Derivative at a point, Standard differentiation Questions in English

(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.
Multiply and divide by $(x-1)$:
$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{x^{16}-1}{x-1}\right) = \frac{(16x^{15})(x-1) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.
Comparing this with the given expression $\frac{15x^p - 16x^q + 1}{(x-1)^2}$,we get $p = 16$ and $q = 15$.
Thus,$(p, q) = (16, 15)$.

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.
Multiply and divide by $(1-x)$:
$y = \frac{(1-x)(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x}$
Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:
$y = \frac{(1-x^2)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x} = \frac{(1-x^4)(1+x^4) \dots (1+x^{2^n})}{1-x}$
Continuing this process,we get:
$y = \frac{1-x^{2^{n+1}}}{1-x}$
Now,differentiate using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(1-x)(-2^{n+1}x^{2^{n+1}-1}) - (1-x^{2^{n+1}})(-1)}{(1-x)^2}$
Substitute $x=0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{(1-0)(0) - (1-0)(-1)}{(1-0)^2} = \frac{0 + 1}{1} = 1$.

(A) Given function is $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$.
We can rewrite the function as $f(x) = \sqrt{a} \cdot x^{1/2} + a^2 \cdot \sqrt{a}^{-1} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2}$.
Differentiating with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f^{\prime}(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^{3/2} \cdot \left(-\frac{1}{2}\right) x^{-3/2}$.
$f^{\prime}(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x\sqrt{x}}$.
Now,substitute $x = a$ into the derivative:
$f^{\prime}(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a\sqrt{a}}$.
$f^{\prime}(a) = \frac{1}{2} - \frac{a^{3/2}}{2a^{3/2}}$.
$f^{\prime}(a) = \frac{1}{2} - \frac{1}{2} = 0$.

(C) Given $y = x \sin x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = x \cos x + \sin x$.
Now,consider the expression $\frac{\frac{dy}{dx} - \frac{y}{x}}{x \frac{dy}{dx} - y}$.
Substituting $\frac{dy}{dx} = x \cos x + \sin x$ and $y = x \sin x$:
Numerator: $\frac{dy}{dx} - \frac{y}{x} = (x \cos x + \sin x) - \frac{x \sin x}{x} = x \cos x + \sin x - \sin x = x \cos x$.
Denominator: $x \frac{dy}{dx} - y = x(x \cos x + \sin x) - x \sin x = x^2 \cos x + x \sin x - x \sin x = x^2 \cos x$.
Thus,the expression becomes $\frac{x \cos x}{x^2 \cos x} = \frac{1}{x}$.
Given that at $x = \alpha$,the expression equals $1$,we have $\frac{1}{\alpha} = 1$,which implies $\alpha = 1$.

(A) Given $f(x) = \sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$.
Applying the chain rule,we differentiate with respect to $x$:
$f^{\prime}(x) = \cos \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right) \cdot \sinh \left(\frac{x^2+1}{x^2+2}\right) \cdot \frac{d}{dx} \left(\frac{x^2+1}{x^2+2}\right)$.
Using the quotient rule for the inner derivative:
$\frac{d}{dx} \left(\frac{x^2+1}{x^2+2}\right) = \frac{(x^2+2)(2x) - (x^2+1)(2x)}{(x^2+2)^2} = \frac{2x^3 + 4x - 2x^3 - 2x}{(x^2+2)^2} = \frac{2x}{(x^2+2)^2}$.
Thus,$f^{\prime}(x) = \cos \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right) \cdot \sinh \left(\frac{x^2+1}{x^2+2}\right) \cdot \frac{2x}{(x^2+2)^2}$.
Evaluating at $x = 1$:
$f^{\prime}(1) = \cos \left(\cosh \left(\frac{1^2+1}{1^2+2}\right)\right) \cdot \sinh \left(\frac{1^2+1}{1^2+2}\right) \cdot \frac{2(1)}{(1^2+2)^2} = \frac{2}{9} \sinh \left(\frac{2}{3}\right) \cos \left(\cosh \left(\frac{2}{3}\right)\right)$.

(A) Let $u = x^{\frac{5}{2}}-x^{\frac{3}{2}}+1$ and $v = x^2-3 x+5$. Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$:
$u' = \frac{5}{2} x^{\frac{3}{2}}-\frac{3}{2} x^{\frac{1}{2}}$
$v' = 2x-3$
$\frac{d}{dx}(uv) = (\frac{5}{2} x^{\frac{3}{2}}-\frac{3}{2} x^{\frac{1}{2}})(x^2-3 x+5) + (x^{\frac{5}{2}}-x^{\frac{3}{2}}+1)(2 x-3)$
$= (\frac{5}{2} x^{\frac{7}{2}} - \frac{15}{2} x^{\frac{5}{2}} + \frac{25}{2} x^{\frac{3}{2}} - \frac{3}{2} x^{\frac{5}{2}} + \frac{9}{2} x^{\frac{3}{2}} - \frac{15}{2} x^{\frac{1}{2}}) + (2 x^{\frac{7}{2}} - 3 x^{\frac{5}{2}} - 2 x^{\frac{5}{2}} + 3 x^{\frac{3}{2}} + 2 x - 3)$
$= (\frac{5}{2} + 2) x^{\frac{7}{2}} + (-\frac{15}{2} - \frac{3}{2} - 3 - 2) x^{\frac{5}{2}} + (\frac{25}{2} + \frac{9}{2} + 3) x^{\frac{3}{2}} - \frac{15}{2} x^{\frac{1}{2}} + 2 x - 3$
$= \frac{9}{2} x^{\frac{7}{2}} - 14 x^{\frac{5}{2}} + 20 x^{\frac{3}{2}} - \frac{15}{2} x^{\frac{1}{2}} + 2 x - 3$

(C) Let $y = \log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot \cos \sqrt{\frac{x^2+1}{x^2+2}} \cdot \frac{1}{2\sqrt{\frac{x^2+1}{x^2+2}}} \cdot \frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right)$.
Calculating the derivative of the inner quotient: $\frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right) = \frac{2x(x^2+2) - 2x(x^2+1)}{(x^2+2)^2} = \frac{2x^3+4x-2x^3-2x}{(x^2+2)^2} = \frac{2x}{(x^2+2)^2}$.
Substituting $x = \sqrt{2}$,we have $\frac{x^2+1}{x^2+2} = \frac{2+1}{2+2} = \frac{3}{4}$.
Thus,$\frac{dy}{dx} = \cot \left( \sqrt{\frac{3}{4}} \right) \cdot \frac{1}{2 \sqrt{3/4}} \cdot \frac{2\sqrt{2}}{(2+2)^2} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{2 \cdot \frac{\sqrt{3}}{2}} \cdot \frac{2\sqrt{2}}{16} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{8} = \frac{\sqrt{2} \cot \left( \frac{\sqrt{3}}{2} \right)}{8 \sqrt{3}}$.

(B) Given,$x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.
Applying componendo and dividendo,we get:
$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})}$
$\frac{1+x}{1-x} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$
Squaring both sides,we get $\sqrt{y} = \frac{1-x}{1+x}$,so $y = \left(\frac{1-x}{1+x}\right)^2$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v-uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$
Therefore,$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(1+x)^3}$.

(C) Given $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right]$.
Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$,$v = \frac{2-1}{2+1} = \frac{1}{3}$.
Using the quotient rule,$\frac{dv}{dx} = \frac{(2^x+1)(2^x \log 2) - (2^x-1)(2^x \log 2)}{(2^x+1)^2} = \frac{2^x \log 2 (2^x+1-2^x+1)}{(2^x+1)^2} = \frac{2 \cdot 2^x \log 2}{(2^x+1)^2} = \frac{2^{x+1} \log 2}{(2^x+1)^2}$.
At $x=1$,$\left(\frac{dv}{dx}\right)_{x=1} = \frac{2^2 \log 2}{(2+1)^2} = \frac{4 \log 2}{9}$.
Now,$y = \log(\tan \sqrt{v})$. Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan \sqrt{v}} \cdot \sec^2 \sqrt{v} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}$.
Using $\frac{1}{\tan \sqrt{v}} \cdot \sec^2 \sqrt{v} = \frac{\cos \sqrt{v}}{\sin \sqrt{v}} \cdot \frac{1}{\cos^2 \sqrt{v}} = \frac{1}{\sin \sqrt{v} \cos \sqrt{v}} = \frac{2}{\sin(2\sqrt{v})}$.
So,$\frac{dy}{dx} = \frac{2}{\sin(2\sqrt{v})} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx} = \frac{1}{\sin(2\sqrt{v}) \cdot \sqrt{v}} \cdot \frac{dv}{dx}$.
At $x=1$,$v = \frac{1}{3}$,so $\sqrt{v} = \frac{1}{\sqrt{3}}$.
$\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sin(2/\sqrt{3}) \cdot (1/\sqrt{3})} \cdot \frac{4 \log 2}{9} = \frac{\sqrt{3} \cdot 4 \log 2}{9 \sin(2/\sqrt{3})} = \frac{4 \sqrt{3} \log 2}{9 \sin(2/\sqrt{3})}$.

(D) Given $f(x) = \sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}$.
Applying the chain rule,$f'(x) = \frac{1}{2\sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}} \cdot \left( \frac{2x+1}{x^2+x+1} + \frac{2\sinh(2x-3)}{2\sqrt{\cosh(2x-3)}} \right)$.
At $x=0$,$f'(0) = \frac{1}{2\sqrt{\log(1) + \sqrt{\cosh(-3)}}} \cdot \left( \frac{1}{1} + \frac{\sinh(-3)}{\sqrt{\cosh(-3)}} \right)$.
Since $\log(1) = 0$,$\cosh(-3) = \cosh(3)$,and $\sinh(-3) = -\sinh(3)$:
$f'(0) = \frac{1 - \frac{\sinh(3)}{\sqrt{\cosh(3)}}}{2\sqrt{\sqrt{\cosh(3)}}} = \frac{\sqrt{\cosh(3)} - \sinh(3)}{2(\cosh(3))^{1/2} \cdot (\cosh(3))^{1/4}} = \frac{\sqrt{\cosh(3)} - \sinh(3)}{2(\cosh(3))^{3/4}}$.

(A) Note: Assuming the base of the logarithm is $e$.
Given:
Assertion: For $x < 0$,$\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$,$|x| = -x$.
Let $f(x) = \log |x|$.
For $x < 0$,we have $|x| = -x$,so $f(x) = \log(-x)$.
Now,differentiating with respect to $x$:
$\frac{d}{d x}(\log(-x)) = \frac{1}{-x} \cdot \frac{d}{d x}(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.
Now,differentiating again with respect to $x$:
$\frac{d^2}{d x^2}(\log |x|) = \frac{d}{d x}(\frac{1}{x}) = -\frac{1}{x^2}$.
Since $|x|^2 = (-x)^2 = x^2$,the Assertion states $\frac{d^2}{d x^2}(\log |x|) = \frac{1}{x^2}$,but we found $-\frac{1}{x^2}$.
Therefore,the Assertion is false and the Reason is true.

(A) Given the function $f(x) = \log \left[e^x \left(\frac{x-2}{x+2}\right)^{3/4}\right]$.
Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$,we can simplify the expression:
$f(x) = \log(e^x) + \log \left(\left(\frac{x-2}{x+2}\right)^{3/4}\right)$
$f(x) = x + \frac{3}{4} [\log(x-2) - \log(x+2)]$.
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx}(\log(x-2)) - \frac{d}{dx}(\log(x+2)) \right]$
$f'(x) = 1 + \frac{3}{4} \left( \frac{1}{x-2} - \frac{1}{x+2} \right)$.
Simplify the term inside the parenthesis:
$\frac{1}{x-2} - \frac{1}{x+2} = \frac{(x+2) - (x-2)}{(x-2)(x+2)} = \frac{4}{x^2-4}$.
Substituting this back into the derivative:
$f'(x) = 1 + \frac{3}{4} \left( \frac{4}{x^2-4} \right) = 1 + \frac{3}{x^2-4}$.
Finally,evaluate at $x = 0$:
$f'(0) = 1 + \frac{3}{0^2-4} = 1 - \frac{3}{4} = \frac{1}{4}$.

(C) Given that $g(x)$ is the anti-derivative of $f(x)$,we have $g^{\prime}(x) = f(x)$.
We need to find the function $h(x)$ such that $\int h(x) \, dx = \log _e(1+(g(x))^2) + c$.
By the definition of an anti-derivative,$h(x) = \frac{d}{dx} [\log _e(1+(g(x))^2) + c]$.
Using the chain rule,$\frac{d}{dx} [\log _e(1+(g(x))^2)] = \frac{1}{1+(g(x))^2} \cdot \frac{d}{dx} (1+(g(x))^2)$.
$= \frac{1}{1+(g(x))^2} \cdot (2 g(x) \cdot g^{\prime}(x))$.
Since $g^{\prime}(x) = f(x)$,we substitute this into the expression:
$h(x) = \frac{2 g(x) f(x)}{1+(g(x))^2}$.
Thus,the correct option is $C$.

(D) Given $y = \frac{e^{\sin x} + \sinh^3 x}{\cosh x - \tan x}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$,where $u = e^{\sin x} + \sinh^3 x$ and $v = \cosh x - \tan x$.
First,find the derivatives at $x = 0$:
$u(0) = e^{\sin 0} + \sinh^3 0 = e^0 + 0 = 1$.
$u'(x) = e^{\sin x} \cos x + 3 \sinh^2 x \cosh x$.
$u'(0) = e^0 \cos 0 + 3 \sinh^2 0 \cosh 0 = 1 \cdot 1 + 0 = 1$.
$v(0) = \cosh 0 - \tan 0 = 1 - 0 = 1$.
$v'(x) = \sinh x - \sec^2 x$.
$v'(0) = \sinh 0 - \sec^2 0 = 0 - 1 = -1$.
Now,substitute these values into the quotient rule formula at $x = 0$:
$y'(0) = \frac{u'(0)v(0) - u(0)v'(0)}{(v(0))^2} = \frac{(1)(1) - (1)(-1)}{(1)^2} = \frac{1 + 1}{1} = 2$.
Thus,the correct option is $D$.

(A) Let $u = \cosh^{-1} x$. The derivative is $\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$.
Let $v = \log x$. The derivative is $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/\sqrt{x^2-1}}{1/x} = \frac{x}{\sqrt{x^2-1}}$.
At $x=5$,$\frac{du}{dv} = \frac{5}{\sqrt{5^2-1}} = \frac{5}{\sqrt{25-1}} = \frac{5}{\sqrt{24}}$.

(A) Given $e^x = y + \sqrt{y^2 - 1}$.
Rearranging the terms,we get $e^x - y = \sqrt{y^2 - 1}$.
Squaring both sides,we have $(e^x - y)^2 = y^2 - 1$.
Expanding the square: $e^{2x} + y^2 - 2ye^x = y^2 - 1$.
Simplifying,we get $e^{2x} + 1 = 2ye^x$,which implies $y = \frac{e^{2x} + 1}{2e^x} = \frac{e^x + e^{-x}}{2} = \cosh x$.
Now,differentiating $y = \cosh x$ with respect to $x$,we get $\frac{dy}{dx} = \sinh x$.

(A) Given $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(\tanh x)) + \frac{d}{dx}(\sinh(\sin 6x))$
$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - \tanh^2 x}} \cdot \frac{d}{dx}(\tanh x) + \cosh(\sin 6x) \cdot \frac{d}{dx}(\sin 6x)$
Using the identities $1 - \tanh^2 x = \text{sech}^2 x$ and $\frac{d}{dx}(\tanh x) = \text{sech}^2 x$:
$\frac{dy}{dx} = \frac{-1}{\sqrt{\text{sech}^2 x}} \cdot \text{sech}^2 x + \cosh(\sin 6x) \cdot (6 \cos 6x)$
$\frac{dy}{dx} = \frac{-1}{\text{sech} x} \cdot \text{sech}^2 x + 6 \cos 6x \cosh(\sin 6x)$
$\frac{dy}{dx} = -\text{sech} x \cdot \text{sech} x \cdot \cosh x + 6 \cos 6x \cosh(\sin 6x)$ (Note: $\text{sech} x = \frac{1}{\cosh x}$)
$\frac{dy}{dx} = \frac{-1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x)$.
Thus,the correct option is $A$.

(D) Given $y = \tan(\cos^{-1} x)$.
Let $\cos^{-1} x = \theta$,then $\cos \theta = x$.
We know that $\tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x}$.
So,$y = \frac{\sqrt{1 - x^2}}{x}$.
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1 - x^2}) - \sqrt{1 - x^2} \cdot \frac{d}{dx}(x)}{x^2}$
$\frac{dy}{dx} = \frac{x \cdot \frac{-2x}{2\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$
$\frac{dy}{dx} = \frac{\frac{-x^2}{\sqrt{1 - x^2}} - \sqrt{1 - x^2}}{x^2}$
$\frac{dy}{dx} = \frac{-x^2 - (1 - x^2)}{x^2 \sqrt{1 - x^2}} = \frac{-1}{x^2 \sqrt{1 - x^2}}$.

(B) To find the derivative of $\tan t + t^2 \operatorname{cosech} t$ with respect to $t$,we use the sum rule and the product rule.
The derivative of $\tan t$ is $\sec^2 t$.
Using the product rule for $t^2 \operatorname{cosech} t$,we have:
$\frac{d}{dt}(t^2 \operatorname{cosech} t) = \frac{d}{dt}(t^2) \cdot \operatorname{cosech} t + t^2 \cdot \frac{d}{dt}(\operatorname{cosech} t)$.
Since $\frac{d}{dt}(t^2) = 2t$ and $\frac{d}{dt}(\operatorname{cosech} t) = -\operatorname{cosech} t \operatorname{coth} t$,we get:
$2t \operatorname{cosech} t + t^2(-\operatorname{cosech} t \operatorname{coth} t) = 2t \operatorname{cosech} t - t^2 \operatorname{cosech} t \operatorname{coth} t$.
Combining these,the final result is $\sec^2 t + 2t \operatorname{cosech} t - t^2 \operatorname{cosech} t \operatorname{coth} t$.

(D) The derivatives of the given inverse functions are as follows:
$(A)$ $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$ for $|x| > 1$. This matches $III$.
$(B)$ $\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1-x^2}$ for $x \in (-1, 1)$. This matches $I$.
$(C)$ $\frac{d}{dx}(\coth^{-1} x) = \frac{1}{1-x^2}$ for $x \in R - [-1, 1]$. This matches $IV$.
$(D)$ $\frac{d}{dx}(\operatorname{cosech}^{-1} x) = \frac{-1}{|x|\sqrt{x^2+1}}$ for $x \neq 0$. This matches $II$.
Thus,the correct matching is $A-III, B-I, C-IV, D-II$.

(A) Given $f(x) = \cosh^{-1}\left(\frac{1-x}{1+x}\right)$.
Using the derivative formula $\frac{d}{dx}(\cosh^{-1}(u)) = \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx}$:
$f'(x) = \frac{1}{\sqrt{\left(\frac{1-x}{1+x}\right)^2 - 1}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$
$f'(x) = \frac{1}{\sqrt{\frac{(1-x)^2 - (1+x)^2}{(1+x)^2}}} \cdot \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}$
$f'(x) = \frac{1+x}{\sqrt{1 - 2x + x^2 - (1 + 2x + x^2)}} \cdot \frac{-1 - x - 1 + x}{(1+x)^2}$
$f'(x) = \frac{1+x}{\sqrt{-4x}} \cdot \frac{-2}{(1+x)^2}$
$f'(x) = \frac{1+x}{2\sqrt{-x}} \cdot \frac{-2}{(1+x)^2} = \frac{-1}{(1+x)\sqrt{-x}}$.

(B) Given $y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)$.
Applying the chain rule for differentiation:
$\frac{d}{dx}(\tan^{-1}(\sin \sqrt{x})) = \frac{1}{1+(\sin \sqrt{x})^2} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} = \frac{\cos \sqrt{x}}{2\sqrt{x}(1+\sin^2 \sqrt{x})}$.
For the second term,using $\frac{d}{dx}(\operatorname{cosec}^{-1}(u)) = -\frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx}$:
$\frac{d}{dx}(\operatorname{cosec}^{-1}(e^{2x+1})) = -\frac{1}{e^{2x+1}\sqrt{(e^{2x+1})^2-1}} \cdot e^{2x+1} \cdot 2 = -\frac{2}{\sqrt{e^{4x+2}-1}}$.
Combining these,we get $\frac{dy}{dx} = \frac{\cos \sqrt{x}}{2 \sqrt{x}(1+\sin ^2 \sqrt{x})} - \frac{2}{\sqrt{e^{4 x+2}-1}}$.

(D) Given the functional equation $f(x+y)=f(x) f(y)$.
By the definition of the derivative,$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$.
Using the given relation,$f(x+h) = f(x)f(h)$.
So,$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h)-1}{h}$.
Since $f(0+0) = f(0)f(0)$,we have $f(0) = f(0)^2$,so $f(0)=1$ (assuming $f(x) \neq 0$).
Thus,$f^{\prime}(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{f(h)-1}{h} = 11$.
Therefore,$f^{\prime}(x) = f(x) \cdot 11$.
At $x=3$,$f^{\prime}(3) = f(3) \cdot 11 = 3 \cdot 11 = 33$.

(D) Given $g(x) = (f(2f(x)+2))^2$.
Applying the chain rule,we differentiate $g(x)$ with respect to $x$:
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot \frac{d}{dx}(f(2f(x)+2))$
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot \frac{d}{dx}(2f(x)+2)$
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot 2f^{\prime}(x)$
$g^{\prime}(x) = 4 \cdot f(2f(x)+2) \cdot f^{\prime}(2f(x)+2) \cdot f^{\prime}(x)$.
Now,substitute $x=0$:
$g^{\prime}(0) = 4 \cdot f(2f(0)+2) \cdot f^{\prime}(2f(0)+2) \cdot f^{\prime}(0)$.
Given $f(0)=-1$ and $f^{\prime}(0)=1$:
$g^{\prime}(0) = 4 \cdot f(2(-1)+2) \cdot f^{\prime}(2(-1)+2) \cdot (1)$
$g^{\prime}(0) = 4 \cdot f(0) \cdot f^{\prime}(0) \cdot 1$
$g^{\prime}(0) = 4 \cdot (-1) \cdot (1) \cdot 1 = -4$.

(B) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx} = 6x^2+10x-9$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int (6x^2+10x-9) dx = 2x^3+5x^2-9x+C$.
Given that $f(2)=0$,we substitute $x=2$ into the equation:
$f(2) = 2(2)^3 + 5(2)^2 - 9(2) + C = 0$
$16 + 20 - 18 + C = 0$
$18 + C = 0 \Rightarrow C = -18$.
Thus,the function is $f(x) = 2x^3+5x^2-9x-18$.
Now,we find $f(-2)$:
$f(-2) = 2(-2)^3 + 5(-2)^2 - 9(-2) - 18$
$f(-2) = 2(-8) + 5(4) + 18 - 18$
$f(-2) = -16 + 20 + 0 = 4$.

(B) Given,$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$.
Integrating both sides with respect to $x$,we get:
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = \int \frac{1}{x^4-1} dx$.
We know that $\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]$.
So,$\int \frac{1}{x^4-1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
Using standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|$ and $\int \frac{1}{x^2+1} dx = \tan^{-1} x$:
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| \right) - \frac{1}{2} \tan^{-1} x$.
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = -\frac{1}{2} \tan^{-1} x + \frac{1}{4} \log \left| \frac{x-1}{x+1} \right|$.
Comparing the coefficients,we get $a = -\frac{1}{2}$ and $b = \frac{1}{4}$.
Therefore,$a - 2b = -\frac{1}{2} - 2(\frac{1}{4}) = -\frac{1}{2} - \frac{1}{2} = -1$.

(C) Given that $f(x)$ is an odd differentiable function.
By definition of an odd function,$f(-x) = -f(x)$.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[-f(x)]$
$-f^{\prime}(-x) = -f^{\prime}(x)$
$f^{\prime}(-x) = f^{\prime}(x)$
This shows that the derivative of an odd function is an even function.
Now,substitute $x = 3$ into the equation $f^{\prime}(-x) = f^{\prime}(x)$:
$f^{\prime}(-3) = f^{\prime}(3)$
Since it is given that $f^{\prime}(3) = 2$,we have:
$f^{\prime}(-3) = 2$.

(C) Given $f(x) = x^m$,where $m \geq 0$ and $m \in \mathbb{Z}$.
First,find the derivative $f^{\prime}(x) = m x^{m-1}$.
The given condition is $f^{\prime}(a+b) = f^{\prime}(a) + f^{\prime}(b)$.
Substituting the derivative,we get $m(a+b)^{m-1} = m a^{m-1} + m b^{m-1}$.
If $m=0$,$f(x) = 1$,so $f^{\prime}(x) = 0$. The equation becomes $0 = 0+0$,which is true,but usually,we consider non-trivial cases for $m$. Let's check other values.
If $m=1$,$f^{\prime}(x) = 1$. The equation becomes $1 = 1+1$,which is $1=2$ (False).
If $m=2$,$f^{\prime}(x) = 2x$. The equation becomes $2(a+b) = 2a + 2b$,which simplifies to $2a+2b = 2a+2b$ (True).
If $m=3$,$f^{\prime}(x) = 3x^2$. The equation becomes $3(a+b)^2 = 3a^2 + 3b^2$,which simplifies to $a^2 + 2ab + b^2 = a^2 + b^2$,implying $2ab = 0$. Since $a, b > 0$,this is False.
Thus,the value of $m$ is $2$.

(A) The differential $df$ is given by $df = f'(x) dx$.
First,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - \tan^{-1}(e^{5x})]$
$f'(x) = \frac{1}{1 + e^{10x}} \cdot (10e^{10x}) - \frac{1}{1 + (e^{5x})^2} \cdot (5e^{5x})$
$f'(x) = \frac{10e^{10x}}{1 + e^{10x}} - \frac{5e^{5x}}{1 + e^{10x}}$
At $x = 0$,$e^{10(0)} = e^0 = 1$ and $e^{5(0)} = e^0 = 1$.
$f'(0) = \frac{10(1)}{1 + 1} - \frac{5(1)}{1 + 1} = \frac{10}{2} - \frac{5}{2} = \frac{5}{2} = 2.5$.
Now,calculate the differential for $dx = 0.2$:
$df = f'(0) \cdot dx = 2.5 \cdot 0.2 = 0.5$.

(C) Given functions are $F(x)=e^{x}$ and $G(x)=e^{-x}$.
We define $H(x) = G(F(x))$.
Substituting $F(x)$ into $G(x)$,we get $H(x) = G(e^{x}) = e^{-(e^{x})}$.
Now,we differentiate $H(x)$ with respect to $x$ using the chain rule:
$\frac{dH}{dx} = \frac{d}{dx}(e^{-e^{x}}) = e^{-e^{x}} \cdot \frac{d}{dx}(-e^{x}) = e^{-e^{x}} \cdot (-e^{x}) = -e^{x} \cdot e^{-e^{x}}$.
To find the value at $x=0$,we substitute $x=0$ into the derivative:
$\left. \frac{dH}{dx} \right|_{x=0} = -e^{0} \cdot e^{-e^{0}} = -1 \cdot e^{-1} = -\frac{1}{e}$.

(C) Given $f_1(x) = e^x$ and $f_{n+1}(x) = e^{f_n(x)}$.
Taking the natural logarithm on both sides of $f_n(x) = e^{f_{n-1}(x)}$,we get $\ln(f_n(x)) = f_{n-1}(x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{f_n(x)} \cdot f_n'(x) = f_{n-1}'(x)$
$\Rightarrow f_n'(x) = f_n(x) \cdot f_{n-1}'(x) \quad \dots (i)$
For $n=1$,$f_1'(x) = e^x = f_1(x)$.
For $n=2$,$f_2'(x) = f_2(x) \cdot f_1'(x) = f_2(x) \cdot f_1(x)$.
For $n=3$,$f_3'(x) = f_3(x) \cdot f_2'(x) = f_3(x) \cdot f_2(x) \cdot f_1(x)$.
By induction,for any $n \geq 1$,$\frac{d}{dx} f_n(x) = f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_1(x)$.

(C) Given,$y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}})$.
Taking the natural logarithm on both sides:
$\log y = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \ldots + \log(1+x^{2^{n}})$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}}$.
Thus,$\frac{d y}{d x} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}} \right]$.
At $x=0$,$y = (1+0)(1+0) \ldots (1+0) = 1$.
Substituting $x=0$ in the derivative expression:
$\left(\frac{d y}{d x}\right)_{x=0} = 1 \left[ \frac{1}{1+0} + 0 + 0 + \ldots + 0 \right] = 1 \times 1 = 1$.

(B) Given $f(x) = x^{n}$.
Taking the derivative,we get $f^{\prime}(x) = n x^{n-1}$.
Substituting this into the given equation $f^{\prime}(\alpha+\beta) = f^{\prime}(\alpha) + f^{\prime}(\beta)$,we have:
$n(\alpha+\beta)^{n-1} = n\alpha^{n-1} + n\beta^{n-1}$.
For $n \neq 0$,we can divide by $n$:
$(\alpha+\beta)^{n-1} = \alpha^{n-1} + \beta^{n-1}$.
If $n=1$,then $(\alpha+\beta)^{0} = \alpha^{0} + \beta^{0} \Rightarrow 1 = 1 + 1$,which is $1 = 2$ (False).
If $n=2$,then $(\alpha+\beta)^{2-1} = \alpha^{2-1} + \beta^{2-1} \Rightarrow \alpha+\beta = \alpha+\beta$ (True).
If $n=0$,$f(x) = x^{0} = 1$,so $f^{\prime}(x) = 0$. Then $0 = 0 + 0$ (True).
However,checking the options provided,$n=2$ is the standard solution for this type of problem.

(A) Given,$y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$.
Applying the product rule to the first term and the chain rule to the second term:
$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) + \frac{1}{1+x}$.
Using the quotient rule for $\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right)$:
$\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) = \frac{(3^{x}+1)(3^{x} \ln 3) - (3^{x}-1)(3^{x} \ln 3)}{(3^{x}+1)^{2}} = \frac{3^{x} \ln 3 (3^{x} + 1 - 3^{x} + 1)}{(3^{x}+1)^{2}} = \frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}$.
Substituting this back into the derivative expression:
$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x + \sin x \left(\frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}\right) + \frac{1}{1+x}$.
Evaluating at $x = 0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \left(\frac{3^{0}-1}{3^{0}+1}\right) \cos(0) + \sin(0) \left(\frac{2 \cdot 3^{0} \ln 3}{(3^{0}+1)^{2}}\right) + \frac{1}{1+0}$.
$= \left(\frac{1-1}{1+1}\right) \cdot 1 + 0 \cdot \left(\frac{2 \ln 3}{4}\right) + 1 = 0 + 0 + 1 = 1$.

(A) Let the second degree polynomial be $f(x) = ax^2 + bx + c$,where $a \neq 0$.
Given $f(1) = f(-1)$,we have $a(1)^2 + b(1) + c = a(-1)^2 + b(-1) + c$,which simplifies to $a + b + c = a - b + c$,implying $2b = 0$,so $b = 0$.
Thus,$f(x) = ax^2 + c$.
The derivative is $f^{\prime}(x) = 2ax$.
Since $p, q, r$ are in $A$.$P$.,we have $2q = p + r$.
Multiplying by $2a$,we get $2a(2q) = 2a(p + r)$,which means $2(2aq) = 2ap + 2ar$.
Substituting $f^{\prime}(x) = 2ax$,we get $2f^{\prime}(q) = f^{\prime}(p) + f^{\prime}(r)$.
This condition implies that $f^{\prime}(p), f^{\prime}(q), f^{\prime}(r)$ are in $A$.$P$.

(C) First,convert the degree measure to radians: $x^{\circ} = \frac{\pi x}{180}$.
We use the identity $\cos(30^{\circ} + x^{\circ}) = \sin(90^{\circ} - (30^{\circ} + x^{\circ})) = \sin(60^{\circ} - x^{\circ})$.
Let $A = 60^{\circ} - x^{\circ}$. The expression becomes $3 \sin A - 4 \sin^3 A$.
Using the trigonometric identity $3 \sin A - 4 \sin^3 A = \sin(3A)$,we get $\sin(3(60^{\circ} - x^{\circ})) = \sin(180^{\circ} - 3x^{\circ}) = \sin(3x^{\circ})$.
Now,express this in terms of radians: $f(x) = \sin(3 \cdot \frac{\pi x}{180}) = \sin(\frac{\pi x}{60})$.
Differentiating with respect to $x$ using the chain rule: $\frac{d}{dx} [\sin(\frac{\pi x}{60})] = \cos(\frac{\pi x}{60}) \cdot \frac{\pi}{60} = \frac{\pi}{60} \cos(3x^{\circ})$.

(B) Given $f''(x) = g''(x)$. Integrating both sides with respect to $x$,we get $f'(x) = g'(x) + C_1$.
From the given conditions,$f'(1) = 4$ and $2g'(1) = 4 \implies g'(1) = 2$.
Substituting these into the derivative equation: $4 = 2 + C_1 \implies C_1 = 2$.
Thus,$f'(x) - g'(x) = 2$. Integrating again,we get $f(x) - g(x) = 2x + C_2$.
Given $g(2) = 9$ and $3f(2) = 9 \implies f(2) = 3$.
Substituting $x = 2$ into the equation $f(x) - g(x) = 2x + C_2$: $3 - 9 = 2(2) + C_2 \implies -6 = 4 + C_2 \implies C_2 = -10$.
Therefore,$f(x) - g(x) = 2x - 10$.
For $x = 25$,$f(25) - g(25) = 2(25) - 10 = 50 - 10 = 40$.