Differentiation of Exponential and Logarithmic Questions in English

(B) Let $y = \log(\tan x)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x)$
$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x$
Since $\tan x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we have:
$\frac{dy}{dx} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x}$
Multiplying the numerator and denominator by $2$:
$\frac{dy}{dx} = \frac{2}{2 \sin x \cos x}$
Using the identity $\sin 2x = 2 \sin x \cos x$:
$\frac{dy}{dx} = \frac{2}{\sin 2x} = 2 \csc 2x$.
Thus,the correct option is $B$.

(A) Given the function: $y = \log_{10}x + \log_x 10 + \log_x x + \log_{10} 10$
Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we simplify the expression:
$y = \frac{\ln x}{\ln 10} + \frac{\ln 10}{\ln x} + 1 + 1$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\ln x}{\ln 10} \right) + \frac{d}{dx} \left( \frac{\ln 10}{\ln x} \right) + \frac{d}{dx}(2)$
$\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{x} + \ln 10 \cdot \frac{d}{dx} ((\ln x)^{-1}) + 0$
$\frac{dy}{dx} = \frac{1}{x \ln 10} + \ln 10 \cdot (-1)(\ln x)^{-2} \cdot \frac{1}{x}$
$\frac{dy}{dx} = \frac{1}{x \ln 10} - \frac{\ln 10}{x (\ln x)^2}$
Thus,the correct option is $A$.

(B) Given $f(x) = \log_{x}(\log x)$.
Using the base change formula,we can write $f(x) = \frac{\log(\log x)}{\log x}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{\left(\frac{d}{dx}(\log(\log x))\right)(\log x) - (\log(\log x))(\frac{d}{dx}(\log x))}{(\log x)^2}$
$f'(x) = \frac{\left(\frac{1}{\log x} \cdot \frac{1}{x}\right)(\log x) - (\log(\log x))(\frac{1}{x})}{(\log x)^2}$
$f'(x) = \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} = \frac{1 - \log(\log x)}{x(\log x)^2}$.
Now,evaluate $f'(x)$ at $x = e$:
$f'(e) = \frac{1 - \log(\log e)}{e(\log e)^2}$.
Since $\log e = 1$ and $\log(1) = 0$,we have:
$f'(e) = \frac{1 - 0}{e(1)^2} = \frac{1}{e}$.

(C) We need to find the derivative of $y = \log_7(\log_7 x)$.
Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write:
$y = \frac{\log_e(\log_7 x)}{\log_e 7} = \frac{1}{\log_e 7} \cdot \log_e(\log_7 x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{d}{dx}(\log_e(\log_7 x))$.
Using the chain rule,$\frac{d}{dx}(\log_e u) = \frac{1}{u} \cdot \frac{du}{dx}$,where $u = \log_7 x = \frac{\log_e x}{\log_e 7}$:
$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{d}{dx}(\log_7 x)$.
Since $\frac{d}{dx}(\log_7 x) = \frac{d}{dx}(\frac{\log_e x}{\log_e 7}) = \frac{1}{x \log_e 7}$:
$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{1}{x \log_e 7} = \frac{1}{x \cdot \log_7 x \cdot (\log_e 7)^2}$.
Alternatively,using $\log_7 x = \frac{\log_e x}{\log_e 7}$ and $\log_7 e = \frac{1}{\log_e 7}$:
$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{\log_7 e}{x} = \frac{\log_7 e}{x \log_e x}$.
Thus,the correct option is $C$.

(B) Given $y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right)$.
Using logarithmic properties,$y = \log(1 + \sqrt{x}) - \log(1 - \sqrt{x})$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\log(1 + \sqrt{x})] - \frac{d}{dx} [\log(1 - \sqrt{x})]$
$\frac{dy}{dx} = \frac{1}{1 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1 - \sqrt{x}} \cdot \left( -\frac{1}{2\sqrt{x}} \right)$
$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{1}{1 + \sqrt{x}} + \frac{1}{1 - \sqrt{x}} \right)$
$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{1 - \sqrt{x} + 1 + \sqrt{x}}{(1 + \sqrt{x})(1 - \sqrt{x})} \right)$
$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \frac{2}{1 - x} \right)$
$\frac{dy}{dx} = \frac{1}{\sqrt{x}(1 - x)}$.

(A) Given the function $y = e^{x + 3\log x}$.
Using the properties of exponents and logarithms,we can simplify the expression:
$y = e^x \cdot e^{3\log x} = e^x \cdot e^{\log(x^3)} = e^x \cdot x^3$.
Now,differentiate $y$ with respect to $x$ using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$:
$\frac{dy}{dx} = \frac{d}{dx}(e^x \cdot x^3) = e^x \cdot \frac{d}{dx}(x^3) + x^3 \cdot \frac{d}{dx}(e^x)$.
$\frac{dy}{dx} = e^x \cdot (3x^2) + x^3 \cdot (e^x)$.
Factor out $e^x \cdot x^2$:
$\frac{dy}{dx} = e^x \cdot x^2(3 + x)$.
Thus,the correct option is $A$.

(C) Let $y = \log \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \sec^2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} \right)$
$\frac{dy}{dx} = \frac{1}{\tan \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \sec^2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \cdot \frac{1}{2}$
Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$:
$\frac{dy}{dx} = \frac{\cos \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\sin \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \frac{1}{\cos^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)} \cdot \frac{1}{2}$
$\frac{dy}{dx} = \frac{1}{2 \sin \left( \frac{\pi}{4} + \frac{x}{2} \right) \cos \left( \frac{\pi}{4} + \frac{x}{2} \right)}$
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:
$\frac{dy}{dx} = \frac{1}{\sin \left( 2 \left( \frac{\pi}{4} + \frac{x}{2} \right) \right)} = \frac{1}{\sin \left( \frac{\pi}{2} + x \right)}$
Since $\sin \left( \frac{\pi}{2} + x \right) = \cos x$:
$\frac{dy}{dx} = \frac{1}{\cos x} = \sec x$.

(B) Let $y = \log (\sqrt{x - a} + \sqrt{x - b})$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\sqrt{x - a} + \sqrt{x - b}} \cdot \frac{d}{dx}(\sqrt{x - a} + \sqrt{x - b})$
$= \frac{1}{\sqrt{x - a} + \sqrt{x - b}} \cdot \left( \frac{1}{2\sqrt{x - a}} + \frac{1}{2\sqrt{x - b}} \right)$
$= \frac{1}{\sqrt{x - a} + \sqrt{x - b}} \cdot \frac{1}{2} \left( \frac{\sqrt{x - b} + \sqrt{x - a}}{\sqrt{x - a}\sqrt{x - b}} \right)$
$= \frac{1}{\sqrt{x - a} + \sqrt{x - b}} \cdot \frac{\sqrt{x - a} + \sqrt{x - b}}{2\sqrt{(x - a)(x - b)}}$
$= \frac{1}{2\sqrt{(x - a)(x - b)}}$.

(D) Let $y = \log_e \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$.
Using logarithmic properties,$y = \frac{1}{2} [\log_e(1 + \sin x) - \log_e(1 - \sin x)]$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{\cos x}{1 + \sin x} - \frac{-\cos x}{1 - \sin x} \right]$
$= \frac{\cos x}{2} \left[ \frac{1 - \sin x + 1 + \sin x}{(1 + \sin x)(1 - \sin x)} \right]$
$= \frac{\cos x}{2} \left[ \frac{2}{1 - \sin^2 x} \right]$
$= \frac{\cos x}{\cos^2 x} = \frac{1}{\cos x} = \sec x$.

(B) Let $y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.
Using the trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:
$y = \log \sqrt{\frac{2 \sin^2 (x/2)}{2 \cos^2 (x/2)}} = \log \sqrt{\tan^2 \frac{x}{2}} = \log \left( \tan \frac{x}{2} \right)$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{\tan (x/2)} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2}$.
$\frac{dy}{dx} = \frac{\cos (x/2)}{\sin (x/2)} \cdot \frac{1}{\cos^2 (x/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin (x/2) \cos (x/2)}$.
Using the identity $\sin x = 2 \sin (x/2) \cos (x/2)$,we have:
$\frac{dy}{dx} = \frac{1}{\sin x} = \text{cosec } x$.

(A) Given $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$.
Squaring both sides,we get $y^2 = \frac{1 + e^x}{1 - e^x}$.
Differentiating both sides with respect to $x$ using the quotient rule:
$2y \frac{dy}{dx} = \frac{(1 - e^x)(e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$
$2y \frac{dy}{dx} = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1 - e^x)^2} = \frac{2e^x}{(1 - e^x)^2}$
$\frac{dy}{dx} = \frac{e^x}{y(1 - e^x)^2}$
Substituting $y = \sqrt{\frac{1 + e^x}{1 - e^x}}$:
$\frac{dy}{dx} = \frac{e^x}{\sqrt{\frac{1 + e^x}{1 - e^x}} (1 - e^x)^2} = \frac{e^x}{\sqrt{1 + e^x} \sqrt{1 - e^x} (1 - e^x)} = \frac{e^x}{(1 - e^x) \sqrt{(1 + e^x)(1 - e^x)}}$
$\frac{dy}{dx} = \frac{e^x}{(1 - e^x) \sqrt{1 - e^{2x}}}$.

(D) We need to find the derivative of the product of two logarithmic functions: $\frac{d}{dx}[(\log_e x)(\log_a x)]$.
First,use the change of base formula for logarithms: $\log_a x = \frac{\log_e x}{\log_e a}$.
Substituting this into the expression: $\frac{d}{dx}[(\log_e x) \cdot \frac{\log_e x}{\log_e a}] = \frac{1}{\log_e a} \cdot \frac{d}{dx}[(\log_e x)^2]$.
Using the chain rule,the derivative of $(\log_e x)^2$ is $2(\log_e x) \cdot \frac{d}{dx}(\log_e x) = 2(\log_e x) \cdot \frac{1}{x}$.
Thus,the expression becomes: $\frac{1}{\log_e a} \cdot \frac{2\log_e x}{x} = \frac{2}{\log_e a} \cdot \frac{\log_e x}{x}$.
Since $\frac{\log_e x}{\log_e a} = \log_a x$,the final result is $\frac{2\log_a x}{x}$.

(D) Let $y = \log \left( \frac{e^x}{1 + e^x} \right)$.
Using the property of logarithms,$\log \left( \frac{a}{b} \right) = \log a - \log b$,we get:
$y = \log(e^x) - \log(1 + e^x) = x - \log(1 + e^x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\log(1 + e^x))$.
$\frac{dy}{dx} = 1 - \frac{1}{1 + e^x} \cdot \frac{d}{dx}(1 + e^x)$.
$\frac{dy}{dx} = 1 - \frac{e^x}{1 + e^x}$.
$\frac{dy}{dx} = \frac{1 + e^x - e^x}{1 + e^x} = \frac{1}{1 + e^x}$.
Since $\frac{1}{1 + e^x}$ is not among the given options,the correct answer is $(D)$.

(A) Let $y = \log \sqrt{\sin \sqrt{e^x}}$.
Using the property of logarithms,$\log(a^n) = n \log a$,we can write:
$y = \frac{1}{2} \log(\sin \sqrt{e^x})$.
Now,differentiate with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin \sqrt{e^x}} \cdot \frac{d}{dx}(\sin \sqrt{e^x})$.
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin \sqrt{e^x}} \cdot \cos \sqrt{e^x} \cdot \frac{d}{dx}(\sqrt{e^x})$.
Since $\sqrt{e^x} = (e^x)^{1/2} = e^{x/2}$,we have:
$\frac{dy}{dx} = \frac{1}{2} \cot \sqrt{e^x} \cdot \frac{d}{dx}(e^{x/2})$.
$\frac{dy}{dx} = \frac{1}{2} \cot(e^{x/2}) \cdot e^{x/2} \cdot \frac{1}{2}$.
$\frac{dy}{dx} = \frac{1}{4} e^{x/2} \cot(e^{x/2})$.

(B) Given $y = \log(\log x)$.
Taking exponential on both sides,we get $e^y = \log x$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(e^y) = \frac{d}{dx}(\log x)$.
Using the chain rule,$e^y \frac{dy}{dx} = \frac{1}{x}$.
Thus,the correct option is $B$.

(C) Given $y = \log_{\sin x}(\tan x)$.
Using the change of base formula,we can write $y = \frac{\log(\tan x)}{\log(\sin x)}$.
Differentiating with respect to $x$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(\log(\sin x)) \cdot \frac{d}{dx}(\log(\tan x)) - (\log(\tan x)) \cdot \frac{d}{dx}(\log(\sin x))}{(\log(\sin x))^2}$.
Since $\frac{d}{dx}(\log(\tan x)) = \frac{\sec^2 x}{\tan x} = \frac{1}{\sin x \cos x}$ and $\frac{d}{dx}(\log(\sin x)) = \cot x = \frac{\cos x}{\sin x}$,
At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$\cos x = \frac{1}{\sqrt{2}}$,$\tan x = 1$,$\log(\tan x) = 0$,and $\log(\sin x) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$.
Substituting these values:
$\left( \frac{dy}{dx} \right)_{\pi/4} = \frac{(-\frac{1}{2} \log 2) \cdot (\frac{1}{(1/\sqrt{2})(1/\sqrt{2})}) - (0) \cdot (1)}{(-\frac{1}{2} \log 2)^2} = \frac{-\frac{1}{2} \log 2 \cdot 2}{\frac{1}{4} (\log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

(A) Given $y = \log_2(\log_2(x))$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:
$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2}$
$y = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$
Since $\frac{1}{\log_e 2} = \log_2 e$,we get:
$\frac{dy}{dx} = \frac{\log_2 e}{x \log_e x}$.

(B) Given the function $f(x) = 3e^{x^2}$.
Differentiating with respect to $x$,we get $f'(x) = 3 \cdot e^{x^2} \cdot (2x) = 6xe^{x^2}$.
Now,evaluate the function and its derivative at $x = 0$:
$f(0) = 3e^{0^2} = 3(1) = 3$.
$f'(0) = 6(0)e^{0^2} = 0$.
Substitute these values into the expression $f'(x) - 2xf(x) + \frac{1}{3}f(0) - f'(0)$:
$= 6xe^{x^2} - 2x(3e^{x^2}) + \frac{1}{3}(3) - 0$
$= 6xe^{x^2} - 6xe^{x^2} + 1 - 0$
$= 1$.

(A) Let $y = \log \left\{ e^x \left( \frac{x + 2}{x - 2} \right)^{3/4} \right\}$.
Using the property $\log(ab) = \log a + \log b$,we get $y = \log(e^x) + \log \left( \frac{x + 2}{x - 2} \right)^{3/4}$.
Since $\log(e^x) = x$ and $\log(a^n) = n \log a$,we have $y = x + \frac{3}{4} \log \left( \frac{x + 2}{x - 2} \right)$.
Using $\log \left( \frac{a}{b} \right) = \log a - \log b$,we get $y = x + \frac{3}{4} [ \log(x + 2) - \log(x - 2) ]$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx} \log(x + 2) - \frac{d}{dx} \log(x - 2) \right]$.
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x + 2} - \frac{1}{x - 2} \right]$.
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{(x - 2) - (x + 2)}{(x + 2)(x - 2)} \right] = 1 + \frac{3}{4} \left[ \frac{-4}{x^2 - 4} \right]$.
$\frac{dy}{dx} = 1 - \frac{3}{x^2 - 4} = \frac{x^2 - 4 - 3}{x^2 - 4} = \frac{x^2 - 7}{x^2 - 4}$.

(A) Given the function $y = e^{(1 + \log_e x)}$.
Using the property of exponents $a^{m+n} = a^m \cdot a^n$,we can rewrite the expression as:
$y = e^1 \cdot e^{\log_e x}$
Since $e^{\log_e x} = x$,the equation simplifies to:
$y = e \cdot x$
Now,differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(ex)$
Since $e$ is a constant,we get:
$\frac{dy}{dx} = e \cdot \frac{d}{dx}(x) = e \cdot 1 = e$
Therefore,the correct option is $A$.

(B) Given $y = \log_{10}(x^2)$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can rewrite the expression as:
$y = \frac{\log_e(x^2)}{\log_e 10}$
Using the logarithmic property $\log(a^n) = n \log a$,we get:
$y = \frac{2 \log_e x}{\log_e 10}$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{2 \log_e x}{\log_e 10} \right)$
Since $\frac{2}{\log_e 10}$ is a constant:
$\frac{dy}{dx} = \frac{2}{\log_e 10} \cdot \frac{d}{dx}(\log_e x)$
$\frac{dy}{dx} = \frac{2}{\log_e 10} \cdot \frac{1}{x} = \frac{2}{x \log_e 10}$.

(C) Given $y = 3^{x^2}$.
We know that the derivative of an exponential function is given by $\frac{d}{dx}(a^u) = a^u \cdot \ln(a) \cdot \frac{du}{dx}$,where $a$ is a constant and $u$ is a function of $x$.
Applying the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(3^{x^2})$
$\frac{dy}{dx} = 3^{x^2} \cdot \ln(3) \cdot \frac{d}{dx}(x^2)$
Since $\frac{d}{dx}(x^2) = 2x$,we get:
$\frac{dy}{dx} = 3^{x^2} \cdot \ln(3) \cdot 2x$
Rearranging the terms,we get:
$\frac{dy}{dx} = 3^{x^2} \cdot 2x \cdot \log_e 3$.

(C) Let $y = \log f(e^x + 2x)$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = \frac{1}{f(e^x + 2x)} \cdot f'(e^x + 2x) \cdot \frac{d}{dx}(e^x + 2x)$
$\frac{dy}{dx} = \frac{f'(e^x + 2x) \cdot (e^x + 2)}{f(e^x + 2x)}$
Now,evaluate at $x = 0$:
At $x = 0$,the argument of the function is $e^0 + 2(0) = 1 + 0 = 1$.
Substituting $x = 0$ into the derivative expression:
$\left. \frac{dy}{dx} \right|_{x=0} = \frac{f'(1) \cdot (e^0 + 2)}{f(1)}$
Given $f(1) = 3$ and $f'(1) = 2$:
$\left. \frac{dy}{dx} \right|_{x=0} = \frac{2 \cdot (1 + 2)}{3} = \frac{2 \cdot 3}{3} = 2$.

(C) Let $y = \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\}$.
Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:
$y = \log(e^x) + \log \left( \frac{x - 2}{x + 2} \right)^{3/4}$
$y = x + \frac{3}{4} [\log(x - 2) - \log(x + 2)]$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx}(\log(x - 2)) - \frac{d}{dx}(\log(x + 2)) \right]$
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x - 2} - \frac{1}{x + 2} \right]$
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{(x + 2) - (x - 2)}{(x - 2)(x + 2)} \right]$
$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{4}{x^2 - 4} \right]$
$\frac{dy}{dx} = 1 + \frac{3}{x^2 - 4} = \frac{x^2 - 4 + 3}{x^2 - 4} = \frac{x^2 - 1}{x^2 - 4}$.

(A) Given the function $f(x) = \log_5(\log_7 x)$.
Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we can write:
$f(x) = \frac{\ln(\log_7 x)}{\ln 5}$
$f(x) = \frac{\ln(\frac{\ln x}{\ln 7})}{\ln 5}$
$f(x) = \frac{\ln(\ln x) - \ln(\ln 7)}{\ln 5}$
Now,differentiating with respect to $x$:
$f'(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx}(\ln(\ln x) - \ln(\ln 7))$
$f'(x) = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{d}{dx}(\ln x)$
$f'(x) = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$
$f'(x) = \frac{1}{x(\ln 5)(\ln x)}$
Since $\ln x = \ln 7 \cdot \log_7 x$,we have:
$f'(x) = \frac{1}{x(\ln 5)(\ln 7)(\log_7 x)}$
Thus,the correct option is $A$.

(C) Given $f(x) = \log x$.
We need to find the derivative of $f[\log (x)]$ with respect to $x$.
First,substitute $\log x$ into the function $f(x)$:
$f[\log (x)] = \log(\log x)$.
Now,differentiate $\log(\log x)$ with respect to $x$ using the chain rule:
$\frac{d}{dx} [\log(\log x)] = \frac{1}{\log x} \cdot \frac{d}{dx} (\log x)$.
Since $\frac{d}{dx} (\log x) = \frac{1}{x}$,we have:
$\frac{d}{dx} [\log(\log x)] = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.
Thus,the correct option is $C$.

(B) Given $y = \log(x^x)$.
Using the property of logarithms,$\log(a^b) = b \log a$,we have:
$y = x \log x$.
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x)$
$\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}$
$\frac{dy}{dx} = \log x + 1$.
Since $\log e = 1$,we can write:
$\frac{dy}{dx} = \log x + \log e$.
Using the property $\log a + \log b = \log(ab)$,we get:
$\frac{dy}{dx} = \log(ex)$.
Thus,the correct option is $B$.

(B) The curves are $y = a^x$ and $y = b^x$.
Setting $a^x = b^x$,we get $(a/b)^x = 1$,which implies $x = 0$.
At $x = 0$,$y = a^0 = 1$. So,the point of intersection is $(0, 1)$.
For the first curve $y = a^x$,the slope of the tangent is $m_1 = \frac{dy}{dx} = a^x \ln a$.
At $(0, 1)$,$m_1 = a^0 \ln a = \ln a$.
For the second curve $y = b^x$,the slope of the tangent is $m_2 = \frac{dy}{dx} = b^x \ln b$.
At $(0, 1)$,$m_2 = b^0 \ln b = \ln b$.
The angle $\alpha$ between the curves is given by $\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \alpha = \frac{\ln a - \ln b}{1 + \ln a \ln b} = \frac{\log a - \log b}{1 + \log a \log b}$.

(D) To find the intersection point,set $y = a^x$ and $y = b^x$ equal to each other:
$a^x = b^x \implies (a/b)^x = 1 \implies x = 0$.
For $x = 0$,$y = a^0 = 1$. Thus,the intersection point is $(0, 1)$.
Now,find the slopes of the tangents at $(0, 1)$:
For $y = a^x$,$\frac{dy}{dx} = a^x \ln a$. At $x = 0$,$m_1 = \ln a$.
For $y = b^x$,$\frac{dy}{dx} = b^x \ln b$. At $x = 0$,$m_2 = \ln b$.
The angle $\alpha$ between the curves is given by $\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values of $m_1$ and $m_2$:
$\tan \alpha = \left| \frac{\ln a - \ln b}{1 + \ln a \ln b} \right|$.

(B) Let $y = a^{\log_{10}(\csc^{-1}x)}$.
Using the derivative formula $\frac{d}{dx}(a^u) = a^u \cdot \ln a \cdot \frac{du}{dx}$,where $u = \log_{10}(\csc^{-1}x)$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx} \left( \frac{\ln(\csc^{-1}x)}{\ln 10} \right) = \frac{1}{\ln 10} \cdot \frac{1}{\csc^{-1}x} \cdot \frac{d}{dx}(\csc^{-1}x)$.
Since $\frac{d}{dx}(\csc^{-1}x) = -\frac{1}{|x|\sqrt{x^2 - 1}}$,we have:
$\frac{du}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{\csc^{-1}x} \cdot \left( -\frac{1}{|x|\sqrt{x^2 - 1}} \right)$.
Now,substitute back into the derivative of $y$:
$\frac{dy}{dx} = a^{\log_{10}(\csc^{-1}x)} \cdot \ln a \cdot \left( \frac{1}{\ln 10} \cdot \frac{1}{\csc^{-1}x} \cdot \left( -\frac{1}{|x|\sqrt{x^2 - 1}} \right) \right)$.
Since $\frac{\ln a}{\ln 10} = \log_{10}a$,the expression simplifies to:
$\frac{dy}{dx} = - a^{\log_{10}(\csc^{-1}x)} \cdot \frac{1}{\csc^{-1}x} \cdot \frac{1}{|x|\sqrt{x^2 - 1}} \cdot \log_{10}a$.

(A) Given $f(x) = e^{g(x)}$.
By differentiating with respect to $x$,we get $f'(x) = e^{g(x)} \cdot g'(x)$.
Using the Leibniz integral rule,$g'(x) = \frac{d}{dx} \int_{2}^{x} \frac{t}{1 + t^4} \, dt = \frac{x}{1 + x^4}$.
Now,evaluate $g(2) = \int_{2}^{2} \frac{t}{1 + t^4} \, dt = 0$.
Therefore,$f'(2) = e^{g(2)} \cdot g'(2) = e^0 \cdot \frac{2}{1 + 2^4} = 1 \cdot \frac{2}{1 + 16} = \frac{2}{17}$.

(C) Given $y = \log_{\sin x} (\tan x)$. Using the change of base formula,we can write $y = \frac{\ln(\tan x)}{\ln(\sin x)}$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \ln(\tan x)$ and $v = \ln(\sin x)$:
$u' = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$
$v' = \frac{1}{\sin x} \cdot \cos x = \cot x$
At $x = \frac{\pi}{4}$,$\tan x = 1$,$\sin x = \frac{1}{\sqrt{2}}$,$\cos x = \frac{1}{\sqrt{2}}$,$\sec^2 x = 2$,and $\cot x = 1$.
$u = \ln(1) = 0$
$v = \ln(1/\sqrt{2}) = -\frac{1}{2} \ln 2$
$u' = \frac{1}{(1/\sqrt{2})(1/\sqrt{2})} = 2$
$v' = 1$
Substituting these values into the derivative formula:
$\left( \frac{dy}{dx} \right)_{\pi/4} = \frac{v(x) u'(x) - u(x) v'(x)}{(v(x))^2} = \frac{(-\frac{1}{2} \ln 2)(2) - (0)(1)}{(-\frac{1}{2} \ln 2)^2} = \frac{-\ln 2}{\frac{1}{4} (\ln 2)^2} = \frac{-4}{\ln 2}$.

(B) Given $\phi(x) = \log_{5} (\log_{3} x)$.
Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$,we can write:
$\phi(x) = \frac{\ln(\log_{3} x)}{\ln 5} = \frac{\ln(\frac{\ln x}{\ln 3})}{\ln 5} = \frac{\ln(\ln x) - \ln(\ln 3)}{\ln 5}$.
Differentiating with respect to $x$:
$\phi'(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx} (\ln(\ln x) - \ln(\ln 3))$.
$\phi'(x) = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$.
Now,evaluating at $x = e$:
$\phi'(e) = \frac{1}{\ln 5} \cdot \frac{1}{\ln e} \cdot \frac{1}{e}$.
Since $\ln e = 1$,we get:
$\phi'(e) = \frac{1}{e \ln 5}$.

(A) Let $y = \log(\cos x)$.
Using the chain rule for differentiation:
$\frac{dy}{dx} = \frac{d}{dx} [\log(\cos x)] = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$.
Since $\frac{d}{dx}(\cos x) = -\sin x$,we have:
$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x$.

(A) Let $y = \log(\cos e^{x})$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[\log(\cos e^{x})]$
$= \frac{1}{\cos e^{x}} \cdot \frac{d}{dx}(\cos e^{x})$
$= \frac{1}{\cos e^{x}} \cdot (-\sin e^{x}) \cdot \frac{d}{dx}(e^{x})$
$= \frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x}$
$= -e^{x} \tan e^{x}$

(A) Let $y = \log (\log x)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\log (\log x)]$
Applying the chain rule,where the derivative of $\log(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx} (\log x)$
Since the derivative of $\log x$ is $\frac{1}{x}$:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}$
Thus,the final result is:
$\frac{dy}{dx} = \frac{1}{x \log x}$,for $x > 1$.

(N/A) Let $y = \log _{7}(\log x) = \frac{\log (\log x)}{\log 7}$ (by change of base formula).
The function is defined for all real numbers $x > 1$. Therefore,
$\frac{dy}{dx} = \frac{1}{\log 7} \frac{d}{dx}(\log (\log x))$
$= \frac{1}{\log 7} \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$
$= \frac{1}{\log 7} \cdot \frac{1}{\log x} \cdot \frac{1}{x}$
$= \frac{1}{x \log x \log 7}$

(D) Let $y = \log _{\cos x} \operatorname{cosec} x$.
Using the change of base formula for logarithms,$y = \frac{\ln(\operatorname{cosec} x)}{\ln(\cos x)} = \frac{-\ln(\sin x)}{\ln(\cos x)}$.
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \frac{d}{dx}(\ln(\sin x)) - \ln(\sin x) \cdot \frac{d}{dx}(\ln(\cos x))}{(\ln(\cos x))^2} \right]$.
$\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \cot x - \ln(\sin x) \cdot (-\tan x)}{(\ln(\cos x))^2} \right]$.
At $x = \frac{\pi}{4}$,$\sin x = \cos x = \frac{1}{\sqrt{2}}$,so $\ln(\sin x) = \ln(\cos x) = \ln(2^{-1/2}) = -\frac{1}{2} \ln 2$.
Substituting these values:
$\left. \frac{dy}{dx} \right|_{x=\frac{\pi}{4}} = -\left[ \frac{(-\frac{1}{2} \ln 2)(1) - (-\frac{1}{2} \ln 2)(-1)}{(-\frac{1}{2} \ln 2)^2} \right] = -\left[ \frac{-\frac{1}{2} \ln 2 - \frac{1}{2} \ln 2}{\frac{1}{4} (\ln 2)^2} \right] = -\left[ \frac{-\ln 2}{\frac{1}{4} (\ln 2)^2} \right] = \frac{4}{\ln 2}$.
Finally,the value of $\log _{e} 2 \cdot \frac{dy}{dx}$ at $x = \frac{\pi}{4}$ is $\ln 2 \cdot \frac{4}{\ln 2} = 4$.

(D) The function $f^{\prime}(x) = \ln(x^2-1)$ is defined when $x^2-1 > 0$,which implies $x \in (-\infty, -1) \cup (1, \infty)$.
Since $S = (-\infty, -1) \cup (1, \infty)$ is a disconnected set consisting of two disjoint intervals,the constant of integration can be chosen independently for each interval.
Let $f(x) = \int \ln(x^2-1) dx + C_1$ for $x \in (1, \infty)$ and $f(x) = \int \ln(x^2-1) dx + C_2$ for $x \in (-\infty, -1)$.
Given $f(2) = 0$,we can determine $C_1$ uniquely.
However,there is no condition linking the value of the function on the interval $(1, \infty)$ to the interval $(-\infty, -1)$.
Since the constant $C_2$ can be any real number,there are infinitely many such functions $f$.

(B) Given $y = \log_{10} x + \log_x 10 + \log_x x + \log_{10} 10$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we have:
$y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 1 + 1$.
$y = \frac{\log_e x}{\log_e 10} + \frac{\log_e 10}{\log_e x} + 2$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{d}{dx}(\log_e x) + \log_e 10 \cdot \frac{d}{dx}((\log_e x)^{-1}) + 0$.
$\frac{dy}{dx} = \frac{1}{\log_e 10} \cdot \frac{1}{x} + \log_e 10 \cdot (-1)(\log_e x)^{-2} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x \log_e 10} - \frac{\log_e 10}{x(\log_e x)^2}$.

(C) Given,$y = e^{m \sin^{-1} x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{-1} x) = y \cdot \frac{m}{\sqrt{1 - x^2}}$.
Squaring both sides,we get:
$(\frac{dy}{dx})^2 = \frac{m^2 y^2}{1 - x^2}$.
Multiplying both sides by $(1 - x^2)$,we obtain:
$(1 - x^2) (\frac{dy}{dx})^2 = m^2 y^2$.
Comparing this with the given equation $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,we find that $A = m^2$.

(C) Given $y = \log_3(\log_3 x)$.
Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write $y = \frac{\log_e(\log_3 x)}{\log_e 3} = \frac{\log_e(\frac{\log_e x}{\log_e 3})}{\log_e 3}$.
$y = \frac{\log_e(\log_e x) - \log_e(\log_e 3)}{\log_e 3}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 3)]$.
$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.
At $x = 3$:
$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{1}{\log_e 3} \cdot \frac{1}{3} = \frac{1}{3(\log_e 3)^2} = \frac{1}{3}(\log_e 3)^{-2}$.
Since the provided options were incorrect,the correct value is $\frac{1}{3}(\log_e 3)^{-2}$.

(D) Given $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$.
Note that $(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} - \sqrt{x-1}) = (x+1) - (x-1) = 2$.
Thus,$\sqrt{x+1} - \sqrt{x-1} = \frac{2}{v}$.
However,the argument of the log is $\sqrt{x-1} - \sqrt{x+1} = -(\sqrt{x+1} - \sqrt{x-1}) = -\frac{2}{v}$.
Since the domain of $\log$ requires a positive argument,we consider the magnitude or the complex form. Assuming the standard derivative approach:
$u = \log(-2) - \log(v)$.
Taking the derivative with respect to $v$:
$\frac{du}{dv} = \frac{d}{dv}(\log(-2) - \log(v)) = 0 - \frac{1}{v} = -\frac{1}{v}$.

(C) Given $f(x)=e^x$ and $g(x)=\sin ^{-1} x$.
$h(x)=f(g(x))=e^{\sin ^{-1} x}$.
Differentiating $h(x)$ with respect to $x$:
$h^{\prime}(x)=\frac{d}{dx}(e^{\sin ^{-1} x}) = e^{\sin ^{-1} x} \cdot \frac{d}{dx}(\sin ^{-1} x) = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$.
Now,consider the ratio $\frac{h^{\prime}(x)}{h(x)}$:
$\frac{h^{\prime}(x)}{h(x)} = \frac{e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin ^{-1} x}} = \frac{1}{\sqrt{1-x^2}}$.
Finally,squaring the result:
$\left(\frac{h^{\prime}(x)}{h(x)}\right)^2 = \left(\frac{1}{\sqrt{1-x^2}}\right)^2 = \frac{1}{1-x^2}$.

(C) Given $y = \log_{\sin x} \tan x$. Using the change of base formula,we can write $y = \frac{\log \tan x}{\log \sin x}$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \log \tan x$ and $v = \log \sin x$:
$u' = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.
$v' = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Thus,$\frac{dy}{dx} = \frac{(\log \sin x)(2 \csc 2x) - (\log \tan x)(\cot x)}{(\log \sin x)^2}$.
At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$\tan x = 1$,$\log \tan x = 0$,$\csc 2x = \csc \frac{\pi}{2} = 1$,and $\cot x = 1$.
Substituting these values:
$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = \frac{(\log \frac{1}{\sqrt{2}})(2 \cdot 1) - (0)(1)}{(\log \frac{1}{\sqrt{2}})^2} = \frac{2 \log(2^{-1/2})}{(\log 2^{-1/2})^2} = \frac{2 \cdot (-\frac{1}{2} \log 2)}{(-\frac{1}{2} \log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

(B) Given $y=e^{\sin \left(\operatorname{cosec}^{-1} x\right)}$.
Since $\operatorname{cosec}^{-1} x = \sin^{-1} \left(\frac{1}{x}\right)$,we have $\sin \left(\operatorname{cosec}^{-1} x\right) = \sin \left(\sin^{-1} \frac{1}{x}\right) = \frac{1}{x}$.
Therefore,the function simplifies to $y = e^{\frac{1}{x}}$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x} = \frac{d}{d x} \left(e^{\frac{1}{x}}\right) = e^{\frac{1}{x}} \cdot \frac{d}{d x} \left(\frac{1}{x}\right)$.
Since $\frac{d}{d x} \left(\frac{1}{x}\right) = -\frac{1}{x^2}$,we get $\frac{d y}{d x} = e^{\frac{1}{x}} \left(-\frac{1}{x^2}\right) = -\frac{e^{\frac{1}{x}}}{x^2}$.

(D) Given $y = \log \left[a^{3x} \left(\frac{5-x}{x+4}\right)^{\frac{3}{4}}\right]$.
Using the properties of logarithms,$\log(mn) = \log m + \log n$ and $\log(m^n) = n \log m$:
$y = \log(a^{3x}) + \log\left(\left(\frac{5-x}{x+4}\right)^{\frac{3}{4}}\right)$
$y = 3x \log a + \frac{3}{4} \log(5-x) - \frac{3}{4} \log(x+4)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(3x \log a) + \frac{3}{4} \frac{d}{dx}(\log(5-x)) - \frac{3}{4} \frac{d}{dx}(\log(x+4))$
$\frac{dy}{dx} = 3 \log a + \frac{3}{4} \left(\frac{1}{5-x}\right)(-1) - \frac{3}{4} \left(\frac{1}{x+4}\right)(1)$
$\frac{dy}{dx} = 3 \log a - \frac{3}{4(5-x)} - \frac{3}{4(x+4)}$

(C) Given $f(x) = \log_{x^2}(\log_{e} x)$.
Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(\log_{e} x)}{\log(x^2)} = \frac{\log(\log_{e} x)}{2 \log x}$.
Now,differentiate with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{d}{dx}(\log(\log_{e} x)) \cdot \log x - \log(\log_{e} x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \right]$
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{\log_{e} x} \cdot \frac{1}{x} \cdot \log x - \log(\log_{e} x) \cdot \frac{1}{x}}{(\log x)^2} \right]$
Since $\log_{e} x = \log x$,the expression simplifies to:
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{\log(\log_{e} x)}{x}}{(\log x)^2} \right] = \frac{1 - \log(\log_{e} x)}{2x(\log x)^2}$.
At $x = e$,$\log_{e} e = 1$ and $\log(\log_{e} e) = \log(1) = 0$.
$f^{\prime}(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

(D) Given $f(x) = \log_{x^2}(\log x)$.
Using the change of base formula,$\log_a b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(\log x)}{\log(x^2)} = \frac{\log(\log x)}{2 \log x}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{1}{2} \cdot \frac{\frac{d}{dx}(\log(\log x)) \cdot \log x - \log(\log x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2}$.
$f'(x) = \frac{1}{2} \cdot \frac{(\frac{1}{\log x} \cdot \frac{1}{x}) \cdot \log x - \log(\log x) \cdot \frac{1}{x}}{(\log x)^2}$.
$f'(x) = \frac{1}{2} \cdot \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} = \frac{1 - \log(\log x)}{2x(\log x)^2}$.
At $x = e$,$\log x = \log e = 1$ and $\log(\log x) = \log(1) = 0$.
Substituting these values:
$f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

(A) Given $f(x) = 3^x$,the derivative is $f^{\prime}(x) = 3^x \log 3$. Evaluating at $x = 0$,we get $f^{\prime}(0) = 3^0 \log 3 = \log 3$.
Given $g(x) = 4^x$,the derivative is $g^{\prime}(x) = 4^x \log 4$. Evaluating at $x = 0$,we get $g^{\prime}(0) = 4^0 \log 4 = \log 4$.
Substituting these values into the expression,we have $\frac{f^{\prime}(0)-g^{\prime}(0)}{1+f^{\prime}(0) g^{\prime}(0)} = \frac{\log 3 - \log 4}{1 + (\log 3)(\log 4)}$.
Using the logarithmic property $\log a - \log b = \log \left(\frac{a}{b}\right)$,the numerator becomes $\log \left(\frac{3}{4}\right)$.
Thus,the expression is $\frac{\log \left(\frac{3}{4}\right)}{1 + (\log 3)(\log 4)}$.