Logarithmic Differentiation Questions in English

Let $y = (\log x)^{\log x}$.
Taking the logarithm on both sides,we get:
$\log y = \log x \cdot \log(\log x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[\log x] \cdot \log(\log x) + \log x \cdot \frac{d}{dx}[\log(\log x)]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \cdot \log(\log x) + \log x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = \frac{\log(\log x)}{x} + \frac{1}{x}$.
Therefore,the derivative is:
$\frac{dy}{dx} = y \left[ \frac{\log(\log x) + 1}{x} \right] = (\log x)^{\log x} \left[ \frac{1 + \log(\log x)}{x} \right]$.

Let $y = (\sin x - \cos x)^{(\sin x - \cos x)}$.
Taking the natural logarithm on both sides,we get:
$\log y = \log [(\sin x - \cos x)^{(\sin x - \cos x)}]$
$\log y = (\sin x - \cos x) \cdot \log (\sin x - \cos x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x - \cos x) \cdot \log (\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{d}{dx} \log (\sin x - \cos x)$.
Since $\frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x$,we have:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \cdot \log (\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} \cdot (\cos x + \sin x)$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) \cdot \log (\sin x - \cos x) + (\cos x + \sin x)$.
Factoring out $(\cos x + \sin x)$:
$\frac{1}{y} \frac{dy}{dx} = (\cos x + \sin x) [1 + \log (\sin x - \cos x)]$.
Multiplying by $y$:
$\frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \log (\sin x - \cos x)]$.

Let $y = x^{x^{2}-3} + (x-3)^{x^{2}}$.
Let $u = x^{x^{2}-3}$ and $v = (x-3)^{x^{2}}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ... $(1)$.
For $u = x^{x^{2}-3}$,taking log on both sides: $\log u = (x^{2}-3) \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = (x^{2}-3) \cdot \frac{1}{x} + \log x \cdot (2x)$.
Thus,$\frac{du}{dx} = x^{x^{2}-3} \left( \frac{x^{2}-3}{x} + 2x \log x \right)$.
For $v = (x-3)^{x^{2}}$,taking log on both sides: $\log v = x^{2} \log (x-3)$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x^{2} \cdot \frac{1}{x-3} + \log (x-3) \cdot (2x)$.
Thus,$\frac{dv}{dx} = (x-3)^{x^{2}} \left( \frac{x^{2}}{x-3} + 2x \log (x-3) \right)$.
Substituting these into $(1)$,we get $\frac{dy}{dx} = x^{x^{2}-3} \left( \frac{x^{2}-3}{x} + 2x \log x \right) + (x-3)^{x^{2}} \left( \frac{x^{2}}{x-3} + 2x \log (x-3) \right)$.

(C) Given $y(x) = x^x$. Taking the natural logarithm on both sides,$\ln y = x \ln x$.
Differentiating with respect to $x$,$\frac{1}{y} y^{\prime} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Thus,$y^{\prime} = y(1 + \ln x) = x^x(1 + \ln x)$.
Now,differentiate $y^{\prime}$ with respect to $x$ using the product rule:
$y^{\prime \prime} = \frac{d}{dx}[x^x] \cdot (1 + \ln x) + x^x \cdot \frac{d}{dx}[1 + \ln x]$
$y^{\prime \prime} = x^x(1 + \ln x)(1 + \ln x) + x^x \cdot \frac{1}{x} = x^x(1 + \ln x)^2 + x^{x-1}$.
At $x = 2$:
$y^{\prime}(2) = 2^2(1 + \ln 2) = 4(1 + \ln 2)$.
$y^{\prime \prime}(2) = 2^2(1 + \ln 2)^2 + 2^{2-1} = 4(1 + \ln 2)^2 + 2$.
Now calculate $y^{\prime \prime}(2) - 2y^{\prime}(2)$:
$= 4(1 + \ln 2)^2 + 2 - 2[4(1 + \ln 2)]$
$= 4(1 + 2\ln 2 + (\ln 2)^2) + 2 - 8 - 8\ln 2$
$= 4 + 8\ln 2 + 4(\ln 2)^2 + 2 - 8 - 8\ln 2$
$= 4(\ln 2)^2 - 2$.

(D) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^4$.
Taking logarithm on both sides:
$\log y = 4[\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 4 \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.
At $x=0$,$y = [1 \cdot 1 \cdot 1 \ldots 1]^4 = 1$.
Substituting $x=0$ into the derivative expression:
$\frac{1}{1} \left( \frac{dy}{dx} \right)_{x=0} = 4 [1 + 2 + 3 + \ldots + n]$.
Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:
$\left( \frac{dy}{dx} \right)_{x=0} = 4 \left( \frac{n(n+1)}{2} \right) = 2n(n+1)$.

(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$
Taking $\log$ on both sides,we get
$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$
Differentiating with respect to $x$,we get
$\frac{1}{y} \frac{dy}{dx} = \frac{3}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$
$\therefore \frac{dy}{dx} = \frac{3y}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$
Now at $x=0$,$y = [(1)(1)(1) \ldots (1)]^{\frac{3}{2}} = 1$
$\left. \frac{dy}{dx} \right|_{x=0} = \frac{3(1)}{2} \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + \ldots + \frac{n}{0+1} \right]$
$= \frac{3}{2} (1 + 2 + 3 + \ldots + n)$
$= \frac{3}{2} \times \frac{n(n+1)}{2}$
$= \frac{3n(n+1)}{4}$

(C) Given $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2n})$.
Taking $\log$ on both sides,we get $\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \dots + \log(1+x^{2n})$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2n \cdot x^{2n-1}}{1+x^{2n}}$.
At $x=0$,the value of $y = (1+0)(1+0) \dots (1+0) = 1$.
Substituting $x=0$ and $y=1$ in the derivative equation,we get $\frac{1}{1} \left. \frac{dy}{dx} \right|_{x=0} = \frac{1}{1+0} + 0 + 0 + \dots + 0$.
Therefore,$\left. \frac{dy}{dx} \right|_{x=0} = 1$.

(C) Given $y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$.
Taking logarithm on both sides:
$\log y = \frac{1}{3} \log (1+3 x) + \frac{1}{4} \log (1+4 x) + \frac{1}{5} \log (1+5 x) - \frac{1}{7} \log (1+7 x) - \frac{1}{8} \log (1+8 x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{3} \cdot \frac{3}{1+3 x} + \frac{1}{4} \cdot \frac{4}{1+4 x} + \frac{1}{5} \cdot \frac{5}{1+5 x} - \frac{1}{7} \cdot \frac{7}{1+7 x} - \frac{1}{8} \cdot \frac{8}{1+8 x}$.
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+3 x} + \frac{1}{1+4 x} + \frac{1}{1+5 x} - \frac{1}{1+7 x} - \frac{1}{1+8 x}$.
At $x=0$,$y = \frac{\sqrt[3]{1} \sqrt[4]{1} \sqrt[5]{1}}{\sqrt[7]{1} \sqrt[8]{1}} = 1$.
Substituting $x=0$ and $y=1$ into the derivative equation:
$\frac{1}{1} \left(\frac{d y}{d x}\right)_{x=0} = \frac{1}{1+0} + \frac{1}{1+0} + \frac{1}{1+0} - \frac{1}{1+0} - \frac{1}{1+0} = 1 + 1 + 1 - 1 - 1 = 1$.
Thus,$\left(\frac{d y}{d x}\right)_{x=0} = 1$.

(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^n$.
Taking the natural logarithm on both sides:
$\log y = n [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = n \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.
At $x=0$,$y = [(1)(1)(1) \ldots (1)]^n = 1^n = 1$.
Substituting $x=0$ and $y=1$ into the derivative expression:
$\frac{1}{1} \left( \frac{dy}{dx} \right)_{x=0} = n [1 + 2 + 3 + \ldots + n]$.
Using the sum formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:
$\left( \frac{dy}{dx} \right)_{x=0} = n \left[ \frac{n(n+1)}{2} \right] = \frac{n^2(n+1)}{2}$.

(D) Given $y = \sqrt{e^{\sqrt{x}}}$.
Taking the natural logarithm on both sides:
$\ln y = \ln(e^{\sqrt{x}})^{1/2} = \frac{1}{2} \sqrt{x} \ln e = \frac{\sqrt{x}}{2}$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{4\sqrt{x}}$.
Therefore,$\frac{dy}{dx} = \frac{y}{4\sqrt{x}}$.
Substituting $y = \sqrt{e^{\sqrt{x}}}$:
$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

(C) Given $x^y = e^{x-y}$.
Taking logarithm on both sides,we get:
$y \log x = x - y$
$y(1 + \log x) = x$
$y = \frac{x}{1 + \log x} \quad \dots(1)$
Differentiating both sides with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(1 + \log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \log x)}{(1 + \log x)^2}$
$\frac{dy}{dx} = \frac{(1 + \log x)(1) - x \cdot (\frac{1}{x})}{(1 + \log x)^2}$
$\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2}$
$\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$

(A) Given,$x^{p} y^{q}=(x+y)^{p+q}$.
Taking the natural logarithm on both sides,we get:
$p \ln x + q \ln y = (p+q) \ln (x+y)$.
Differentiating both sides with respect to $x$:
$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p+q}{x+y} \left(1 + \frac{dy}{dx}\right)$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{q}{y} \frac{dy}{dx} - \frac{p+q}{x+y} \frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$.
$\frac{dy}{dx} \left( \frac{q(x+y) - y(p+q)}{y(x+y)} \right) = \frac{x(p+q) - p(x+y)}{x(x+y)}$.
$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}$.
$\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}$.
Canceling the common term $(qx - py)$ and $(x+y)$ from both sides:
$\frac{dy}{dx} = \frac{y}{x}$.

(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.
Taking natural logarithm on both sides:
$\ln y = \ln \left( e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}} \right)$
$\ln y = \ln(e^{4x}) + \ln \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$
$\ln y = 4x + \frac{3}{4} [\ln(x-4) - \ln(x+3)]$
Differentiating with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{1}{x-4} - \frac{1}{x+3} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{(x+3) - (x-4)}{(x-4)(x+3)} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{7}{(x-4)(x+3)} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{21}{4(x-4)(x+3)}$
Therefore,$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$.

(A) Let $y = u + v$,where $u = x^x$ and $v = x^{\frac{1}{x}}$.
Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
For $u = x^x$,taking $\log$ on both sides: $\log u = x \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
So,$\frac{du}{dx} = x^x(1 + \log x)$.
For $v = x^{\frac{1}{x}}$,taking $\log$ on both sides: $\log v = \frac{1}{x} \log x$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = (-\frac{1}{x^2}) \log x + \frac{1}{x} (\frac{1}{x}) = \frac{1 - \log x}{x^2}$.
So,$\frac{dv}{dx} = x^{\frac{1}{x}} \frac{1 - \log x}{x^2}$.
Therefore,$\frac{dy}{dx} = x^x(1 + \log x) + x^{\frac{1}{x}} \frac{1 - \log x}{x^2}$.

(A) Given the function $f(\theta) = \cos \theta_1 \cdot \cos \theta_2 \cdot \cos \theta_3 \cdots \cos \theta_n$.
Taking the natural logarithm on both sides,we get:
$\ln(f(\theta)) = \ln(\cos \theta_1) + \ln(\cos \theta_2) + \cdots + \ln(\cos \theta_n)$.
Differentiating both sides with respect to $\theta$ (assuming $\theta$ represents the set of variables $\theta_1, \theta_2, \dots, \theta_n$):
$\frac{1}{f(\theta)} \cdot f^{\prime}(\theta) = \frac{1}{\cos \theta_1} \cdot (-\sin \theta_1) \cdot \frac{d\theta_1}{d\theta} + \cdots + \frac{1}{\cos \theta_n} \cdot (-\sin \theta_n) \cdot \frac{d\theta_n}{d\theta}$.
If we consider the derivative with respect to the individual variables or the sum of logarithmic derivatives,we observe that $\frac{f^{\prime}(\theta)}{f(\theta)} = -(\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n)$.
Therefore,$\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n = -\frac{f^{\prime}(\theta)}{f(\theta)}$.

(C) Let $y = x^{\left(x^x\right)}$. Taking the natural logarithm on both sides,we get: $\log y = x^x \log x$.
Now,differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (x^x) \cdot \log x + x^x \cdot \frac{d}{dx} (\log x)$.
To find $\frac{d}{dx} (x^x)$,let $u = x^x$. Then $\log u = x \log x$. Differentiating gives $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
So,$\frac{du}{dx} = x^x (1 + \log x)$.
Substituting this back into the main equation: $\frac{1}{y} \frac{dy}{dx} = [x^x (1 + \log x)] \log x + x^x \cdot \frac{1}{x}$.
$\frac{dy}{dx} = y \left[ x^x \log x (1 + \log x) + x^x \cdot \frac{1}{x} \right]$.
$\frac{dy}{dx} = x^{\left(x^x\right)} \cdot x^x \left( \frac{1}{x} + \log x (1 + \log x) \right)$.

(A) Given $y = \log \left[e^{5x} \left(\frac{3x-4}{x+5}\right)^{\frac{4}{3}}\right]$.
Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:
$y = \log(e^{5x}) + \log\left(\left(\frac{3x-4}{x+5}\right)^{\frac{4}{3}}\right)$
$y = 5x \log e + \frac{4}{3} \log\left(\frac{3x-4}{x+5}\right)$
$y = 5x + \frac{4}{3} [\log(3x-4) - \log(x+5)]$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(5x) + \frac{4}{3} \left[ \frac{d}{dx}(\log(3x-4)) - \frac{d}{dx}(\log(x+5)) \right]$
$\frac{dy}{dx} = 5 + \frac{4}{3} \left[ \frac{1}{3x-4} \cdot 3 - \frac{1}{x+5} \cdot 1 \right]$
$\frac{dy}{dx} = 5 + \frac{4}{3x-4} - \frac{4}{3(x+5)}$.

(D) Given $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$.
Taking the natural logarithm on both sides:
$\log y = 2[\log(x+1) + \log(4x+1) + \log(9x+1) + \ldots + \log(n^2x+1)]$.
Differentiating with respect to $x$ using the chain rule:
$\frac{1}{y} \frac{dy}{dx} = 2 \left[ \frac{1}{x+1} + \frac{4}{4x+1} + \frac{9}{9x+1} + \ldots + \frac{n^2}{n^2x+1} \right]$.
Thus,$\frac{dy}{dx} = 2y \left[ \frac{1^2}{x+1} + \frac{2^2}{4x+1} + \frac{3^2}{9x+1} + \ldots + \frac{n^2}{n^2x+1} \right]$.
At $x=0$,$y(0) = ((0+1)(0+1) \ldots (0+1))^2 = 1^2 = 1$.
Substituting $x=0$ into the derivative expression:
$\left. \frac{dy}{dx} \right|_{x=0} = 2(1) [1^2 + 2^2 + 3^2 + \ldots + n^2]$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:
$\left. \frac{dy}{dx} \right|_{x=0} = 2 \times \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$.

(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.
Taking the natural logarithm on both sides:
$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.
Differentiating both sides with respect to $x$:
$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$.
At $x=1$,$f(1) = (1+1)(1+1)(1+1)(1+1) = 2 \times 2 \times 2 \times 2 = 16$.
Substituting $x=1$ into the derivative expression:
$\frac{f^{\prime}(1)}{f(1)} = \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = \frac{1+2+4+8}{2} = \frac{15}{2}$.
Therefore,$f^{\prime}(1) = f(1) \times \frac{15}{2} = 16 \times \frac{15}{2} = 8 \times 15 = 120$.

(A) Given $y=(\sin x)^{\tan x}$.
Taking the natural logarithm on both sides,we get $\log y = \tan x \cdot \log(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x) \cdot \log(\sin x) + \tan x \cdot \frac{d}{dx}(\log(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + \tan x \cdot \frac{1}{\sin x} \cdot \cos x$.
Since $\frac{\cos x}{\sin x} = \cot x$,we have $\tan x \cdot \cot x = 1$.
Thus,$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1$.
Multiplying by $y$,we get $\frac{dy}{dx} = y[1 + \sec^2 x \log(\sin x)]$.
Substituting $y = (\sin x)^{\tan x}$,we get $\frac{dy}{dx} = (\sin x)^{\tan x}[1 + \sec^2 x \log(\sin x)]$.

(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^2$.
Taking the natural logarithm on both sides,we get:
$\ln y = 2[\ln(x+1) + \ln(2x+1) + \ln(3x+1) + \ldots + \ln(nx+1)]$.
Differentiating both sides with respect to $x$,we get:
$\frac{1}{y} \cdot \frac{dy}{dx} = 2 \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.
Therefore,$\frac{dy}{dx} = 2y \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.
At $x=0$,$y = [(0+1)(0+1)(0+1) \ldots (0+1)]^2 = 1^2 = 1$.
Substituting $x=0$ and $y=1$ into the expression for $\frac{dy}{dx}$:
$\left( \frac{dy}{dx} \right)_{x=0} = 2(1) \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + \ldots + \frac{n}{0+1} \right]$.
$= 2(1 + 2 + 3 + \ldots + n)$.
Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get:
$= 2 \times \frac{n(n+1)}{2} = n(n+1)$.

(A) Given $y = \left(\frac{x^{2}}{x+1}\right)^{x}$.
Taking natural logarithm on both sides:
$\log y = x \log \left(\frac{x^{2}}{x+1}\right) = x [\log(x^{2}) - \log(x+1)] = x [2 \log x - \log(x+1)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x (2 \log x - \log(x+1))]$.
Using the product rule:
$\frac{1}{y} \frac{dy}{dx} = 1 \cdot [2 \log x - \log(x+1)] + x \left[ \frac{2}{x} - \frac{1}{x+1} \right]$.
$\frac{1}{y} \frac{dy}{dx} = \log \left(\frac{x^{2}}{x+1}\right) + x \left[ \frac{2(x+1) - x}{x(x+1)} \right]$.
$\frac{1}{y} \frac{dy}{dx} = \log \left(\frac{x^{2}}{x+1}\right) + x \left[ \frac{2x + 2 - x}{x(x+1)} \right] = \log \left(\frac{x^{2}}{x+1}\right) + \frac{x+2}{x+1}$.
Thus,$\frac{dy}{dx} = y \left[ \frac{x+2}{x+1} + \log \left(\frac{x^{2}}{x+1}\right) \right]$.
Comparing this with the given form $\frac{dy}{dx} = y [g(x) + \log \left(\frac{x^{2}}{x+1}\right)]$,we get $g(x) = \frac{x+2}{x+1}$.

(C) Given $y = x^{x e^{x}}$.
Taking logarithm on both sides,we get $\log y = x e^{x} \log x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{d y}{d x} = \frac{d}{d x}(x e^{x}) \cdot \log x + (x e^{x}) \cdot \frac{d}{d x}(\log x)$.
$\frac{1}{y} \frac{d y}{d x} = (e^{x} + x e^{x}) \log x + (x e^{x}) \cdot \frac{1}{x}$.
$\frac{1}{y} \frac{d y}{d x} = e^{x} \log x + x e^{x} \log x + e^{x}$.
$\frac{1}{y} \frac{d y}{d x} = e^{x}(1 + \log x + x \log x)$.
Since $\frac{d y}{d x} = y \cdot g(x)$,we have $g(x) = e^{x}(1 + \log x + x \log x)$.

(C) Let $u = (\log x)^{x}$.
Taking logarithm on both sides,$\log u = x \log(\log x)$.
Differentiating with respect to $x$,we get $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1$.
So,$\frac{du}{dx} = (\log x)^{x} \left[ \frac{1}{\log x} + \log(\log x) \right]$.
Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^{x} [ \frac{1}{\log x} + \log(\log x) ]}{1/x}$.
Therefore,$\frac{du}{dv} = x(\log x)^{x} \left[ \frac{1}{\log x} + \log(\log x) \right]$.

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.
Now,differentiating with respect to $x$,we have $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x)$.
Thus,$\frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.
Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^x [\frac{1}{\log x} + \log(\log x)]}{1/x} = x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.

(B) Given $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$.
Taking the natural logarithm on both sides:
$\log y = \frac{1}{2} [\log(1-\sin^{-1}x) - \log(1+\sin^{-1}x)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1-\sin^{-1}x} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) - \frac{1}{1+\sin^{-1}x} \cdot \left(\frac{1}{\sqrt{1-x^2}}\right) \right]$.
Simplifying the expression:
$\frac{dy}{dx} = -\frac{y}{2\sqrt{1-x^2}} \left( \frac{1}{1-\sin^{-1}x} + \frac{1}{1+\sin^{-1}x} \right)$.
At $x=0$,$\sin^{-1}(0)=0$ and $y=1$:
$\left(\frac{dy}{dx}\right)_{(0,1)} = -\frac{1}{2\sqrt{1-0}} \left( \frac{1}{1-0} + \frac{1}{1+0} \right) = -\frac{1}{2} (1+1) = -1$.

(C) Given $y = e^{\cos ^{-1}\left(\sqrt{1-x^2}\right)}$.
Taking the natural logarithm on both sides,we get $\log _e y = \cos ^{-1} \sqrt{1-x^2}$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{d}{d x} \left( \cos ^{-1} \sqrt{1-x^2} \right)$.
Using the chain rule,$\frac{d}{d x} \cos ^{-1}(u) = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx}$,where $u = \sqrt{1-x^2}$.
$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{\sqrt{1-(1-x^2)}} \cdot \frac{d}{dx} \left( (1-x^2)^{1/2} \right)$.
$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{\sqrt{x^2}} \cdot \left( \frac{1}{2} (1-x^2)^{-1/2} \cdot (-2x) \right)$.
$\frac{1}{y} \frac{d y}{d x} = \frac{-1}{|x|} \cdot \frac{-x}{\sqrt{1-x^2}}$.
Assuming $x > 0$,$|x| = x$,so $\frac{1}{y} \frac{d y}{d x} = \frac{x}{x \sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}}$.

(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$.
Using $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$,we have:
$y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}}\right)\right] + x^x = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{\sqrt{2}} \cdot \frac{d}{dx}((1+x)^{1/2}) + \frac{d}{dx}(x^x)$.
For $x^x$,let $u = x^x$,then $\ln u = x \ln x$.
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$,so $\frac{du}{dx} = x^x(1 + \ln x)$.
Thus,$\frac{dy}{dx} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$.
At $x = 1$:
$\left[\frac{dy}{dx}\right]_{x=1} = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

(D) Given $y = (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4$.
Taking the natural logarithm on both sides:
$\ln(y) = 2 \ln(x+3) + 3 \ln(x+4) + 4 \ln(x+5)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$.
Thus,$\frac{dy}{dx} = y \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)$.
We can express the sum as $\sum_{i=2}^4 \frac{i}{x+i+1}$.
For $i=2$: $\frac{2}{x+2+1} = \frac{2}{x+3}$.
For $i=3$: $\frac{3}{x+3+1} = \frac{3}{x+4}$.
For $i=4$: $\frac{4}{x+4+1} = \frac{4}{x+5}$.
Therefore,the derivative is $y \sum_{i=2}^4 \left( \frac{i}{x+i+1} \right)$,which matches option $D$.

(A) Let $y = x^{2 x}$.
Taking the natural logarithm on both sides,we get $\log y = \log(x^{2 x})$.
Using the property $\log(a^b) = b \log a$,we have $\log y = 2 x \log x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{d y}{d x} = 2 \cdot \frac{d}{d x}(x \log x)$.
$\frac{1}{y} \frac{d y}{d x} = 2 \left( x \cdot \frac{1}{x} + \log x \cdot 1 \right)$.
$\frac{1}{y} \frac{d y}{d x} = 2(1 + \log x)$.
Therefore,$\frac{d y}{d x} = y \cdot 2(1 + \log x)$.
Substituting $y = x^{2 x}$,we get $\frac{d y}{d x} = 2 x^{2 x}(1 + \log x)$.
Thus,the correct option is $A$.

(A) Given $y=(1+x)(1+x^{2})(1+x^{4})$.
We can write $y = \frac{(1-x)(1+x)(1+x^{2})(1+x^{4})}{1-x} = \frac{1-x^{8}}{1-x}$ for $x \neq 1$.
Using the quotient rule,$\frac{dy}{dx} = \frac{(1-x)(-8x^{7}) - (1-x^{8})(-1)}{(1-x)^{2}} = \frac{-8x^{7}+8x^{8}+1-x^{8}}{(1-x)^{2}} = \frac{7x^{8}-8x^{7}+1}{(1-x)^{2}}$.
Applying $L$'Hopital's rule or expanding the series,we evaluate the limit as $x \to 1$.
Alternatively,using logarithmic differentiation: $\ln y = \ln(1+x) + \ln(1+x^{2}) + \ln(1+x^{4})$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}}$.
At $x=1$,$y = (2)(2)(2) = 8$.
$\frac{1}{8} \frac{dy}{dx} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} = \frac{1}{2} + 1 + 2 = 3.5$.
$\frac{dy}{dx} = 8 \times 3.5 = 28$.

(B) Given,$y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$.
Taking the natural logarithm on both sides:
$\ln y = \ln(\sin x) + \ln(\sin 2x) + \ln(\sin 3x) + \ldots + \ln(\sin nx)$
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{1}{y} \cdot y^{\prime} = \frac{d}{dx}(\ln \sin x) + \frac{d}{dx}(\ln \sin 2x) + \ldots + \frac{d}{dx}(\ln \sin nx)$
$\frac{y^{\prime}}{y} = \frac{\cos x}{\sin x} + 2 \cdot \frac{\cos 2x}{\sin 2x} + 3 \cdot \frac{\cos 3x}{\sin 3x} + \ldots + n \cdot \frac{\cos nx}{\sin nx}$
$\frac{y^{\prime}}{y} = \sum_{k=1}^{n} k \cot kx$
Therefore,$y^{\prime} = y \cdot \sum_{k=1}^{n} k \cot kx$.

(D) Given that,$x^{m} y^{n}=(x+y)^{m+n}$.
Taking the natural logarithm on both sides,we get:
$m \ln x + n \ln y = (m+n) \ln (x+y)$
Differentiating both sides with respect to $x$,we have:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = (m+n) \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx}$
$\left(\frac{n}{y} - \frac{m+n}{x+y}\right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}$
$\left(\frac{nx + ny - my - ny}{y(x+y)}\right) \frac{dy}{dx} = \frac{mx + nx - mx - my}{x(x+y)}$
$\left(\frac{nx - my}{y(x+y)}\right) \frac{dy}{dx} = \frac{nx - my}{x(x+y)}$
Canceling the common term $(nx - my)$ from both sides:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x}$
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(B) Given $y = 2x^{3x}$.
Taking natural logarithm on both sides: $\ln y = \ln(2x^{3x}) = \ln 2 + 3x \ln x$.
Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = 0 + 3 \ln x + 3x \cdot \frac{1}{x} = 3 \ln x + 3$.
Thus,$\frac{dy}{dx} = y(3 \ln x + 3) = 2x^{3x}(3 \ln x + 3)$.
At $x = 1$,$\frac{dy}{dx} = 2(1)^{3(1)}(3 \ln 1 + 3)$.
Since $\ln 1 = 0$,we have $\frac{dy}{dx} = 2(1)(0 + 3) = 2 \times 3 = 6$.

(C) Given,$y=x^{\sin x}+(\sin x)^x$.
Let $u=x^{\sin x}$ and $v=(\sin x)^x$.
Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
For $u=x^{\sin x}$,taking log on both sides: $\log u = \sin x \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \sin x \cdot \frac{1}{x} + \cos x \log x$.
So,$\frac{du}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$.
For $v=(\sin x)^x$,taking log on both sides: $\log v = x \log(\sin x)$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{\cos x}{\sin x} + \log(\sin x) = x \cot x + \log(\sin x)$.
So,$\frac{dv}{dx} = (\sin x)^x (x \cot x + \log(\sin x))$.
At $x = \frac{\pi}{2}$:
$\frac{du}{dx} = (\frac{\pi}{2})^{\sin(\pi/2)} (\frac{\sin(\pi/2)}{\pi/2} + \cos(\pi/2) \log(\pi/2)) = (\frac{\pi}{2})^1 (\frac{1}{\pi/2} + 0) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1$.
$\frac{dv}{dx} = (\sin(\pi/2))^{\pi/2} (\frac{\pi}{2} \cot(\pi/2) + \log(\sin(\pi/2))) = (1)^{\pi/2} (\frac{\pi}{2} \cdot 0 + \log(1)) = 1 \cdot (0 + 0) = 0$.
Therefore,$\frac{dy}{dx} = 1 + 0 = 1$.

(D) Given $y=(x-1)^{2}(x-2)^{3}(x-3)^{5}$.
Taking the natural logarithm on both sides:
$\log y = \log [(x-1)^{2}(x-2)^{3}(x-3)^{5}]$
Using logarithmic properties,$\log y = 2 \log (x-1) + 3 \log (x-2) + 5 \log (x-3)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3}$.
Therefore,$\frac{dy}{dx} = y \left[ \frac{2}{x-1} + \frac{3}{x-2} + \frac{5}{x-3} \right]$.
Substituting $x=4$:
$y(4) = (4-1)^{2}(4-2)^{3}(4-3)^{5} = 3^{2} \times 2^{3} \times 1^{5} = 9 \times 8 \times 1 = 72$.
$\left( \frac{dy}{dx} \right)_{x=4} = 72 \left[ \frac{2}{4-1} + \frac{3}{4-2} + \frac{5}{4-3} \right] = 72 \left[ \frac{2}{3} + \frac{3}{2} + 5 \right]$.
$= 72 \left[ \frac{4 + 9 + 30}{6} \right] = 72 \times \frac{43}{6} = 12 \times 43 = 516$.

(C) Let $y = u + v$,where $u = (\log x)^{1/x}$ and $v = x^{\log x}$.
Taking log on both sides for $u$: $\log u = \frac{1}{x} \log(\log x)$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{1}{x} \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot (-\frac{1}{x^2}) = \frac{1}{x^2 \log x} - \frac{\log(\log x)}{x^2}$.
At $x=e$,$\log u = \frac{1}{e} \log(\log e) = \frac{1}{e} \log(1) = 0$,so $u = e^0 = 1$.
Thus,$\frac{du}{dx} = u \left[ \frac{1}{e^2 \log e} - \frac{\log(\log e)}{e^2} \right] = 1 \left[ \frac{1}{e^2} - 0 \right] = \frac{1}{e^2}$.
For $v = x^{\log x}$,taking log: $\log v = \log x \cdot \log x = (\log x)^2$.
Differentiating: $\frac{1}{v} \frac{dv}{dx} = 2 \log x \cdot \frac{1}{x}$.
At $x=e$,$\log v = (\log e)^2 = 1$,so $v = e^1 = e$.
Thus,$\frac{dv}{dx} = v \left[ \frac{2 \log e}{e} \right] = e \left[ \frac{2}{e} \right] = 2$.
Therefore,$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = \frac{1}{e^2} + 2$.

(D) Given $y = \sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}$.
At $x = 2$,we calculate the value of $y$:
$y = \sqrt{\frac{2^4 \sqrt{3(2)-5}}{(2^2-3)(2(2)-3)}} = \sqrt{\frac{16 \sqrt{1}}{(4-3)(4-3)}} = \sqrt{\frac{16}{1 \times 1}} = \sqrt{16} = 4$.
Taking the natural logarithm on both sides:
$\ln(y) = \frac{1}{2} [4 \ln(x) + \frac{1}{2} \ln(3x-5) - \ln(x^2-3) - \ln(2x-3)]$.
Differentiating with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} [\frac{4}{x} + \frac{1}{2} \cdot \frac{3}{3x-5} - \frac{2x}{x^2-3} - \frac{2}{2x-3}]$.
Substitute $x = 2$ and $y = 4$:
$\frac{1}{4} \left(\frac{dy}{dx}\right)_{x=2} = \frac{1}{2} [\frac{4}{2} + \frac{3}{2(1)} - \frac{4}{4-3} - \frac{2}{4-3}] = \frac{1}{2} [2 + 1.5 - 4 - 2] = \frac{1}{2} [-2.5] = -1.25$.
Therefore,$\left(\frac{dy}{dx}\right)_{x=2} = 4 \times (-1.25) = -5$.

(C) Given $f(x) = x^{\operatorname{Sec}^{-1} x}$. Taking the natural logarithm on both sides,we get $\log f(x) = \operatorname{Sec}^{-1} x \cdot \log x$.
Differentiating both sides with respect to $x$,we have $\frac{1}{f(x)} f^{\prime}(x) = \frac{d}{dx} (\operatorname{Sec}^{-1} x) \cdot \log x + \operatorname{Sec}^{-1} x \cdot \frac{d}{dx} (\log x)$.
Using the derivative formula $\frac{d}{dx} \operatorname{Sec}^{-1} x = \frac{1}{|x| \sqrt{x^2 - 1}}$,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{\log x}{x \sqrt{x^2 - 1}} + \frac{\operatorname{Sec}^{-1} x}{x}$.
Thus,$f^{\prime}(x) = x^{\operatorname{Sec}^{-1} x} \left( \frac{\log x}{x \sqrt{x^2 - 1}} + \frac{\operatorname{Sec}^{-1} x}{x} \right)$.
At $x = 2$,$\operatorname{Sec}^{-1} 2 = \frac{\pi}{3}$.
So,$f^{\prime}(2) = 2^{\pi / 3} \left( \frac{\log 2}{2 \sqrt{2^2 - 1}} + \frac{\pi / 3}{2} \right) = 2^{\pi / 3} \left( \frac{\log 2}{2 \sqrt{3}} + \frac{\pi}{6} \right)$.
Simplifying,$f^{\prime}(2) = 2^{\pi / 3} \left( \frac{\sqrt{3} \log 2}{6} + \frac{\pi}{6} \right) = \frac{2^{\pi / 3}}{6} (\pi + \sqrt{3} \log 2)$.

(A) Given $f(x) = x^{\tan ^{-1} x}$. Taking logarithm on both sides,$\log f(x) = \tan ^{-1} x \cdot \log x$.
Differentiating with respect to $x$,$\frac{1}{f(x)} \frac{df}{dx} = \frac{1}{1+x^2} \log x + \frac{\tan ^{-1} x}{x}$.
Thus,$\frac{df}{dx} = x^{\tan ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{\tan ^{-1} x}{x} \right]$.
Now,$g(x) = \sec ^{-1} \left( \frac{1}{2x^2-1} \right) = \cos ^{-1} (2x^2-1)$.
Using the substitution $x = \cos \theta$,$g(x) = \cos ^{-1} (2 \cos^2 \theta - 1) = \cos ^{-1} (\cos 2\theta) = 2\theta = 2 \cos ^{-1} x$.
Differentiating $g(x)$ with respect to $x$,$\frac{dg}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{2}{\sqrt{1-x^2}}$.
Finally,the derivative of $f(x)$ with respect to $g(x)$ is $\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{x^{\tan ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{\tan ^{-1} x}{x} \right]}{-2/\sqrt{1-x^2}} = -\frac{1}{2} \sqrt{1-x^2} x^{\tan ^{-1} x} \left[ \frac{\log x}{1+x^2} + \frac{\tan ^{-1} x}{x} \right]$.

(A) Let $y = x^{\sin x} + (\sin x)^x$.
Let $U = x^{\sin x}$ and $V = (\sin x)^x$.
Then $\frac{dy}{dx} = \frac{dU}{dx} + \frac{dV}{dx}$.
For $U = x^{\sin x}$,taking $\log$ on both sides: $\log U = \sin x \log x$.
Differentiating with respect to $x$: $\frac{1}{U} \frac{dU}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{\sin x}{x} + \cos x \log x$.
So,$\frac{dU}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x]$.
For $V = (\sin x)^x$,taking $\log$ on both sides: $\log V = x \log(\sin x)$.
Differentiating with respect to $x$: $\frac{1}{V} \frac{dV}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log(\sin x) + x \cot x$.
So,$\frac{dV}{dx} = (\sin x)^x [\log(\sin x) + x \cot x]$.
Thus,$\frac{dy}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x] + (\sin x)^x [\log(\sin x) + x \cot x]$.

(A) Let $u = x^{\sin x}$. Taking logarithm on both sides,we get $\log u = \sin x \log x$.
Differentiating with respect to $x$,we have:
$\frac{1}{u} \frac{du}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \cos x \log x + \frac{\sin x}{x}$.
Thus,$\frac{du}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)$.
Let $v = (\sin x)^x$. Taking logarithm on both sides,we get $\log v = x \log \sin x$.
Differentiating with respect to $x$,we have:
$\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log \sin x + x \cdot \frac{1}{\sin x} \cdot \cos x = \log \sin x + x \cot x$.
Thus,$\frac{dv}{dx} = (\sin x)^x (x \cot x + \log \sin x)$.
The rate of change of $u$ with respect to $v$ is $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)}{(\sin x)^x (x \cot x + \log \sin x)}$.

(C) Assertion $(A)$: Let $y = \frac{x^2 \sin x}{\log x}$. Taking $\log$ on both sides,$\log y = \log(x^2) + \log(\sin x) - \log(\log x) = 2 \log x + \log(\sin x) - \log(\log x)$.
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{1}{\log x} \cdot \frac{1}{x} = \frac{2}{x} + \cot x - \frac{1}{x \log x}$.
Thus,$\frac{dy}{dx} = y \left(\cot x + \frac{2}{x} - \frac{1}{x \log x}\right) = \frac{x^2 \sin x}{\log x} \left(\cot x + \frac{2}{x} - \frac{1}{x \log x}\right)$. So,$A$ is true.
Reason $(R)$: Using logarithmic differentiation,$\frac{d}{dx} \left(\frac{uv}{w}\right) = \frac{uv}{w} \frac{d}{dx} (\log u + \log v - \log w) = \frac{uv}{w} \left(\frac{u'}{u} + \frac{v'}{v} - \frac{w'}{w}\right)$.
The given formula in $R$ has a plus sign instead of a minus sign for the $w$ term,so $R$ is false.

(C) Let $y = f(x) = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.
Taking the natural logarithm on both sides:
$\log y = \log \left[ \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right]$
$\log y = 2 \log (x+1) + \frac{1}{2} \log (x-1) - 3 \log (x+4) - x$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1$.
Thus,$\frac{d y}{d x} = y \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$.
Comparing this with the given equation,we identify $f(x) = y = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.
Now,substitute $x = 5$ into $f(x)$:
$f(5) = \frac{(5+1)^2 \sqrt{5-1}}{(5+4)^3 e^5} = \frac{6^2 \sqrt{4}}{9^3 e^5} = \frac{36 \times 2}{729 e^5} = \frac{72}{729 e^5} = \frac{8}{81 e^5}$.

(A) Let $y = \frac{x^2+1}{(x^2+5)(x^2+9)}$.
Taking the natural logarithm on both sides:
$\log y = \log(x^2+1) - \log(x^2+5) - \log(x^2+9)$.
Differentiating with respect to $x$:
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2x}{x^2+1} - \frac{2x}{x^2+5} - \frac{2x}{x^2+9}$.
$\frac{dy}{dx} = y \cdot 2x \left[ \frac{1}{x^2+1} - \frac{1}{x^2+5} - \frac{1}{x^2+9} \right]$.
Substituting $y = \frac{x^2+1}{(x^2+5)(x^2+9)}$:
$\frac{dy}{dx} = \frac{2x(x^2+1)}{(x^2+5)(x^2+9)} \left[ \frac{1}{x^2+1} - \frac{1}{x^2+5} - \frac{1}{x^2+9} \right]$.
Comparing this with the given expression,we get:
$f(x) = x^2+1, g(x) = x^2+5, h(x) = x^2+9$.
Now,calculate $2h(x) - f(x) - g(x)$:
$2(x^2+9) - (x^2+1) - (x^2+5) = 2x^2 + 18 - x^2 - 1 - x^2 - 5 = 12$.

(C) Given $y = (\log_{x} \sin x)^{x}$.
Taking $\log$ on both sides: $\log y = x \log(\log_{x} \sin x)$.
Using the change of base formula,$\log_{x} \sin x = \frac{\log \sin x}{\log x}$.
So,$\log y = x \log \left( \frac{\log \sin x}{\log x} \right) = x [\log(\log \sin x) - \log(\log x)]$.
Differentiating with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 1 \cdot [\log(\log \sin x) - \log(\log x)] + x \left[ \frac{1}{\log \sin x} \cdot \frac{1}{\sin x} \cdot \cos x - \frac{1}{\log x} \cdot \frac{1}{x} \right]$.
$\frac{1}{y} \frac{dy}{dx} = \log(\log_{x} \sin x) + x \left[ \frac{\cot x}{\log \sin x} - \frac{1}{x \log x} \right]$.
$\frac{dy}{dx} = y \left[ \log(\log_{x} \sin x) + \frac{x \cot x}{\log \sin x} - \frac{1}{\log x} \right]$.

(A) Given $y = (\tan x)^{\sin x}$.
Taking the natural logarithm on both sides,we get $\log y = \sin x \log(\tan x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \cos x \log(\tan x) + \sin x \cdot \frac{1}{\tan x} \cdot \sec^2 x$.
Since $\frac{\sin x}{\tan x} = \cos x$,the expression simplifies to:
$\frac{1}{y} \frac{dy}{dx} = \cos x \log(\tan x) + \cos x \cdot \sec^2 x$.
Note that $\cos x \cdot \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \sec x$.
Therefore,$\frac{dy}{dx} = y \{\sec x + \cos x \log(\tan x)\}$.
Substituting $y = (\tan x)^{\sin x}$,we get $\frac{dy}{dx} = (\tan x)^{\sin x} \{\sec x + \cos x \log(\tan x)\}$.

(B) Given the function $f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x}$ for $a>0$.
Taking the natural logarithm on both sides:
$\log f(x) = (a+1+2x) \log \left(\frac{a+x}{1+x}\right)$.
Differentiating both sides with respect to $x$:
$\frac{f^{\prime}(x)}{f(x)} = 2 \log \left(\frac{a+x}{1+x}\right) + (a+1+2x) \left(\frac{1}{a+x} - \frac{1}{1+x}\right)$.
Evaluating at $x=0$:
$\frac{f^{\prime}(0)}{f(0)} = 2 \log \left(\frac{a}{1}\right) + (a+1) \left(\frac{1}{a} - 1\right)$.
$\frac{f^{\prime}(0)}{f(0)} = 2 \log a + (a+1) \left(\frac{1-a}{a}\right)$.
Since $f(0) = a^{a+1}$,we have:
$f^{\prime}(0) = a^{a+1} \left[ 2 \log a + \frac{(a+1)(1-a)}{a} \right]$.
$f^{\prime}(0) = a^{a+1} \left[ 2 \log a + \frac{1-a^2}{a} \right]$.
Thus,the correct option is $B$.

(C) Given $h(x) = x^{x^x}$.
Taking the natural logarithm on both sides,we get $\log h(x) = x^x \log x$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$\frac{h'(x)}{h(x)} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$.
We know that $\frac{d}{dx}(x^x) = x^x(1 + \log x)$.
Substituting this into the equation:
$\frac{h'(x)}{h(x)} = x^x(1 + \log x) \log x + x^x \cdot \frac{1}{x} = x^x(1 + \log x) \log x + x^{x-1}$.
At $x = 1$,we have $h(1) = 1^{1^1} = 1$,so $\log h(1) = \log 1 = 0$.
Substituting $x = 1$ into the expression for $\frac{h'(x)}{h(x)}$:
$\frac{h'(1)}{h(1)} = 1^1(1 + \log 1) \log 1 + 1^{1-1} = 1(1 + 0)(0) + 1^0 = 0 + 1 = 1$.
Since $\log h(1) = 0$,the expression $1 + \log h(1) = 1 + 0 = 1$.
Thus,at $x = 1$,$\frac{h'(x)}{h(x)} = 1 + \log h(x)$.