cdot \frac{1}{\sqrt{(4 - y^2)/4}} \cdot \frac{1}{2} = 3 \cdot \frac{1}{\frac{\sqrt{4 - y^2}}{2}} \cdot \frac{1}{2} = \frac{3}{\sqrt{4 - y^2}}$.

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}\right)$.
Let $3^x = \tan \theta$,then $\theta = \tan^{-1}(3^x)$.
$f(x) = \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(3^x)$.
Differentiating with respect to $x$:
$f'(x) = 2 \cdot \frac{1}{1 + (3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2}{1 + 9^x} \cdot 3^x \ln 3$.
Now,evaluate at $x = -\frac{1}{2}$:
$f'(-\frac{1}{2}) = \frac{2 \cdot 3^{-1/2}}{1 + 9^{-1/2}} \ln 3 = \frac{2 / \sqrt{3}}{1 + 1/3} \ln 3 = \frac{2 / \sqrt{3}}{4/3} \ln 3 = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} \ln 3 = \frac{\sqrt{3}}{2} \ln 3$.
Since $\ln 3 = 2 \ln \sqrt{3}$,we have $f'(-\frac{1}{2}) = \frac{\sqrt{3}}{2} \cdot 2 \ln \sqrt{3} = \sqrt{3} \ln \sqrt{3}$.

(C) Given $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$.
Using the chain rule,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = f'\left(\frac{2x + 3}{3 - 2x}\right) \cdot \frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right)$.
First,calculate the derivative of the inner function using the quotient rule:
$\frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right) = \frac{(3 - 2x)(2) - (2x + 3)(-2)}{(3 - 2x)^2} = \frac{6 - 4x + 4x + 6}{(3 - 2x)^2} = \frac{12}{(3 - 2x)^2}$.
Now,substitute this back into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right] \cdot \frac{12}{(3 - 2x)^2}$.
Thus,$\frac{dy}{dx} = \frac{12}{(3 - 2x)^2} \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right]$.

(A) Given $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$.
Divide numerator and denominator by $2$ inside the bracket:
$\frac{{\frac{{\sqrt 3 }}{2}\cos x + \frac{1}{2}\sin x}}{{\frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x}} = \frac{{\sin(x + \frac{\pi }{3})}}{{\cos(x + \frac{\pi }{3})}} = \tan(x + \frac{\pi }{3})$.
Thus,$2y = {\left( {{{\cot }^{ - 1}}(\tan(x + \frac{\pi }{3}))} \right)^2}$.
Using ${{\cot }^{ - 1}}(\tan \theta ) = \frac{\pi }{2} - \theta$:
$2y = {\left( {\frac{\pi }{2} - (x + \frac{\pi }{3})} \right)^2} = {\left( {\frac{\pi }{6} - x} \right)^2}$.
Differentiating with respect to $x$:
$2\frac{{dy}}{{dx}} = 2(\frac{\pi }{6} - x) \cdot (-1)$.
$\frac{{dy}}{{dx}} = -(\frac{\pi }{6} - x) = x - \frac{\pi }{6}$.

(C) Let $L = \lim_{x \to 2} \frac{1}{x - 2} \int_{6}^{f(x)} 2t \, dt$.
Since $f(2) = 6$,the integral becomes $\int_{6}^{6} 2t \, dt = 0$,leading to a $\frac{0}{0}$ indeterminate form.
Applying $L'\text{H\^opital's Rule}$,we differentiate the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx} \int_{6}^{f(x)} 2t \, dt = 2f(x) \cdot f'(x)$ (by Leibniz Integral Rule).
Denominator: $\frac{d}{dx} (x - 2) = 1$.
Thus,$L = \lim_{x \to 2} \frac{2f(x)f'(x)}{1} = 2f(2)f'(2)$.
Given $f(2) = 6$,we get $L = 2(6)f'(2) = 12f'(2)$.

(A) Let $y = {\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$.
Dividing numerator and denominator by $\cos x$,we get:
$y = {\tan ^{ - 1}}\left( {\frac{{\tan x - 1}}{{\tan x + 1}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\tan x - \tan(\frac{\pi}{4})}}{{1 + \tan x \cdot \tan(\frac{\pi}{4})}}} \right)$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have:
$y = {\tan ^{ - 1}}\left( {\tan \left( {x - \frac{\pi }{4}} \right)} \right)$.
Since $x \in \left( {0, \frac{\pi }{2}} \right)$,then $x - \frac{\pi }{4} \in \left( {-\frac{\pi }{4}, \frac{\pi }{4}} \right)$,which is within the principal range of $\tan^{-1}$.
Thus,$y = x - \frac{\pi }{4}$.
Now,we need to find the derivative of $y$ with respect to $u = \frac{x}{2}$.
Since $u = \frac{x}{2}$,we have $x = 2u$.
Substituting $x$ in $y$,we get $y = 2u - \frac{\pi }{4}$.
Therefore,$\frac{dy}{du} = \frac{d}{du}(2u - \frac{\pi }{4}) = 2$.

(A) Let $f(x) = \cos ^{-1}(\sin x)$.
We know that $\sin x = \cos \left(\frac{\pi}{2} - x\right)$.
Substituting this into the function,we get:
$f(x) = \cos ^{-1}\left[\cos \left(\frac{\pi}{2} - x\right)\right]$.
Using the property $\cos ^{-1}(\cos \theta) = \theta$,we have:
$f(x) = \frac{\pi}{2} - x$.
Now,differentiating $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}\left(\frac{\pi}{2} - x\right) = 0 - 1 = -1$.
Thus,the derivative is $-1$.

(B) Let $f(x) = \tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$.
Using trigonometric identities $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$ and $1 + \cos x = 2 \cos ^{2} \left(\frac{x}{2}\right)$,we get:
$f(x) = \tan ^{-1}\left[\frac{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{2 \cos ^{2} \left(\frac{x}{2}\right)}\right]$
$f(x) = \tan ^{-1}\left[\tan \left(\frac{x}{2}\right)\right]$
$f(x) = \frac{x}{2}$
Now,differentiating with respect to $x$:
$f'(x) = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2}$.

(A) Let $f(x) = \sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$.
We can rewrite the expression inside the sine inverse as:
$f(x) = \sin ^{-1}\left(\frac{2 \cdot 2^{x}}{1+(2^{x})^{2}}\right)$.
Let $2^{x} = \tan \theta$,then $\theta = \tan ^{-1}(2^{x})$.
Substituting this into the function,we get:
$f(x) = \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) = \sin ^{-1}(\sin 2\theta) = 2\theta = 2 \tan ^{-1}(2^{x})$.
Now,differentiating with respect to $x$ using the chain rule:
$f'(x) = 2 \cdot \frac{1}{1+(2^{x})^{2}} \cdot \frac{d}{dx}(2^{x})$.
Since $\frac{d}{dx}(a^{x}) = a^{x} \log a$,we have $\frac{d}{dx}(2^{x}) = 2^{x} \log 2$.
Therefore,$f'(x) = 2 \cdot \frac{1}{1+4^{x}} \cdot 2^{x} \log 2 = \frac{2^{x+1} \log 2}{1+4^{x}}$.

(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$.
We know that $1+\sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.
Similarly,$1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Since $0 < x < \frac{\pi}{2}$,we have $0 < \frac{x}{2} < \frac{\pi}{4}$,which implies $\cos \frac{x}{2} > \sin \frac{x}{2}$.
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:
$y = \cot ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}\right]$
$y = \cot ^{-1}\left[\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}\right]$
$y = \cot ^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

(B) Let $f = \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$.
Put $x = \tan \theta$,so $\theta = \tan ^{-1} x$.
Then $f = \tan ^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2} \tan ^{-1} x$.
Thus,$\frac{df}{dx} = \frac{1}{2(1+x^{2})}$.
Let $g = \tan ^{-1}\left(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right)$.
Put $x = \sin \theta$,so $\theta = \sin ^{-1} x$.
Then $g = \tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin ^{2} \theta}\right) = \tan ^{-1}(\tan 2\theta) = 2\theta = 2 \sin ^{-1} x$.
Thus,$\frac{dg}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.
Therefore,$\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{1}{2(1+x^{2})} \cdot \frac{\sqrt{1-x^{2}}}{2} = \frac{\sqrt{1-x^{2}}}{4(1+x^{2})}$.
At $x = \frac{1}{2}$,$\frac{df}{dg} = \frac{\sqrt{1-(1/2)^{2}}}{4(1+(1/2)^{2})} = \frac{\sqrt{3/4}}{4(5/4)} = \frac{\sqrt{3}/2}{5} = \frac{\sqrt{3}}{10}$.

(A) Given $y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)$.
Since $1 \pm \sin x = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2}\right)^2$,we have $\sqrt{1+\sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$ and $\sqrt{1-\sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$.
For $x \in \left(\frac{\pi}{2}, \pi\right)$,$\frac{x}{2} \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. In this interval,$\cos \frac{x}{2} > 0$,$\sin \frac{x}{2} > 0$,and $\sin \frac{x}{2} > \cos \frac{x}{2}$.
Thus,$\sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1-\sin x} = \sin \frac{x}{2} - \cos \frac{x}{2}$.
Substituting these into the expression for $y(x)$:
$y(x) = \cot^{-1}\left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\sin \frac{x}{2} - \cos \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\sin \frac{x}{2} - \cos \frac{x}{2})}\right) = \cot^{-1}\left(\frac{2\sin \frac{x}{2}}{2\cos \frac{x}{2}}\right) = \cot^{-1}(\tan \frac{x}{2})$.
Using $\cot^{-1}(\tan \theta) = \frac{\pi}{2} - \theta$,we get $y(x) = \frac{\pi}{2} - \frac{x}{2}$.
Therefore,$\frac{dy}{dx} = -\frac{1}{2}$.

(B) Given $y = \sin^3\left(\frac{\pi}{3} \cos(g(x))\right)$ where $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{3/2}$.
At $x=1$,$g(1) = \frac{\pi}{3\sqrt{2}}(-4+5+1)^{3/2} = \frac{\pi}{3\sqrt{2}}(2)^{3/2} = \frac{\pi}{3\sqrt{2}}(2\sqrt{2}) = \frac{2\pi}{3}$.
$y(1) = \sin^3\left(\frac{\pi}{3} \cos\left(\frac{2\pi}{3}\right)\right) = \sin^3\left(\frac{\pi}{3} \cdot \left(-\frac{1}{2}\right)\right) = \sin^3\left(-\frac{\pi}{6}\right) = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$.
Now,$y' = 3\sin^2\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \cos\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \left(-\frac{\pi}{3}\sin(g(x))\right) \cdot g'(x)$.
$g'(x) = \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2}(-4x^3 + 5x^2 + 1)^{1/2} \cdot (-12x^2 + 10x) = \frac{\pi}{2\sqrt{2}} \sqrt{-4x^3 + 5x^2 + 1} (-12x^2 + 10x)$.
$g'(1) = \frac{\pi}{2\sqrt{2}} \cdot \sqrt{2} \cdot (-2) = -\pi$.
Substituting $x=1$ into $y'$:
$y'(1) = 3\sin^2(-\pi/6) \cdot \cos(-\pi/6) \cdot \left(-\frac{\pi}{3}\sin(2\pi/3)\right) \cdot (-\pi) = 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \left(-\frac{\pi}{3} \cdot \frac{\sqrt{3}}{2}\right) \cdot (-\pi) = \frac{3\pi^2}{16}$.
Checking option $B$: $2y'(1) + 3\pi^2 y(1) = 2\left(\frac{3\pi^2}{16}\right) + 3\pi^2\left(-\frac{1}{8}\right) = \frac{3\pi^2}{8} - \frac{3\pi^2}{8} = 0$.

(A) Let $f(x) = \int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt$. Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's Rule.
Using Leibniz Rule,$f'(x) = -(\sin(2x) + \cos(x)) \cdot 3x^2$.
The limit becomes $\lim _{x \rightarrow \frac{\pi}{2}} \frac{-3x^2(\sin(2x) + \cos(x))}{2(x - \frac{\pi}{2})}$.
As $x \rightarrow \frac{\pi}{2}$,let $h = x - \frac{\pi}{2}$,so $x = h + \frac{\pi}{2}$.
$= \lim _{h \rightarrow 0} \frac{-3(h + \frac{\pi}{2})^2(\sin(2h + \pi) + \cos(h + \frac{\pi}{2}))}{2h}$.
$= \lim _{h \rightarrow 0} \frac{-3(h + \frac{\pi}{2})^2(-\sin(2h) - \sin(h))}{2h}$.
$= \lim _{h \rightarrow 0} \frac{3(h + \frac{\pi}{2})^2(\sin(2h) + \sin(h))}{2h}$.
$= \frac{3(\frac{\pi}{2})^2}{2} \cdot \lim _{h}$ ${\rightarrow 0} (\frac{\sin(2h)}{h} + \frac{\sin(h)}{h}) = \frac{3\pi^2}{8} \cdot (2 + 1) = \frac{9\pi^2}{8}$.

(A) Given,$f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$.
We know that $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 1 - 2\sin^2 \theta$.
Let $\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) = \alpha$. Then $\tan \alpha = \frac{\sin \theta}{\sqrt{\cos 2\theta}}$.
Using the identity $1 + \tan^2 \alpha = \sec^2 \alpha$,we have $\sec^2 \alpha = 1 + \frac{\sin^2 \theta}{\cos 2\theta} = \frac{\cos 2\theta + \sin^2 \theta}{\cos 2\theta} = \frac{\cos^2 \theta - \sin^2 \theta + \sin^2 \theta}{\cos 2\theta} = \frac{\cos^2 \theta}{\cos 2\theta}$.
Thus,$\cos^2 \alpha = \frac{\cos 2\theta}{\cos^2 \theta}$,which implies $\sin^2 \alpha = 1 - \frac{\cos 2\theta}{\cos^2 \theta} = \frac{\cos^2 \theta - \cos 2\theta}{\cos^2 \theta} = \frac{\cos^2 \theta - (2\cos^2 \theta - 1)}{\cos^2 \theta} = \frac{1 - \cos^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$.
Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,$\tan \theta$ lies in $(-1, 1)$,so $\sin \alpha = \tan \theta$.
Therefore,$f(\theta) = \sin \alpha = \tan \theta$.
Now,$\frac{d}{d(\tan \theta)}(\tan \theta) = 1$.

(A) Given $y = \cot^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$.
Using the identities $1-\sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$ and $1+\sin x = (\cos\frac{x}{2} + \sin\frac{x}{2})^2$,we get:
$y = \cot^{-1}\left(\frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\right)$
Dividing numerator and denominator by $\cos\frac{x}{2}$:
$y = \cot^{-1}\left(\frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}}\right)$
Since $\cot^{-1}(z) = \tan^{-1}(\frac{1}{z})$,we have:
$y = \tan^{-1}\left(\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\right)$
Using $\tan(\frac{\pi}{4} + \theta) = \frac{1 + \tan\theta}{1 - \tan\theta}$:
$y = \tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right) = \frac{\pi}{4} + \frac{x}{2}$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + \frac{x}{2}) = 0 + \frac{1}{2} = \frac{1}{2}$.

(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$
Differentiating with respect to $x$,we get
$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\right)$
$= \frac{1}{2 \sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}} \cdot \frac{d}{dx}\left(\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}\right)$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{d}{dx}(1-\sin ^{-1} x) - (1-\sin ^{-1} x) \cdot \frac{d}{dx}(1+\sin ^{-1} x)}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{-1}{\sqrt{1-x^2}} - (1-\sin ^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \left[ \frac{-1-\sin ^{-1} x - 1 + \sin ^{-1} x}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \frac{-2}{(1+\sin ^{-1} x)^2}$
$= \frac{-\sqrt{1+\sin ^{-1} x}}{\sqrt{1-\sin ^{-1} x} \cdot \sqrt{1-x^2} \cdot (1+\sin ^{-1} x)^2}$
At $x=0$,$\sin^{-1}(0) = 0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{-\sqrt{1+0}}{\sqrt{1-0} \cdot \sqrt{1-0} \cdot (1+0)^2} = \frac{-1}{1 \cdot 1 \cdot 1} = -1$

(C) Given that $t = K^{\cos^{-1} x}$.
Substituting $t$ into the expression for $y$,we get $y = \frac{t}{1+t}$.
To find $\frac{dy}{dt}$,we differentiate $y$ with respect to $t$ using the quotient rule:
$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{t}{1+t} \right)$.
Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}$,where $u = t$ and $v = 1+t$:
$\frac{dy}{dt} = \frac{(1+t)(1) - t(1)}{(1+t)^2}$.
$\frac{dy}{dt} = \frac{1+t-t}{(1+t)^2} = \frac{1}{(1+t)^2}$.
Substituting $t = K^{\cos^{-1} x}$ back into the expression,we get:
$\frac{dy}{dt} = \frac{1}{(1 + K^{\cos^{-1} x})^2}$.

(A) Let $y = \sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ and $z = \sin ^{-1}\left(3 x-4 x^3\right)$.
Substitute $x = \sin \theta$,which implies $\theta = \sin ^{-1} x$.
Then,$y = \sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) = \sin ^{-1}(2 \sin \theta \cos \theta) = \sin ^{-1}(\sin 2 \theta) = 2 \theta = 2 \sin ^{-1} x$.
Similarly,$z = \sin ^{-1}\left(3 \sin \theta - 4 \sin ^3 \theta\right) = \sin ^{-1}(\sin 3 \theta) = 3 \theta = 3 \sin ^{-1} x$.
Now,differentiate $y$ and $z$ with respect to $x$:
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$ and $\frac{dz}{dx} = \frac{3}{\sqrt{1-x^2}}$.
Finally,the derivative of $y$ with respect to $z$ is given by $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/\sqrt{1-x^2}}{3/\sqrt{1-x^2}} = \frac{2}{3}$.

(A) Let $y = \tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ and $z = \cos ^{-1}(x^2)$.
Substitute $x^2 = \cos 2\theta$,which implies $\theta = \frac{1}{2} \cos ^{-1}(x^2)$.
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 the identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$,we get:
$y = \tan ^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right) = \tan ^{-1}\left(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right)$.
Dividing numerator and denominator by $\cos \theta$,we have:
$y = \tan ^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) = \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$.
Substituting $\theta = \frac{1}{2} \cos ^{-1}(x^2)$ back into the equation:
$y = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1}(x^2) = \frac{\pi}{4} - \frac{1}{2} z$.
Now,differentiating $y$ with respect to $z$:
$\frac{dy}{dz} = \frac{d}{dz}\left(\frac{\pi}{4} - \frac{1}{2} z\right) = -\frac{1}{2}$.

(B) Let $u = \tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ and $v = \cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right]$.
Substitute $x = \tan \theta$,then $\theta = \tan ^{-1} x$.
For $u$:
$u = \tan ^{-1}\left[\frac{\sec \theta - 1}{\tan \theta}\right] = \tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] = \tan ^{-1}\left[\frac{2 \sin ^2 (\theta/2)}{2 \sin (\theta/2) \cos (\theta/2)}\right] = \tan ^{-1}(\tan (\theta/2)) = \frac{\theta}{2} = \frac{\tan ^{-1} x}{2}$.
For $v$:
$v = \cos ^{-1}\left[\sqrt{\frac{1+\sec \theta}{2 \sec \theta}}\right] = \cos ^{-1}\left[\sqrt{\frac{\cos \theta + 1}{2}}\right] = \cos ^{-1}\left[\sqrt{\frac{2 \cos ^2 (\theta/2)}{2}}\right] = \cos ^{-1}(\cos (\theta/2)) = \frac{\theta}{2} = \frac{\tan ^{-1} x}{2}$.
Since $u = v$,we have $\frac{du}{dv} = \frac{d}{dv}(v) = 1$.

(A) Let $u = \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $v = \tan ^{-1} x$.
Substitute $x = \tan \theta$,where $\theta = \tan ^{-1} x$.
Since $-1 < x < 1$,we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,which implies $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.
Then $u = \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) = \sin ^{-1}(\sin 2 \theta) = 2 \theta = 2 \tan ^{-1} x$.
Now,we need to find $\frac{du}{dv} = \frac{d}{dv}(2 \tan ^{-1} x)$.
Since $v = \tan ^{-1} x$,we have $u = 2v$.
Therefore,$\frac{du}{dv} = \frac{d}{dv}(2v) = 2$.

(A) Let $y = \sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ and $u = \cos ^{-1} x$.
Substitute $x = \cos \theta$,where $\theta = \cos ^{-1} x$.
Then,$\frac{\sqrt{1+x}+\sqrt{1-x}}{2} = \frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{2} = \frac{\sqrt{2 \cos ^{2} \frac{\theta}{2}}+\sqrt{2 \sin ^{2} \frac{\theta}{2}}}{2} = \frac{\sqrt{2} \cos \frac{\theta}{2} + \sqrt{2} \sin \frac{\theta}{2}}{2} = \frac{1}{\sqrt{2}} \cos \frac{\theta}{2} + \frac{1}{\sqrt{2}} \sin \frac{\theta}{2}$.
Using $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get $\sin \frac{\pi}{4} \cos \frac{\theta}{2} + \cos \frac{\pi}{4} \sin \frac{\theta}{2} = \sin \left(\frac{\pi}{4} + \frac{\theta}{2}\right)$.
Thus,$y = \sin ^{-1} \left(\sin \left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right) = \frac{\pi}{4} + \frac{\theta}{2} = \frac{\pi}{4} + \frac{1}{2} u$.
Now,differentiate $y$ with respect to $u$: $\frac{dy}{du} = \frac{d}{du} \left(\frac{\pi}{4} + \frac{1}{2} u\right) = \frac{1}{2}$.

(D) Let $u = \tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and $v = \sin ^{-1}\left(3 x-4 x^3\right)$.
Substitute $x = \sin \theta$,where $\theta = \sin ^{-1} x$.
Then $u = \tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\right) = \tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan ^{-1}(\tan \theta) = \theta$.
And $v = \sin ^{-1}(3 \sin \theta - 4 \sin ^3 \theta) = \sin ^{-1}(\sin 3 \theta) = 3 \theta$.
We need to find $\frac{du}{dv}$.
Since $u = \theta$ and $v = 3 \theta$,we have $\frac{du}{d\theta} = 1$ and $\frac{dv}{d\theta} = 3$.
Therefore,$\frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{1}{3}$.

(C) Let $4x = \tan \theta$,then $\theta = \tan^{-1}(4x)$.
Substituting this into the expression for $y$:
$y = \tan^{-1}\left(\frac{3(4x) - (4x)^3}{1 - 3(4x)^2}\right)$
Using the identity $\tan(3\theta) = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$,we get:
$y = \tan^{-1}(\tan(3\theta)) = 3\theta$
Substituting $\theta = \tan^{-1}(4x)$ back:
$y = 3 \tan^{-1}(4x)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (4x)^2} \cdot \frac{d}{dx}(4x)$
$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + 16x^2} \cdot 4 = \frac{12}{1 + 16x^2}$

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}\right)$.
Let $3^x = \tan \theta$,then $\theta = \tan^{-1}(3^x)$.
Substituting this into the function,we get $f(x) = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(3^x)$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2 \cdot 3^x \log 3}{1+9^x}$.
Evaluating at $x = \frac{1}{2}$,we have $f^{\prime}\left(\frac{1}{2}\right) = \frac{2 \cdot 3^{1/2} \log 3}{1+9^{1/2}} = \frac{2 \sqrt{3} \log 3}{1+3} = \frac{2 \sqrt{3} \log 3}{4} = \frac{\sqrt{3} \log 3}{2} = \sqrt{3} \log(3^{1/2}) = \sqrt{3} \log(\sqrt{3})$.

(B) Given $y = \tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right)$.
We can rewrite the first term by dividing the numerator and denominator by $2$:
$y = \tan ^{-1}\left(\frac{\frac{3}{2}+x}{1-\frac{3}{2} x}\right)+\tan ^{-1}\left(\frac{4 x-x}{1+(4 x)(x)}\right)$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$ and $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$:
$y = \tan ^{-1}\left(\frac{3}{2}\right) + \tan ^{-1} x + \tan ^{-1} (4 x) - \tan ^{-1} x$
$y = \tan ^{-1}\left(\frac{3}{2}\right) + \tan ^{-1} (4 x)$
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 0 + \frac{1}{1+(4 x)^2} \cdot \frac{d}{d x}(4 x)$
$\frac{d y}{d x} = \frac{4}{1+16 x^2}$

(D) Let $y = \cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)$.
Simplifying the expression inside the inverse cosine function:
$y = \cos ^{-1}\left(\frac{x^2-1}{x^2+1}\right) = \cos ^{-1}\left(-\left(\frac{1-x^2}{1+x^2}\right)\right)$.
Using the property $\cos ^{-1}(-z) = \pi - \cos ^{-1}(z)$,we get:
$y = \pi - \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan ^{-1} x$:
$y = \pi - \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)$.
Since $\cos 2\theta = \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$,we have:
$y = \pi - \cos ^{-1}(\cos 2\theta) = \pi - 2\theta = \pi - 2\tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\pi - 2\tan ^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}\right) = \frac{-2}{1+x^2}$.

(B) Given $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$.
We can rewrite the expression as $y = \sin^{-1}\left(3\left(\frac{x}{2}\right) - 4\left(\frac{x}{2}\right)^3\right)$.
Let $\frac{x}{2} = \sin \theta$,which implies $\theta = \sin^{-1}\left(\frac{x}{2}\right)$.
Substituting this into the equation,we get $y = \sin^{-1}(3\sin \theta - 4\sin^3 \theta)$.
Using the trigonometric identity $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$,we have $y = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting $\theta$ back,$y = 3\sin^{-1}\left(\frac{x}{2}\right)$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1 - (x/2)^2}} \cdot \frac{1}{2}$.
$\frac{dy}{dx} = \frac{3}{2 \cdot \sqrt{\frac{4-x^2}{4}}} = \frac{3}{2 \cdot \frac{\sqrt{4-x^2}}{2}} = \frac{3}{\sqrt{4-x^2}}$.

(D) Let $y = \tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.
For $y$,substitute $x = \cos 2\theta$. Then $\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} = \tan \theta$.
So,$y = \tan^{-1}(\tan \theta) = \theta = \frac{1}{2} \cos^{-1} x$.
For $z$,substitute $x = \cos \theta$. Then $z = \cos^{-1}(\cos 3\theta) = 3\theta = 3 \cos^{-1} x$.
Now,differentiate $y$ and $z$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) = \frac{-1}{2\sqrt{1-x^2}}$.
$\frac{dz}{dx} = 3 \left( \frac{-1}{\sqrt{1-x^2}} \right) = \frac{-3}{\sqrt{1-x^2}}$.
Finally,$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-1/(2\sqrt{1-x^2})}{-3/\sqrt{1-x^2}} = \frac{1}{6}$.

(A) Given $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$.
Let $\theta=\cot ^{-1} \sqrt{\frac{1+x}{1-x}}$,then $\cot \theta = \sqrt{\frac{1+x}{1-x}}$.
Squaring both sides,we get $\cot^2 \theta = \frac{1+x}{1-x}$.
Using the identity $1+\cot^2 \theta = \csc^2 \theta$,we have $\csc^2 \theta = 1 + \frac{1+x}{1-x} = \frac{1-x+1+x}{1-x} = \frac{2}{1-x}$.
Since $\sin^2 \theta = \frac{1}{\csc^2 \theta}$,we get $\sin^2 \theta = \frac{1-x}{2}$.
Substituting this back into the expression for $y$,we get $y = \sin^2 \theta = \frac{1-x}{2}$.
Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{2} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(A) Given $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$.
Let $\log x = \tan \theta$,then $y = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta$.
Substituting back,$y = 2 \tan^{-1}(\log x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \times \frac{1}{1+(\log x)^2} \times \frac{1}{x}$.
At $x = 1$,$\log 1 = 0$,so $\left(\frac{dy}{dx}\right)_{x=1} = 2 \times \frac{1}{1+0^2} \times \frac{1}{1} = 2 \times 1 \times 1 = 2$.

(B) Given $y = \sec(\tan^{-1} x)$.
Let $\tan^{-1} x = \theta$,then $\tan \theta = x$.
We know that $\sec^2 \theta = 1 + \tan^2 \theta = 1 + x^2$,so $\sec \theta = \sqrt{1 + x^2}$.
Thus,$y = \sqrt{1 + x^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$.
At $x = 1$:
$\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

(B) Given $y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)$
We can rewrite the first term as $\tan ^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3} x}\right) = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(x)$
We can rewrite the second term as $\tan ^{-1}\left(\frac{5 x-x}{1+(5 x)(x)}\right) = \tan ^{-1}(5 x) - \tan ^{-1}(x)$
Adding these together,we get $y = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(x) + \tan ^{-1}(5 x) - \tan ^{-1}(x)$
Simplifying,$y = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(5 x)$
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 0 + \frac{1}{1+(5 x)^2} \cdot \frac{d}{d x}(5 x)$
$\frac{d y}{d x} = \frac{5}{1+25 x^2}$

(A) Given $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$.
Let $x^2=a^2 \tan \theta$,then $\theta=\tan ^{-1}\left(\frac{x^2}{a^2}\right)$.
Substituting $x^2$ in the expression for $y$:
$y=\cos ^{-1}\left(\frac{a^2}{\sqrt{(a^2 \tan \theta)^2+a^4}}\right) = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^4 \tan ^2 \theta+a^4}}\right)$
$y=\cos ^{-1}\left(\frac{a^2}{a^2 \sqrt{\tan ^2 \theta+1}}\right) = \cos ^{-1}\left(\frac{1}{\sec \theta}\right)$
$y=\cos ^{-1}(\cos \theta) = \theta$
Thus,$y=\tan ^{-1}\left(\frac{x^2}{a^2}\right)$.
Differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x} = \frac{1}{1+\left(\frac{x^2}{a^2}\right)^2} \cdot \frac{d}{d x}\left(\frac{x^2}{a^2}\right)$
$\frac{d y}{d x} = \frac{1}{1+\frac{x^4}{a^4}} \cdot \frac{2x}{a^2} = \frac{a^4}{a^4+x^4} \cdot \frac{2x}{a^2}$
$\frac{d y}{d x} = \frac{2 a^2 x}{x^4+a^4}$.

(A) Given $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$.
Using properties of logarithms,$\log(\frac{e}{x^2}) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.
So,$y = \tan^{-1}\left(\frac{1-2\log x}{1+2\log x}\right) + \tan^{-1}\left(\frac{4+2\log x}{1-8\log x}\right)$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we have $\tan^{-1}(1) - \tan^{-1}(2\log x) = \tan^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)$.
Also,$\tan^{-1}(4) + \tan^{-1}(2\log x) = \tan^{-1}\left(\frac{4+2\log x}{1-4(2\log x)}\right) = \tan^{-1}\left(\frac{4+2\log x}{1-8\log x}\right)$.
Substituting these back,$y = \tan^{-1}(1) - \tan^{-1}(2\log x) + \tan^{-1}(4) + \tan^{-1}(2\log x)$.
$y = \tan^{-1}(1) + \tan^{-1}(4)$.
Since $y$ is a constant,$\frac{dy}{dx} = 0$.

(D) Let $u = \log x$. Then $f(x) = \sin^{-1}\left(\frac{2u}{1+u^2}\right)$.
Using the substitution $u = \tan \theta$,we have $\frac{2u}{1+u^2} = \sin(2\theta)$.
Thus,$f(x) = \sin^{-1}(\sin(2\theta)) = 2\theta = 2 \tan^{-1}(u) = 2 \tan^{-1}(\log x)$.
Now,differentiate with respect to $x$:
$f^{\prime}(x) = 2 \times \frac{1}{1+(\log x)^2} \times \frac{d}{dx}(\log x)$
$f^{\prime}(x) = \frac{2}{1+(\log x)^2} \times \frac{1}{x} = \frac{2}{x(1+(\log x)^2)}$.
Substitute $x = e$:
$f^{\prime}(e) = \frac{2}{e(1+(\log e)^2)} = \frac{2}{e(1+1^2)} = \frac{2}{e(2)} = \frac{1}{e}$.

(C) Given $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$.
Using the identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$:
$y = \tan^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - \sin^2 x - 6 \sin^2 x} \right) = \tan^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - 7 \sin^2 x} \right)$.
Divide numerator and denominator by $\cos^2 x$:
$y = \tan^{-1} \left( \frac{8 \tan x}{1 - 7 \tan^2 x} \right)$.
We can write $8 \tan x = 7 \tan x + \tan x$:
$y = \tan^{-1} \left( \frac{7 \tan x + \tan x}{1 - (7 \tan x)(\tan x)} \right)$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$:
$y = \tan^{-1}(7 \tan x) + \tan^{-1}(\tan x) = \tan^{-1}(7 \tan x) + x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1 + (7 \tan x)^2} \cdot 7 \sec^2 x + 1 = \frac{7 \sec^2 x}{1 + 49 \tan^2 x} + 1$.
At $x=0$,$\tan 0 = 0$ and $\sec 0 = 1$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{7(1)^2}{1 + 49(0)^2} + 1 = \frac{7}{1} + 1 = 8$.

(B) Let $y = \sin ^{-1}\left(3 x-4 x^3\right)$.
We know that $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$.
Let $x = \sin\theta$,then $\theta = \sin^{-1}x$.
Since $\frac{1}{2} < x < 1$,we have $\frac{\pi}{6} < \theta < \frac{\pi}{2}$.
Then $3\theta$ lies in the range $(\frac{\pi}{2}, \frac{3\pi}{2})$.
In this range,$\sin^{-1}(\sin(3\theta)) = \pi - 3\theta$.
Thus,$y = \pi - 3\sin^{-1}x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\pi - 3\sin^{-1}x) = 0 - 3 \times \frac{1}{\sqrt{1-x^2}} = \frac{-3}{\sqrt{1-x^2}}$.

(B) Let $x = \cos \theta$,then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ in the expression:
$y = \sin \left(2 \tan^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)$
Using the identities $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $1-\cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$y = \sin \left(2 \tan^{-1} \sqrt{\frac{2 \cos^2 \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}}\right) = \sin \left(2 \tan^{-1} \left(\cot \frac{\theta}{2}\right)\right)$
Since $\cot \frac{\theta}{2} = \tan \left(\frac{\pi}{2} - \frac{\theta}{2}\right)$:
$y = \sin \left(2 \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = \sin (\pi - \theta) = \sin \theta$
Since $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$:
$y = \sqrt{1 - x^2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2 \sqrt{1 - x^2}} \times \frac{d}{dx}(1 - x^2) = \frac{1}{2 \sqrt{1 - x^2}} \times (-2x) = \frac{-x}{\sqrt{1 - x^2}}$

(B) Given $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$.
Simplify the expression inside the $\sec^{-1}$ function:
$\frac{x+x^{-1}}{x-x^{-1}} = \frac{x+\frac{1}{x}}{x-\frac{1}{x}} = \frac{x^2+1}{x^2-1} = -\frac{1+x^2}{1-x^2}$.
Let $x = \tan \theta$,then $\theta = \tan^{-1} x$.
The expression becomes $-\frac{1+\tan^2 \theta}{1-\tan^2 \theta} = -\frac{1}{\cos 2\theta} = -\sec 2\theta$.
So,$y = \sec^{-1}(-\sec 2\theta)$.
Using the identity $\sec^{-1}(-z) = \pi - \sec^{-1}(z)$,we get $y = \pi - \sec^{-1}(\sec 2\theta) = \pi - 2\theta$.
Substituting back $\theta = \tan^{-1} x$,we have $y = \pi - 2\tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\pi) - 2\frac{d}{dx}(\tan^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}\right) = -\frac{2}{1+x^2}$.

(B) Given $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$.
We can rewrite the terms inside the inverse tangent functions:
$y = \tan^{-1}\left(\frac{5x-x}{1+5x \cdot x}\right) + \tan^{-1}\left(\frac{\frac{3}{8}+x}{1-\frac{3}{8} \cdot x}\right)$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$y = (\tan^{-1}(5x) - \tan^{-1}(x)) + (\tan^{-1}(\frac{3}{8}) + \tan^{-1}(x))$.
Simplifying the expression,we get:
$y = \tan^{-1}(5x) + \tan^{-1}(\frac{3}{8})$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(5x)) + \frac{d}{dx}(\tan^{-1}(\frac{3}{8}))$.
Since $\tan^{-1}(\frac{3}{8})$ is a constant,its derivative is $0$:
$\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot \frac{d}{dx}(5x) + 0$.
$\frac{dy}{dx} = \frac{5}{1+25x^2}$.

(A) Given $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$.
We can rewrite the expression inside the inverse tangent function as:
$y=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)$.
Using the identity $\tan ^{-1}(A)+\tan ^{-1}(B)=\tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x}=\frac{d}{d x}(\tan ^{-1}(3 x+2))+\frac{d}{d x}(\tan ^{-1}(2 x-1))$.
$\frac{d y}{d x}=\frac{1}{1+(3 x+2)^2} \times 3+\frac{1}{1+(2 x-1)^2} \times 2$.
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+4+1}+\frac{2}{4 x^2-4 x+1+1}$.
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{2}{4 x^2-4 x+2}$.
Simplifying the second term:
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}$.

(A) Let $y = \tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)$.
We can rewrite the expression inside the inverse tangent as $y = \tan ^{-1}\left(\frac{2(3 x^{3/2})}{1-(3 x^{3/2})^2}\right)$.
Using the identity $2 \tan ^{-1}(\theta) = \tan ^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get $y = 2 \tan ^{-1}(3 x^{3/2})$.
Now,differentiate $y$ with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3 x^{3/2})^2} \cdot \frac{d}{dx}(3 x^{3/2})$.
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot \frac{9}{2} \sqrt{x}$.
$\frac{dy}{dx} = \frac{9}{1+9 x^3} \cdot \sqrt{x}$.
Comparing this with $\sqrt{x} \cdot g(x)$,we find $g(x) = \frac{9}{1+9 x^3}$.

(B) We are given $y=\tan ^{-1}\left[\frac{1}{1+x(1+x)}\right]+\tan ^{-1}\left[\frac{1}{1+(x+2)(x+1)}\right]$.
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$,we can rewrite the terms:
$y = \tan ^{-1}\left(\frac{(x+1)-x}{1+(x+1)x}\right) + \tan ^{-1}\left(\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}\right)$.
This simplifies to:
$y = (\tan ^{-1}(x+1) - \tan ^{-1} x) + (\tan ^{-1}(x+2) - \tan ^{-1}(x+1))$.
Canceling the common terms,we get:
$y = \tan ^{-1}(x+2) - \tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan ^{-1}(x+2)) - \frac{d}{dx}(\tan ^{-1} x)$.
$\frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}$.
Thus,the correct option is $B$.

(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 $\frac{a}{b}=\tan \alpha$,then:
$y=\tan ^{-1}\left\{\frac{\tan \alpha-\tan x}{1+\tan \alpha \tan x}\right\}$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$y=\tan ^{-1}[\tan(\alpha-x)] = \alpha-x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\alpha-x) = 0-1 = -1$.

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

(D) Given $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$.
Using the trigonometric identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $1-\cos x = 2 \sin^2 \frac{x}{2}$,we get:
$y=\tan ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}$
$y=\tan ^{-1} \sqrt{\cot^2 \frac{x}{2}} = \tan ^{-1} \left(\cot \frac{x}{2}\right)$
Since $\cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$,we have:
$y=\tan ^{-1} \left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right] = \frac{\pi}{2}-\frac{x}{2}$
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x} \left(\frac{\pi}{2} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2}$