Logarithmic Differentiation Questions in English

(A) Given $y = x \sin x$.
Taking the natural logarithm on both sides,we get $\ln y = \ln(x \sin x)$.
Using the property $\ln(ab) = \ln a + \ln b$,we have $\ln y = \ln x + \ln(\sin x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{\sin x} \cdot \cos x$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \cot x$.
Thus,the correct option is $A$.

(A) Given $y = \frac{e^{2x} \cos x}{x \sin x} = \frac{e^{2x} \cot x}{x}$.
Taking the natural logarithm on both sides:
$\ln y = \ln(e^{2x}) + \ln(\cot x) - \ln(x) = 2x + \ln(\cot x) - \ln x$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 2 + \frac{1}{\cot x} \cdot (-\csc^2 x) - \frac{1}{x}$.
$\frac{1}{y} \frac{dy}{dx} = 2 - \tan x \cdot \csc^2 x - \frac{1}{x} = 2 - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} - \frac{1}{x} = 2 - \frac{1}{\sin x \cos x} - \frac{1}{x} = 2 - 2 \csc(2x) - \frac{1}{x}$.
Alternatively,using the quotient rule on $y = \frac{e^{2x} \cot x}{x}$:
$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(e^{2x} \cot x) - e^{2x} \cot x \cdot \frac{d}{dx}(x)}{x^2}$.
$\frac{dy}{dx} = \frac{x [2e^{2x} \cot x + e^{2x} (-\csc^2 x)] - e^{2x} \cot x}{x^2}$.
$\frac{dy}{dx} = \frac{e^{2x} [2x \cot x - x \csc^2 x - \cot x]}{x^2}$.
$\frac{dy}{dx} = \frac{e^{2x} [(2x - 1) \cot x - x \csc^2 x]}{x^2}$.

(B) Given $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$.
Taking the natural logarithm on both sides:
$\ln y = \ln 2 + \frac{3}{2} \ln(x - \sin x) - \frac{1}{2} \ln x$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 0 + \frac{3}{2} \cdot \frac{1}{x - \sin x} \cdot (1 - \cos x) - \frac{1}{2} \cdot \frac{1}{x}$.
Multiplying by $y$:
$\frac{dy}{dx} = y \left[ \frac{3}{2} \cdot \frac{1 - \cos x}{x - \sin x} - \frac{1}{2x} \right]$.
Substituting $y$ back:
$\frac{dy}{dx} = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}} \left[ \frac{3}{2} \cdot \frac{1 - \cos x}{x - \sin x} - \frac{1}{2x} \right]$.

(A) Given $y = x^x$.
Taking $\log$ on both sides,we get $\log y = x \log x$.
Differentiating both sides with respect to $x$,we use the product rule on the right side:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x)$
$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}$
$\frac{1}{y} \frac{dy}{dx} = \log x + 1$
Therefore,$\frac{dy}{dx} = y(1 + \log x) = x^x(1 + \log x)$.
Since $1 = \log e$,we can write $1 + \log x = \log e + \log x = \log(ex)$.
Thus,$\frac{dy}{dx} = x^x \log(ex)$.

(B) Given $y = \sqrt {\frac{{1 + x}}{{1 - x}}} .$
Taking natural logarithm on both sides:
$\ln y = \frac{1}{2} \ln(1 + x) - \frac{1}{2} \ln(1 - x).$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} - \frac{1}{2} \cdot \frac{1}{(1 - x)} \cdot (-1)$
$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} + \frac{1}{2(1 - x)}$
$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{(1 - x) + (1 + x)}{(1 + x)(1 - x)} \right]$
$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{2}{1 - {x^2}} \right] = \frac{1}{1 - {x^2}}$
$\frac{{dy}}{{dx}} = \frac{y}{1 - {x^2}} = \frac{1}{1 - {x^2}} \cdot \sqrt {\frac{{1 + x}}{{1 - x}}} $
$\frac{{dy}}{{dx}} = \frac{1}{{(1 - x)(1 + x)}} \cdot \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 - x)}^{1/2}}}} = \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 + x)}^1}{{(1 - x)}^{3/2}}}} = \frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}.$

(A) Given $y = \sqrt {\frac{(x - a)(x - b)}{(x - c)(x - d)}} $.
Taking the natural logarithm on both sides:
$\log y = \frac{1}{2} [\log (x - a) + \log (x - b) - \log (x - c) - \log (x - d)]$
Now,differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x - a} + \frac{1}{x - b} - \frac{1}{x - c} - \frac{1}{x - d} \right]$
Multiplying both sides by $y$:
$\frac{dy}{dx} = \frac{y}{2} \left[ \frac{1}{x - a} + \frac{1}{x - b} - \frac{1}{x - c} - \frac{1}{x - d} \right]$
Thus,the correct option is $A$.

(C) Given $y = (1 + x)^x$.
Taking the natural logarithm on both sides,we get $\log y = x \log(1 + x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[x] \cdot \log(1 + x) + x \cdot \frac{d}{dx}[\log(1 + x)]$.
$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log(1 + x) + x \cdot \frac{1}{1 + x}$.
Multiplying both sides by $y$,we get $\frac{dy}{dx} = y \left[ \log(1 + x) + \frac{x}{1 + x} \right]$.
Substituting $y = (1 + x)^x$,we obtain $\frac{dy}{dx} = (1 + x)^x \left[ \frac{x}{1 + x} + \log(1 + x) \right]$.

(A) Given $y = x^{\sqrt{x}}$.
Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\sqrt{x}})$.
Using the property $\ln(a^b) = b \ln a$,we have $\ln y = \sqrt{x} \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x}) \cdot \ln x + \sqrt{x} \cdot \frac{d}{dx}(\ln x)$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \ln x + \sqrt{x} \cdot \frac{1}{x}$.
Since $\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$,we have $\frac{1}{y} \frac{dy}{dx} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}$.
To combine the terms,write $\frac{1}{\sqrt{x}}$ as $\frac{2}{2\sqrt{x}}$:
$\frac{1}{y} \frac{dy}{dx} = \frac{\ln x + 2}{2\sqrt{x}}$.
Therefore,$\frac{dy}{dx} = y \left( \frac{2 + \ln x}{2\sqrt{x}} \right) = x^{\sqrt{x}} \left( \frac{2 + \ln x}{2\sqrt{x}} \right)$.

(C) Given $y = ({x^x})^x$.
Using the property of exponents $(a^m)^n = a^{mn}$,we have $y = x^{x^2}$.
Taking the natural logarithm on both sides: $\ln y = \ln(x^{x^2}) = x^2 \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)$.
$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x$.
$\frac{1}{y} \frac{dy}{dx} = x(1 + 2 \ln x)$.
Therefore,$\frac{dy}{dx} = y \cdot x(1 + 2 \ln x) = x({x^x})^x(1 + 2 \ln x)$.

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

(A) Let $y = x^{\log_e x}$.
Taking the natural logarithm on both sides:
$\log_e y = \log_e(x^{\log_e x})$
$\log_e y = (\log_e x)(\log_e x) = (\log_e x)^2$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 2(\log_e x) \cdot \frac{d}{dx}(\log_e x)$
$\frac{1}{y} \frac{dy}{dx} = 2(\log_e x) \cdot \frac{1}{x}$
$\frac{dy}{dx} = y \cdot \frac{2 \log_e x}{x}$
Substituting $y = x^{\log_e x}$:
$\frac{dy}{dx} = x^{\log_e x} \cdot \frac{2 \log_e x}{x}$
$\frac{dy}{dx} = 2 x^{(\log_e x - 1)} \log_e x$.

(C) Given $y = x^{(x^x)}$.
Taking natural logarithm on both sides: $\log y = x^x \log x$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$.
We know that if $z = x^x$,then $\log z = x \log x$,so $\frac{1}{z} \frac{dz}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{dz}{dx} = x^x(\log x + 1)$.
Substituting this back: $\frac{1}{y} \frac{dy}{dx} = [x^x(\log x + 1)] \log x + x^x \cdot \frac{1}{x}$.
Since $x^x \cdot \frac{1}{x} = x^{x-1}$,we have:
$\frac{dy}{dx} = y [x^x(\log x + 1)\log x + x^{x-1}]$.

(A) Given $y = x^{\sin x}$.
Taking natural logarithm on both sides,we get $\ln y = \sin x \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \cos x \ln x + \sin x \cdot \frac{1}{x}$.
$\frac{1}{y} \frac{dy}{dx} = \frac{x \cos x \ln x + \sin x}{x}$.
Therefore,$\frac{dy}{dx} = y \left( \frac{x \cos x \ln x + \sin x}{x} \right)$.
Substituting $y = x^{\sin x}$,we get $\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x + x \cos x \ln x}{x} \right)$.

(B) Let $y = (\sin x)^x$.
Taking the natural logarithm on both sides,we get $\ln y = x \ln(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{d}{dx}(\ln(\sin x)) + \ln(\sin x) \cdot \frac{d}{dx}(x)$.
$\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{\sin x} \cdot \cos x + \ln(\sin x) \cdot 1$.
$\frac{1}{y} \frac{dy}{dx} = x \cot x + \ln(\sin x)$.
Therefore,$\frac{dy}{dx} = y [x \cot x + \ln(\sin x)]$.
Substituting $y = (\sin x)^x$:
$\frac{dy}{dx} = (\sin x)^x [x \cot x + \ln(\sin x)]$.
Rewriting $\cot x$ as $\frac{\cos x}{\sin x}$:
$\frac{dy}{dx} = (\sin x)^x \left[ \frac{x \cos x + \sin x \ln(\sin x)}{\sin x} \right]$.

(A) Given $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$.
Taking natural logarithm on both sides:
$\ln y = \ln \left( \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}} \right)$
$\ln y = \ln(\sqrt{x}) + \ln((2x + 3)^2) - \ln(\sqrt{x + 1})$
$\ln y = \frac{1}{2} \ln x + 2 \ln(2x + 3) - \frac{1}{2} \ln(x + 1)$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} + 2 \cdot \frac{1}{2x + 3} \cdot 2 - \frac{1}{2} \cdot \frac{1}{x + 1}$
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)}$
Therefore,$\frac{dy}{dx} = y \left[ \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)} \right]$.

(B) Let $y = (\sin x)^{\log x}$.
Taking the natural logarithm on both sides:
$\log y = \log x \cdot \log \sin x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log \sin x + \log x \cdot \frac{d}{dx}(\log \sin x)$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log \sin x + \log x \cdot \frac{1}{\sin x} \cdot \cos x$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log \sin x + \cot x \log x$.
Therefore,$\frac{dy}{dx} = y \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$.
Substituting $y = (\sin x)^{\log x}$:
$\frac{dy}{dx} = (\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$.

(A) Given $y = (\tan x)^{\cot x}$.
Taking natural logarithm on both sides,we get $\log y = \cot x \log(\tan x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cot x) \cdot \log(\tan x) + \cot x \cdot \frac{d}{dx}(\log(\tan x))$.
Since $\frac{d}{dx}(\cot x) = -\csc^2 x$ and $\frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin(2x)} = 2 \csc(2x)$ or simply $\csc^2 x \cdot \cot x$ is not correct,let us re-evaluate: $\frac{d}{dx}(\log(\tan x)) = \frac{\sec^2 x}{\tan x} = \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x} = \csc x \sec x$.
Wait,$\frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.
Actually,$\frac{d}{dx}(\log(\tan x)) = \frac{\sec^2 x}{\tan x} = \frac{1}{\sin x \cos x} = \csc x \sec x$.
Let us re-calculate: $\frac{1}{y} \frac{dy}{dx} = -\csc^2 x \log(\tan x) + \cot x \cdot (\frac{\sec^2 x}{\tan x}) = -\csc^2 x \log(\tan x) + \cot x \cdot \frac{1}{\sin x \cos x} \cdot \cos x = -\csc^2 x \log(\tan x) + \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x \cos x} = -\csc^2 x \log(\tan x) + \csc^2 x = \csc^2 x (1 - \log \tan x)$.
Therefore,$\frac{dy}{dx} = y \csc^2 x (1 - \log \tan x)$.

(C) Given $y = x^2 + x^{\log x}$.
Let $u = x^2$ and $v = x^{\log x}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
For $u = x^2$,$\frac{du}{dx} = 2x$.
For $v = x^{\log x}$,taking log on both sides: $\log v = \log x \cdot \log x = (\log x)^2$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = 2 \log x \cdot \frac{1}{x}$.
So,$\frac{dv}{dx} = v \cdot \frac{2 \log x}{x} = x^{\log x} \cdot \frac{2 \log x}{x}$.
Thus,$\frac{dy}{dx} = 2x + \frac{2 \log x \cdot x^{\log x}}{x}$.
Taking $\frac{2}{x}$ as common: $\frac{dy}{dx} = \frac{2(x^2 + \log x \cdot x^{\log x})}{x}$.

(C) Given $y = (\tan x)^{(\tan x)^{\tan x}}$.
Taking $\log$ on both sides,we get $\log y = (\tan x)^{\tan x} \log(\tan x)$.
Let $u = (\tan x)^{\tan x}$. Then $\log u = \tan x \log(\tan x)$.
Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = \sec^2 x \log(\tan x) + \tan x \cdot \frac{1}{\tan x} \cdot \sec^2 x = \sec^2 x (\log(\tan x) + 1)$.
So,$\frac{du}{dx} = u \sec^2 x (\log(\tan x) + 1) = (\tan x)^{\tan x} \sec^2 x (\log(\tan x) + 1)$.
Now,$\log y = u \log(\tan x)$.
Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = \frac{du}{dx} \log(\tan x) + u \cdot \frac{1}{\tan x} \cdot \sec^2 x$.
At $x = \frac{\pi}{4}$,$\tan x = 1$,so $u = 1^1 = 1$ and $y = 1^1 = 1$.
Also,$\log(\tan x) = \log 1 = 0$ and $\sec^2 x = \sec^2(\frac{\pi}{4}) = 2$.
Substituting these values,$\frac{du}{dx} = 1 \cdot 2 \cdot (0 + 1) = 2$.
Then $\frac{1}{1} \frac{dy}{dx} = 2 \cdot 0 + 1 \cdot \frac{1}{1} \cdot 2 = 2$.
Therefore,$\frac{dy}{dx} = 2$.

(C) Given $y = x^{\ln x}$.
Taking the natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.
Using the property $\ln(a^b) = b \ln a$,we have $\ln y = (\ln x)(\ln x) = (\ln x)^2$.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{d}{dx}(\ln x)$.
$\frac{1}{y} \frac{dy}{dx} = 2(\ln x) \cdot \frac{1}{x}$.
$\frac{dy}{dx} = y \cdot \frac{2 \ln x}{x}$.
Substituting $y = x^{\ln x}$ back into the equation:
$\frac{dy}{dx} = x^{\ln x} \cdot \frac{2 \ln x}{x} = 2 x^{\ln x} \cdot x^{-1} \cdot \ln x$.
$\frac{dy}{dx} = 2 x^{\ln x - 1} \ln x$.

(C) Given the equation: ${x^m}{y^n} = 2{(x + y)^{m + n}}$
Taking the natural logarithm on both sides:
$\ln({x^m}{y^n}) = \ln(2{(x + y)^{m + n}})$
$m\ln x + n\ln y = \ln 2 + (m + n)\ln(x + y)$
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(m\ln x + n\ln y) = \frac{d}{dx}(\ln 2 + (m + n)\ln(x + y))$
$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = 0 + \frac{m + n}{x + y}(1 + \frac{dy}{dx})$
$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m + n}{x + y} + \frac{m + n}{x + y}\frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx}(\frac{n}{y} - \frac{m + n}{x + y}) = \frac{m + n}{x + y} - \frac{m}{x}$
$\frac{dy}{dx}(\frac{nx + ny - my - ny}{y(x + y)}) = \frac{mx + nx - mx - my}{x(x + y)}$
$\frac{dy}{dx}(\frac{nx - my}{y(x + y)}) = \frac{nx - my}{x(x + y)}$
$\frac{dy}{dx} = \frac{nx - my}{x(x + y)} \times \frac{y(x + y)}{nx - my}$
$\frac{dy}{dx} = \frac{y}{x}$

(A) Given $y = (x \log x)^{\log \log x}$.
Taking logarithm on both sides: $\log y = \log \log x \cdot \log(x \log x)$.
Using the property $\log(ab) = \log a + \log b$,we get $\log y = \log \log x (\log x + \log \log x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log \log x) \cdot (\log x + \log \log x) + \log \log x \cdot \frac{d}{dx}(\log x + \log \log x)$.
$\frac{1}{y} \frac{dy}{dx} = \left( \frac{1}{\log x} \cdot \frac{1}{x} \right) (\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{\log x} \cdot \frac{1}{x} \right)$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x \log x}(\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{x \log x} \right)$.
Therefore,$\frac{dy}{dx} = y \left\{ \frac{1}{x \log x}(\log x + \log \log x) + \log \log x \left( \frac{1}{x} + \frac{1}{x \log x} \right) \right\}$.
Substituting $y = (x \log x)^{\log \log x}$,we get the final expression.

(A) Given the equation: $x^py^q=(x+y)^{p+q}$
Taking the natural logarithm on both sides:
$p \ln x + q \ln y = (p+q) \ln(x+y)$
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 $\frac{dy}{dx}$:
$\frac{q}{y} \frac{dy}{dx} - \frac{p+q}{x+y} \frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$
$\frac{dy}{dx} \left( \frac{q(x+y) - y(p+q)}{y(x+y)} \right) = \frac{x(p+q) - p(x+y)}{x(x+y)}$
$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}$
$\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}$
Dividing both sides by $\frac{qx - py}{x+y}$:
$\frac{dy}{dx} = \frac{y}{x}$

(D) Given the equation $x^p \cdot y^q = (x + y)^{p + q}$.
Taking the natural logarithm on both sides:
$\ln(x^p \cdot y^q) = \ln((x + y)^{p + q})$
$p \ln x + q \ln y = (p + q) \ln(x + y)$
Differentiating both sides with respect to $x$:
$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = (p + q) \frac{1}{x + y} (1 + \frac{dy}{dx})$
$\frac{q}{y} \frac{dy}{dx} - \frac{p + q}{x + y} \frac{dy}{dx} = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{dy}{dx} \left( \frac{q(x + y) - y(p + q)}{y(x + y)} \right) = \frac{x(p + q) - p(x + y)}{x(x + y)}$
$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x + y)} \right) = \frac{px + qx - px - py}{x(x + y)}$
$\frac{dy}{dx} \left( \frac{qx - py}{y(x + y)} \right) = \frac{qx - py}{x(x + y)}$
$\frac{dy}{dx} = \frac{y}{x}$
Since $\frac{dy}{dx} = \frac{y}{x}$,it is independent of both $p$ and $q$.

(D) Given $y = \tan x \tan 2x \tan 3x$.
We know that $\tan 3x = \tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan x \tan 2x}$.
Rearranging this,we get $\tan 3x(1 - \tan x \tan 2x) = \tan 2x + \tan x$.
$\tan 3x - \tan x \tan 2x \tan 3x = \tan 2x + \tan x$.
Thus,$y = \tan x \tan 2x \tan 3x = \tan 3x - \tan 2x - \tan x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan 3x) - \frac{d}{dx}(\tan 2x) - \frac{d}{dx}(\tan x)$.
$\frac{dy}{dx} = 3 \sec^2 3x - 2 \sec^2 2x - \sec^2 x$.
This matches option $C$.
Since the expression can be manipulated into other forms,let us check if $A$ and $B$ are equivalent.
Using logarithmic differentiation,$\ln y = \ln(\tan x) + \ln(\tan 2x) + \ln(\tan 3x)$.
$\frac{1}{y} \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2 2x}{\tan 2x} + \frac{3 \sec^2 3x}{\tan 3x}$.
$\frac{dy}{dx} = y (\frac{1}{\sin x \cos x} + \frac{2}{\sin 2x \cos 2x} + \frac{3}{\sin 3x \cos 3x}) = y (2 \csc 2x + 4 \csc 4x + 6 \csc 6x) = 2y (\csc 2x + 2 \csc 4x + 3 \csc 6x)$.
This matches option $B$.
Since both $B$ and $C$ are correct,option $D$ is the correct choice.

(B) Given $y = x^{x^2}$.
Taking natural logarithm on both sides,we get $\ln y = \ln(x^{x^2})$.
Using the property $\ln(a^b) = b \ln a$,we have $\ln y = x^2 \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)$.
$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x}$.
$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x$.
$\frac{dy}{dx} = y(2x \ln x + x)$.
Substituting $y = x^{x^2}$ and factoring $x$:
$\frac{dy}{dx} = x^{x^2} \cdot x(2 \ln x + 1)$.

(D) Given $y = x^{(\ln x)^{\ln(\ln x)}}$.
Taking natural logarithm on both sides: $\ln y = (\ln x)^{\ln(\ln x)} \cdot \ln x$.
Taking natural logarithm again: $\ln(\ln y) = \ln((\ln x)^{\ln(\ln x)} \cdot \ln x) = \ln((\ln x)^{\ln(\ln x)}) + \ln(\ln x)$.
Using the property $\ln(a^b) = b \ln a$,we get: $\ln(\ln y) = \ln(\ln x) \cdot \ln(\ln x) + \ln(\ln x) = (\ln(\ln x))^2 + \ln(\ln x)$.
Differentiating both sides with respect to $x$: $\frac{1}{\ln y} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2 \ln(\ln x) \cdot \frac{1}{\ln x} \cdot \frac{1}{x} + \frac{1}{\ln x} \cdot \frac{1}{x}$.
$\frac{1}{y \ln y} \cdot \frac{dy}{dx} = \frac{2 \ln(\ln x) + 1}{x \ln x}$.
$\frac{dy}{dx} = \frac{y \ln y}{x \ln x} (2 \ln(\ln x) + 1)$. This matches option $(a)$.
Since $\ln y = (\ln x)^{\ln(\ln x)} \cdot \ln x$,substituting this into the expression for $\frac{dy}{dx}$ gives: $\frac{dy}{dx} = \frac{y}{x \ln x} \cdot (\ln x)^{\ln(\ln x)} \cdot \ln x \cdot (2 \ln(\ln x) + 1) = \frac{y}{x} (\ln x)^{\ln(\ln x)} (2 \ln(\ln x) + 1)$. This matches option $(b)$.
Therefore,the correct answer is $(d)$.

(C) First,simplify the expression for $y$ by finding a common denominator:
$y = \frac{x^2 + 2x(x - 1) + 3(x - 1)(x - 2) + (x - 1)(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 3)}$
$y = \frac{x^2 + 2x^2 - 2x + 3(x^2 - 3x + 2) + (x^2 - 3x + 2)(x - 3)}{(x - 1)(x - 2)(x - 3)}$
$y = \frac{3x^2 - 2x + 3x^2 - 9x + 6 + x^3 - 3x^2 - 3x^2 + 9x + 2x - 6}{(x - 1)(x - 2)(x - 3)}$
$y = \frac{x^3}{(x - 1)(x - 2)(x - 3)}$
Taking the natural logarithm on both sides:
$\ln y = 3 \ln x - \ln(x - 1) - \ln(x - 2) - \ln(x - 3)$
Differentiating with respect to $x$:
$\frac{y'}{y} = \frac{3}{x} - \frac{1}{x - 1} - \frac{1}{x - 2} - \frac{1}{x - 3}$
Multiply by $x$:
$\frac{xy'}{y} = 3 - \frac{x}{x - 1} - \frac{x}{x - 2} - \frac{x}{x - 3}$
$= (1 - \frac{x}{x - 1}) + (1 - \frac{x}{x - 2}) + (1 - \frac{x}{x - 3})$
$= \frac{x - 1 - x}{x - 1} + \frac{x - 2 - x}{x - 2} + \frac{x - 3 - x}{x - 3}$
$= \frac{-1}{x - 1} + \frac{-2}{x - 2} + \frac{-3}{x - 3}$
$= \frac{1}{1 - x} + \frac{2}{2 - x} + \frac{3}{3 - x}$

(B) Given $y = \frac{(1 + 3x)^{1/3} (1 + 4x)^{1/4} (1 + 5x)^{1/5}}{(1 + 7x)^{1/7} (1 + 8x)^{1/8}}$.
Taking the natural logarithm on both sides:
$\ln y = \frac{1}{3} \ln(1 + 3x) + \frac{1}{4} \ln(1 + 4x) + \frac{1}{5} \ln(1 + 5x) - \frac{1}{7} \ln(1 + 7x) - \frac{1}{8} \ln(1 + 8x)$.
Differentiating with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{1 + 3x} + \frac{1}{4} \cdot \frac{4}{1 + 4x} + \frac{1}{5} \cdot \frac{5}{1 + 5x} - \frac{1}{7} \cdot \frac{7}{1 + 7x} - \frac{1}{8} \cdot \frac{8}{1 + 8x}$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1 + 3x} + \frac{1}{1 + 4x} + \frac{1}{1 + 5x} - \frac{1}{1 + 7x} - \frac{1}{1 + 8x}$.
At $x = 0$,note that $y(0) = \frac{\sqrt[3]{1} \sqrt[4]{1} \sqrt[5]{1}}{\sqrt[7]{1} \sqrt[8]{1}} = 1$.
Substituting $x = 0$ into the derivative expression:
$\frac{1}{1} \cdot y'(0) = (1 + 1 + 1 - 1 - 1) = 1$.
Therefore,$y'(0) = 1$.

(A) Given $y = x^{\ln x}$.
Taking natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.
Using the property $\ln(a^b) = b \ln a$,we have $\ln y = \ln x \cdot \ln x = (\ln x)^2$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[(\ln x)^2]$.
Using the chain rule,$\frac{d}{dx}[(\ln x)^2] = 2 \ln x \cdot \frac{d}{dx}(\ln x) = 2 \ln x \cdot \frac{1}{x}$.
Therefore,$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2 \ln x}{x}$.
Multiplying by $y$,we get $\frac{dy}{dx} = y \cdot \frac{2 \ln x}{x}$.
Substituting $y = x^{\ln x}$,we get $\frac{dy}{dx} = x^{\ln x} \cdot \frac{2 \ln x}{x} = 2 \ln x \cdot x^{\ln x - 1}$.

(A) Given $f(x) = |x|^{|\sin x|}$.
Taking natural logarithm on both sides,$\ln f(x) = |\sin x| \ln |x|$.
For $x < 0$,$|x| = -x$ and $|\sin x| = -\sin x$ (since $\sin x < 0$ for $x \in [-\pi, 0)$).
So,$\ln f(x) = -\sin x \ln(-x)$.
Differentiating with respect to $x$:
$\frac{f'(x)}{f(x)} = -\cos x \ln(-x) - \sin x \cdot \frac{1}{-x} \cdot (-1) = -\cos x \ln(-x) - \frac{\sin x}{x}$.
At $x = -\frac{\pi}{4}$:
$f(-\frac{\pi}{4}) = |-\frac{\pi}{4}|^{|\sin(-\frac{\pi}{4})|} = (\frac{\pi}{4})^{1/\sqrt{2}}$.
$f'(-\frac{\pi}{4}) = f(-\frac{\pi}{4}) \left[ -\cos(-\frac{\pi}{4}) \ln(\frac{\pi}{4}) - \frac{\sin(-\frac{\pi}{4})}{-\frac{\pi}{4}} \right]$.
$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ -\frac{1}{\sqrt{2}} \ln(\frac{\pi}{4}) - \frac{-1/\sqrt{2}}{\pi/4} \right]$.
$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{1}{\sqrt{2}} \ln(\frac{4}{\pi}) - \frac{4}{\sqrt{2}\pi} \right]$.
$f'(-\frac{\pi}{4}) = (\frac{\pi}{4})^{1/\sqrt{2}} \left[ \frac{\sqrt{2}}{2} \ln(\frac{4}{\pi}) - \frac{2\sqrt{2}}{\pi} \right]$.

Let $y = \sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$.
Taking the natural logarithm on both sides,we get:
$\log y = \frac{1}{2} [\log(x-3) + \log(x^{2}+4) - \log(3x^{2}+4x+5)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx}(\log(x-3)) + \frac{d}{dx}(\log(x^{2}+4)) - \frac{d}{dx}(\log(3x^{2}+4x+5)) \right]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+5} \right]$.
Therefore,the derivative is:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}} \left[ \frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+5} \right]$.

(A) Let $y = x^{\sin x}$. Taking the natural logarithm on both sides,we get:
$\log y = \sin x \log x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\sin x)$
$\frac{1}{y} \frac{dy}{dx} = \sin x \cdot \frac{1}{x} + \log x \cdot \cos x$
Multiplying by $y$ on both sides:
$\frac{dy}{dx} = y \left( \frac{\sin x}{x} + \cos x \log x \right)$
Substituting $y = x^{\sin x}$:
$\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$
$\frac{dy}{dx} = x^{\sin x - 1} \sin x + x^{\sin x} \cos x \log x$

Let $y = \cos x \cdot \cos 2x \cdot \cos 3x$.
Taking the logarithm on both sides,we obtain:
$\log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x)$
$\Rightarrow \log y = \log (\cos x) + \log (\cos 2x) + \log (\cos 3x)$.
Differentiating both sides with respect to $x$,we obtain:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) + \frac{1}{\cos 2x} \cdot \frac{d}{dx}(\cos 2x) + \frac{1}{\cos 3x} \cdot \frac{d}{dx}(\cos 3x)$.
$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \frac{-\sin x}{\cos x} + \frac{-\sin 2x \cdot 2}{\cos 2x} + \frac{-\sin 3x \cdot 3}{\cos 3x}$.
$\Rightarrow \frac{dy}{dx} = y [-\tan x - 2\tan 2x - 3\tan 3x]$.
Substituting the value of $y$,we get:
$\frac{dy}{dx} = -\cos x \cdot \cos 2x \cdot \cos 3x [\tan x + 2\tan 2x + 3\tan 3x]$.

Let $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$.
Taking the logarithm on both sides,we obtain:
$\log y = \log \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
$\Rightarrow \log y = \frac{1}{2} \log \left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]$
$\Rightarrow \log y = \frac{1}{2} [\log \{(x-1)(x-2)\} - \log \{(x-3)(x-4)(x-5)\}]$
$\Rightarrow \log y = \frac{1}{2} [\log (x-1) + \log (x-2) - \log (x-3) - \log (x-4) - \log (x-5)]$
Differentiating both sides with respect to $x$,we obtain:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$
$\Rightarrow \frac{dy}{dx} = \frac{y}{2} \left( \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right)$
$\therefore \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]$

(A) Let $y = (\log x)^{\cos x}$.
Taking the logarithm on both sides,we get:
$\log y = \cos x \cdot \log(\log x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\cos x) \cdot \log(\log x) + \cos x \cdot \frac{d}{dx}(\log(\log x))$.
$\frac{1}{y} \cdot \frac{dy}{dx} = -\sin x \cdot \log(\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$.
$\frac{1}{y} \cdot \frac{dy}{dx} = -\sin x \cdot \log(\log x) + \frac{\cos x}{\log x} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = y \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$.
Substituting $y = (\log x)^{\cos x}$,we obtain:
$\frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x \log x} - \sin x \log(\log x) \right]$.

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

Let $y = (x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$.
Taking the natural logarithm on both sides,we get:
$\log y = \log((x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4})$
$\log y = 2 \log(x+3) + 3 \log(x+4) + 4 \log(x+5)$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}$
$\frac{dy}{dx} = y \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]$
Substituting $y$ back:
$\frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(x+5)^{4} \left[ \frac{2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} [2(x^{2}+9x+20) + 3(x^{2}+8x+15) + 4(x^{2}+7x+12)]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} [2x^{2}+18x+40 + 3x^{2}+24x+45 + 4x^{2}+28x+48]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} (9x^{2}+70x+133)$

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

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

Let $y = (\sin x)^{x} + \sin^{-1} \sqrt{x}$.
Let $u = (\sin x)^{x}$ and $v = \sin^{-1} \sqrt{x}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ............$(1)$
For $u = (\sin x)^{x}$:
Taking logarithm on both sides,$\log u = x \log(\sin x)$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log(\sin x) + x \cdot \frac{d}{dx}(\log(\sin x))$
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x$
$\frac{du}{dx} = (\sin x)^{x} [\log(\sin x) + x \cot x]$ ............$(2)$
For $v = \sin^{-1} \sqrt{x}$:
Differentiating with respect to $x$ using the chain rule:
$\frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$
$\frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x - x^2}}$ ............$(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\frac{dy}{dx} = (\sin x)^{x} (x \cot x + \log(\sin x)) + \frac{1}{2\sqrt{x - x^2}}$.

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

(N/A) Let $y = x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1}$.
Let $u = x^{x \cos x}$ and $v = \frac{x^{2}+1}{x^{2}-1}$.
Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ .............$(1)$
For $u = x^{x \cos x}$,taking log on both sides:
$\log u = (x \cos x) \log x$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \cos x) \cdot \log x + (x \cos x) \cdot \frac{d}{dx}(\log x)$
$\frac{1}{u} \frac{du}{dx} = (\cos x - x \sin x) \log x + (x \cos x) \cdot \frac{1}{x}$
$\frac{du}{dx} = x^{x \cos x} [\cos x \log x - x \sin x \log x + \cos x]$
$\frac{du}{dx} = x^{x \cos x} [\cos x(1 + \log x) - x \sin x \log x]$ .............$(2)$
For $v = \frac{x^{2}+1}{x^{2}-1}$,using the quotient rule:
$\frac{dv}{dx} = \frac{(x^{2}-1)(2x) - (x^{2}+1)(2x)}{(x^{2}-1)^{2}}$
$\frac{dv}{dx} = \frac{2x^{3} - 2x - 2x^{3} - 2x}{(x^{2}-1)^{2}} = \frac{-4x}{(x^{2}-1)^{2}}$ .............$(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\frac{dy}{dx} = x^{x \cos x} [\cos x(1 + \log x) - x \sin x \log x] - \frac{4x}{(x^{2}-1)^{2}}$

Let $y = (x \cos x)^{x} + (x \sin x)^{\frac{1}{x}}$.
Let $u = (x \cos x)^{x}$ and $v = (x \sin x)^{\frac{1}{x}}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ $(1)$.
For $u = (x \cos x)^{x}$,taking log on both sides:
$\log u = x \log(x \cos x) = x(\log x + \log \cos x) = x \log x + x \log \cos x$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \log x) + \frac{d}{dx}(x \log \cos x)$
$= (1 + \log x) + (\log \cos x - x \tan x)$.
$\frac{du}{dx} = (x \cos x)^{x} [1 - x \tan x + \log(x \cos x)]$ $(2)$.
For $v = (x \sin x)^{\frac{1}{x}}$,taking log on both sides:
$\log v = \frac{1}{x} \log(x \sin x) = \frac{1}{x}(\log x + \log \sin x)$.
Differentiating with respect to $x$:
$\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\frac{1}{x} \log x) + \frac{d}{dx}(\frac{1}{x} \log \sin x)$
$= \frac{1 - \log x}{x^2} + \frac{x \cot x - \log(x \sin x)}{x^2} = \frac{1 + x \cot x - \log(x \sin x)}{x^2}$.
$\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left[ \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right]$ $(3)$.
Combining $(1), (2),$ and $(3)$:
$\frac{dy}{dx} = (x \cos x)^{x} [1 - x \tan x + \log(x \cos x)] + (x \sin x)^{\frac{1}{x}} \left[ \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right]$.

(C) Given the function $(\cos x)^{y}=(\cos y)^{x}$.
Taking the natural logarithm on both sides:
$y \log \cos x = x \log \cos y$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(y \log \cos x) = \frac{d}{dx}(x \log \cos y)$
$\frac{dy}{dx} \log \cos x + y \cdot \frac{1}{\cos x} \cdot (-\sin x) = 1 \cdot \log \cos y + x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx}$
$\frac{dy}{dx} \log \cos x - y \tan x = \log \cos y - x \tan y \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \log \cos x + x \tan y \frac{dy}{dx} = \log \cos y + y \tan x$
$\frac{dy}{dx} (\log \cos x + x \tan y) = y \tan x + \log \cos y$
Therefore,$\frac{dy}{dx} = \frac{y \tan x + \log \cos y}{x \tan y + \log \cos x}$.

(A) Given the function $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$.
Taking the natural logarithm on both sides,we get:
$\log f(x) = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \log(1+x^{8})$.
Differentiating both sides with respect to $x$,we obtain:
$\frac{1}{f(x)} \cdot f^{\prime}(x) = \frac{d}{dx} \log(1+x) + \frac{d}{dx} \log(1+x^{2}) + \frac{d}{dx} \log(1+x^{4}) + \frac{d}{dx} \log(1+x^{8})$.
Using the chain rule,we have:
$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \frac{8x^{7}}{1+x^{8}}$.
Thus,$f^{\prime}(x) = f(x) \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \frac{8x^{7}}{1+x^{8}} \right]$.
Now,to find $f^{\prime}(1)$,we substitute $x=1$:
$f(1) = (1+1)(1+1^{2})(1+1^{4})(1+1^{8}) = 2 \times 2 \times 2 \times 2 = 16$.
$f^{\prime}(1) = f(1) \left[ \frac{1}{1+1} + \frac{2(1)}{1+1^{2}} + \frac{4(1)^{3}}{1+1^{4}} + \frac{8(1)^{7}}{1+1^{8}} \right]$.
$f^{\prime}(1) = 16 \left[ \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right]$.
$f^{\prime}(1) = 16 \left[ \frac{1+2+4+8}{2} \right] = 16 \times \frac{15}{2} = 8 \times 15 = 120$.

Let $y = (x^{2}-5x+8)(x^{3}+7x+9)$.
Taking logarithm on both sides,we obtain:
$\log y = \log(x^{2}-5x+8) + \log(x^{3}+7x+9)$.
Differentiating both sides with respect to $x$,we obtain:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \log(x^{2}-5x+8) + \frac{d}{dx} \log(x^{3}+7x+9)$.
Using the chain rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{2}-5x+8} \cdot (2x-5) + \frac{1}{x^{3}+7x+9} \cdot (3x^{2}+7)$.
Therefore:
$\frac{dy}{dx} = y \left[ \frac{2x-5}{x^{2}-5x+8} + \frac{3x^{2}+7}{x^{3}+7x+9} \right]$.
Substituting $y$ back:
$\frac{dy}{dx} = (x^{2}-5x+8)(x^{3}+7x+9) \left[ \frac{2x-5}{x^{2}-5x+8} + \frac{3x^{2}+7}{x^{3}+7x+9} \right]$.
Expanding the terms:
$\frac{dy}{dx} = (2x-5)(x^{3}+7x+9) + (3x^{2}+7)(x^{2}-5x+8)$.
$\frac{dy}{dx} = (2x^{4} + 14x^{2} + 18x - 5x^{3} - 35x - 45) + (3x^{4} - 15x^{3} + 24x^{2} + 7x^{2} - 35x + 56)$.
Combining like terms:
$\frac{dy}{dx} = 5x^{4} - 20x^{3} + 45x^{2} - 52x + 11$.

Let $y = u \cdot v \cdot w = u \cdot (v \cdot w).$
Method $1$: By applying the product rule repeatedly:
$\frac{dy}{dx} = \frac{du}{dx} \cdot (v \cdot w) + u \cdot \frac{d}{dx}(v \cdot w)$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \left[ \frac{dv}{dx} \cdot w + v \cdot \frac{dw}{dx} \right]$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}.$
Method $2$: By logarithmic differentiation:
Taking the natural logarithm on both sides of $y = u \cdot v \cdot w,$ we get
$\log y = \log u + \log v + \log w.$
Differentiating both sides with respect to $x,$ we obtain
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx}.$
Multiplying both sides by $y,$ we get
$\frac{dy}{dx} = y \left( \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx} \right).$
Substituting $y = u \cdot v \cdot w,$ we obtain
$\frac{dy}{dx} = (u \cdot v \cdot w) \left( \frac{1}{u} \cdot \frac{du}{dx} + \frac{1}{v} \cdot \frac{dv}{dx} + \frac{1}{w} \cdot \frac{dw}{dx} \right)$
$= \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}.$

Given the function $f(x) = (\sin x)^{\sin x}$.
Let $y = (\sin x)^{\sin x}$.
Taking the natural logarithm on both sides,we get:
$\ln y = \ln ((\sin x)^{\sin x}) = \sin x \cdot \ln(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \ln(\sin x) + \sin x \cdot \frac{d}{dx}(\ln(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\sin x) + \sin x \cdot \frac{1}{\sin x} \cdot \cos x$.
$\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\sin x) + \cos x$.
Factoring out $\cos x$:
$\frac{1}{y} \frac{dy}{dx} = \cos x (1 + \ln(\sin x))$.
Multiplying by $y$ to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = y \cdot \cos x (1 + \ln(\sin x))$.
Substituting $y = (\sin x)^{\sin x}$ back into the equation:
$f^{\prime}(x) = (\sin x)^{\sin x} \cos x (1 + \ln(\sin x))$.

Let $y = (5x)^{3 \cos 2x}$.
Taking the natural logarithm on both sides,we get:
$\ln y = 3 \cos 2x \cdot \ln(5x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ \ln(5x) \cdot \frac{d}{dx}(\cos 2x) + \cos 2x \cdot \frac{d}{dx}(\ln 5x) \right]$.
Applying the chain rule:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ \ln(5x) \cdot (-2 \sin 2x) + \cos 2x \cdot \frac{1}{5x} \cdot 5 \right]$.
Simplifying the expression:
$\frac{1}{y} \frac{dy}{dx} = 3 \left[ -2 \sin 2x \ln(5x) + \frac{\cos 2x}{x} \right]$.
Multiplying by $y$:
$\frac{dy}{dx} = 3(5x)^{3 \cos 2x} \left[ \frac{\cos 2x}{x} - 2 \sin 2x \ln(5x) \right]$.
Thus,$\frac{dy}{dx} = (5x)^{3 \cos 2x} \left[ \frac{3 \cos 2x}{x} - 6 \sin 2x \ln(5x) \right]$.