Logarithmic Differentiation Questions in English

(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 $x = 1$,$\frac{h'(x)}{h(x)} = 1 + \log h(x)$.

(B) Given $y = f(x)^{g(x)}$. Taking natural logarithm on both sides,we get $\log y = g(x) \log f(x)$.
Differentiating both sides with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = g'(x) \log f(x) + g(x) \frac{f'(x)}{f(x)}$.
Thus,$\frac{dy}{dx} = y \left[ \frac{g(x)}{f(x)} f'(x) + \log f(x) g'(x) \right]$.
Comparing this with the given expression $\frac{dy}{dx} = y[H(x)f'(x) + G(x)g'(x)]$,we identify $H(x) = \frac{g(x)}{f(x)}$ and $G(x) = \log f(x)$.
Now,we need to evaluate the integral $I = \int \frac{G(x)H(x)f'(x)}{g(x)} dx$.
Substituting the identified functions,$I = \int \frac{\log f(x) \cdot \frac{g(x)}{f(x)} \cdot f'(x)}{g(x)} dx = \int \frac{\log f(x) f'(x)}{f(x)} dx$.
Let $u = \log f(x)$,then $du = \frac{f'(x)}{f(x)} dx$.
Substituting these into the integral,$I = \int u du = \frac{u^2}{2} + c = \frac{[\log f(x)]^2}{2} + c$.

(B) Let $y = \frac{x \cdot 2^x-x}{1-\cos x}$. Then the given equation is $\frac{dy}{dx} = y(f(x) + \log 2)$.
This implies $\frac{1}{y} \frac{dy}{dx} = f(x) + \log 2$,or $\frac{d}{dx}(\ln |y|) = f(x) + \log 2$.
Thus,$f(x) = \frac{d}{dx}(\ln |y|) - \log 2$.
Since $y = \frac{x(2^x-1)}{1-\cos x}$,we have $\ln |y| = \ln |x| + \ln |2^x-1| - \ln |1-\cos x|$.
Differentiating with respect to $x$:
$\frac{d}{dx}(\ln |y|) = \frac{1}{x} + \frac{1}{2^x-1} \cdot (2^x \log 2) - \frac{\sin x}{1-\cos x}$.
Substituting this into the expression for $f(x)$:
$f(x) = \frac{1}{x} + \frac{2^x \log 2}{2^x-1} - \frac{\sin x}{1-\cos x} - \log 2$.
$f(x) = \frac{1}{x} + \frac{2^x \log 2 - (2^x-1) \log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.
$f(x) = \frac{1}{x} + \frac{2^x \log 2 - 2^x \log 2 + \log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.
$f(x) = \frac{1}{x} + \frac{\log 2}{2^x-1} - \frac{\sin x}{1-\cos x}$.

(D) Given $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$.
First,evaluate $f\left(\frac{\pi}{4}\right)$:
$f\left(\frac{\pi}{4}\right)=\frac{\cos ^2(\pi/4)}{1+\sin ^2(\pi/4)}=\frac{(1/\sqrt{2})^2}{1+(1/\sqrt{2})^2}=\frac{1/2}{1+1/2}=\frac{1/2}{3/2}=\frac{1}{3}$.
Now,differentiate $f(x)$ using logarithmic differentiation:
$\ln(f(x)) = 2\ln(\cos x) - \ln(1+\sin^2 x)$.
Differentiating with respect to $x$:
$\frac{f'(x)}{f(x)} = 2\left(\frac{-\sin x}{\cos x}\right) - \frac{2\sin x \cos x}{1+\sin^2 x} = -2\tan x - \frac{\sin 2x}{1+\sin^2 x}$.
At $x = \frac{\pi}{4}$:
$\frac{f'(\pi/4)}{f(\pi/4)} = -2\tan(\pi/4) - \frac{\sin(\pi/2)}{1+\sin^2(\pi/4)} = -2(1) - \frac{1}{1+1/2} = -2 - \frac{1}{3/2} = -2 - \frac{2}{3} = -\frac{8}{3}$.
Since $f(\pi/4) = 1/3$,we have $f'(\pi/4) = -\frac{8}{3} \times \frac{1}{3} = -\frac{8}{9}$.
Finally,calculate $f(\pi/4) - 3f'(\pi/4)$:
$\frac{1}{3} - 3\left(-\frac{8}{9}\right) = \frac{1}{3} + \frac{8}{3} = \frac{9}{3} = 3$.

(D) Given $y=(\sin x)^{x^2}$.
Taking the natural logarithm on both sides:
$\log y = x^2 \log(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \cdot \frac{d}{dx}(x^2)$.
$\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{1}{\sin x} \cdot \cos x + \log(\sin x) \cdot 2x$.
$\frac{1}{y} \frac{dy}{dx} = x^2 \cot x + 2x \log(\sin x)$.
Multiplying by $y$:
$\frac{dy}{dx} = y \cdot (x^2 \cot x + 2x \log(\sin x))$.
Substituting $y = (\sin x)^{x^2}$:
$\frac{dy}{dx} = (\sin x)^{x^2} (x^2 \cot x + 2x \log(\sin x))$.
Distributing the term:
$\frac{dy}{dx} = x^2 (\sin x)^{x^2} \frac{\cos x}{\sin x} + 2x (\sin x)^{x^2} \log(\sin x)$.
$\frac{dy}{dx} = x^2 (\sin x)^{x^2-1} \cos x + 2x (\sin x)^{x^2} \log(\sin x)$.

(C) Given $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.
Taking the natural logarithm on both sides:
$\log y = \log \left( \frac{(x+1)^2 (x-1)^{1/2}}{(x+4)^3 e^x} \right)$
Using logarithmic properties $\log(ab) = \log a + \log b$ and $\log(a/b) = \log a - \log b$:
$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x \log e$
Since $\log e = 1$,we have:
$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1$
Therefore,$\frac{dy}{dx} = y \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$
Substituting the value of $y$:
$\frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$

(C) Given equation: $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$
Dividing both sides by $(\cos x)^{\sin x}$,we get: $y = \frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}} = (\tan x)^{\sin x}$
Taking natural logarithm on both sides: $\ln y = \sin x \cdot \ln(\tan x)$
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{1}{\tan x} \cdot \sec^2 x$
Simplifying the second term: $\sin x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x$
So,$\frac{dy}{dx} = y [\cos x \cdot \ln(\tan x) + \sec x]$
At $x = \frac{\pi}{4}$,$\tan x = 1$,$\sin x = \frac{1}{\sqrt{2}}$,$\cos x = \frac{1}{\sqrt{2}}$,$\sec x = \sqrt{2}$,and $y = (1)^{1/\sqrt{2}} = 1$
Substituting these values: $\left. \frac{dy}{dx} \right|_{x=\frac{\pi}{4}} = 1 \cdot [\frac{1}{\sqrt{2}} \cdot \ln(1) + \sqrt{2}] = 1 \cdot [0 + \sqrt{2}] = \sqrt{2}$

(D) Let $y = u + v$,where $u = x^{\log x}$ and $v = (\log x)^x$.
Taking log on both sides for $u$: $\log u = (\log x)(\log x) = (\log x)^2$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 2(\log x) \cdot \frac{1}{x} \implies \frac{du}{dx} = x^{\log x} \cdot \frac{2 \log x}{x}$.
At $x=e$: $\frac{du}{dx} = e^{\log e} \cdot \frac{2 \log e}{e} = e^1 \cdot \frac{2}{e} = 2$.
Now for $v = (\log x)^x$: $\log v = x \log(\log x)$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x}$.
$\frac{dv}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right]$.
At $x=e$: $\frac{dv}{dx} = (\log e)^e \left[ \log(\log e) + \frac{1}{\log e} \right] = 1^e [ \log(1) + 1 ] = 1 \cdot [0 + 1] = 1$.
Thus,$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = 2 + 1 = 3$.

(A) Let $f(x) = f_1(x) + f_2(x)$,where $f_1(x) = x^{\tan x}$ and $f_2(x) = (\tan x)^x$.
For $f_1(x) = x^{\tan x}$,taking log on both sides: $\log f_1 = \tan x \cdot \log x$.
Differentiating with respect to $x$: $\frac{1}{f_1} \frac{df_1}{dx} = \sec^2 x \cdot \log x + \tan x \cdot \frac{1}{x}$.
So,$\frac{df_1}{dx} = x^{\tan x} \left( \sec^2 x \cdot \log x + \frac{\tan x}{x} \right)$.
For $f_2(x) = (\tan x)^x$,taking log on both sides: $\log f_2 = x \cdot \log(\tan x)$.
Differentiating with respect to $x$: $\frac{1}{f_2} \frac{df_2}{dx} = 1 \cdot \log(\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x$.
So,$\frac{df_2}{dx} = (\tan x)^x \left( \log(\tan x) + \frac{x \sec^2 x}{\tan x} \right)$.
Now,$f^{\prime}(x) = \frac{df_1}{dx} + \frac{df_2}{dx}$.
At $x = \frac{\pi}{4}$:
$f_1^{\prime}\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^1 \left( (\sqrt{2})^2 \log \frac{\pi}{4} + \frac{1}{\pi/4} \right) = \frac{\pi}{4} (2 \log \frac{\pi}{4} + \frac{4}{\pi}) = \frac{\pi}{2} \log \frac{\pi}{4} + 1$.
$f_2^{\prime}\left(\frac{\pi}{4}\right) = (1)^{\pi/4} \left( \log 1 + \frac{\pi/4 \cdot 2}{1} \right) = 1 \cdot (0 + \frac{\pi}{2}) = \frac{\pi}{2}$.
Therefore,$f^{\prime}\left(\frac{\pi}{4}\right) = \frac{\pi}{2} \log \frac{\pi}{4} + 1 + \frac{\pi}{2} = 1 + \frac{\pi}{2} (\log \frac{\pi}{4} + 1) = 1 + \frac{\pi}{2} \log \left( \frac{e \pi}{4} \right)$.

(D) Given,$y=x^{\sqrt{x}}$.
Taking $\ln$ on both sides,we get $\ln y = \sqrt{x} \ln x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \sqrt{x} \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(\sqrt{x})$.
$\frac{1}{y} \frac{dy}{dx} = \sqrt{x} \cdot \frac{1}{x} + \ln x \cdot \frac{1}{2\sqrt{x}}$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}}$.
$\frac{1}{y} \frac{dy}{dx} = \frac{2 + \ln x}{2\sqrt{x}}$.
Therefore,$\frac{dy}{dx} = \frac{y(2 + \ln x)}{2\sqrt{x}}$.
Thus,option $D$ is correct.

(A) Given the equation: $x^y = y^{\sin x} (\tan x)^{\cos x}$
Taking the natural logarithm on both sides:
$y \log x = \sin x \log y + \cos x \log (\tan x)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y \log x) = \frac{d}{dx}(\sin x \log y) + \frac{d}{dx}(\cos x \log \tan x)$
$\frac{dy}{dx} \log x + y \cdot \frac{1}{x} = (\cos x \log y + \sin x \cdot \frac{1}{y} \frac{dy}{dx}) + (-\sin x \log \tan x + \cos x \cdot \frac{1}{\tan x} \cdot \sec^2 x)$
Simplifying the derivative term:
$\cos x \cdot \frac{1}{\tan x} \cdot \sec^2 x = \cos x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x} = \operatorname{cosec} x$
Rearranging the terms to isolate $\left(\log x - \frac{\sin x}{y}\right) \frac{dy}{dx}$:
$\left(\log x - \frac{\sin x}{y}\right) \frac{dy}{dx} = \cos x \log y - \sin x \log (\tan x) + \operatorname{cosec} x - \frac{y}{x}$

(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 the chain rule:
$\frac{d}{dx}(\log h(x)) = \frac{d}{dx}(x^x \log x)$
$\frac{h^{\prime}(x)}{h(x)} = \frac{d}{dx}(x^x) \cdot \log x + x^x \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(x^x) = x^x(1 + \log x)$,we have:
$\frac{h^{\prime}(x)}{h(x)} = x^x(1 + \log x) \log x + x^x \cdot \frac{1}{x}$
$\frac{h^{\prime}(x)}{h(x)} = x^x(1 + \log x) \log x + x^{x-1}$
At $x = 1$:
$\frac{h^{\prime}(1)}{h(1)} = 1^1(1 + \log 1) \log 1 + 1^{1-1}$
$\frac{h^{\prime}(1)}{h(1)} = 1(1 + 0)(0) + 1^0 = 0 + 1 = 1$.
Now,check the options at $x = 1$:
$h(1) = 1^{1^1} = 1$.
Option $C$ is $1 + \log h(x) = 1 + \log(1) = 1 + 0 = 1$.
Thus,the value is $1 + \log h(x)$.

(C) Given $f(x) = \frac{e^{-x} \sin x}{\log_e x}$.
Taking the natural logarithm on both sides: $\ln f(x) = \ln(e^{-x}) + \ln(\sin x) - \ln(\ln x) = -x + \ln(\sin x) - \ln(\ln x)$.
Differentiating with respect to $x$: $\frac{f'(x)}{f(x)} = -1 + \frac{\cos x}{\sin x} - \frac{1}{x \ln x}$.
Since $f'(x) = f(x) \cdot g(x)$,we have $g(x) = \frac{f'(x)}{f(x)} = -1 + \cot x - \frac{1}{x \ln x}$.
Differentiating $g(x)$ with respect to $x$: $g'(x) = -\operatorname{cosec}^2 x - \frac{d}{dx} \left( \frac{1}{x \ln x} \right) = -\operatorname{cosec}^2 x - \left( \frac{-(1 \cdot \ln x + x \cdot \frac{1}{x})}{(x \ln x)^2} \right) = -\operatorname{cosec}^2 x + \frac{\ln x + 1}{x^2 (\ln x)^2}$.
At $x = e$: $g'(e) = -\operatorname{cosec}^2(e) + \frac{\ln e + 1}{e^2 (\ln e)^2} = -\operatorname{cosec}^2(e) + \frac{1 + 1}{e^2 (1)^2} = 2e^{-2} - \operatorname{cosec}^2(e)$.

(C) Given $f(x) = \frac{(x+1) \sinh x}{e^{2x} \tan x} = (x+1) \sinh x e^{-2x} \cot x$.
Taking the natural logarithm on both sides: $\ln|f(x)| = \ln|x+1| + \ln|\sinh x| - 2x + \ln|\cot x|$.
Differentiating with respect to $x$: $\frac{f'(x)}{f(x)} = \frac{d}{dx}(\ln|x+1|) + \frac{d}{dx}(\ln|\sinh x|) + \frac{d}{dx}(-2x) + \frac{d}{dx}(\ln|\cot x|)$.
$\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \frac{\cosh x}{\sinh x} - 2 + \frac{-\operatorname{cosec}^2 x}{\cot x}$.
Since $\frac{\cosh x}{\sinh x} = \operatorname{coth} x$ and $\frac{\operatorname{cosec}^2 x}{\cot x} = \frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \operatorname{cosec} 2x$.
Thus,$\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \operatorname{coth} x - 2 - 2 \operatorname{cosec} 2x$.
Comparing this with the given expression $\frac{1}{x+1} + \operatorname{coth} x + g(x)$,we get $g(x) = -2 - 2 \operatorname{cosec} 2x = -2(1 + \operatorname{cosec} 2x)$.