The value of frac{d}{{d(ln x)}}({e^x}{ln ^2}x | vedclass

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(A) Let $u = \ln x$. Then $x = e^u$ and $dx = e^u du$. We want to find the derivative of $f(x) = e^x (\ln x)^2$ with respect to $u = \ln x$. Using the

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$e^e (e + 2)$

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(A) Let $u = \ln x$. Then $x = e^u$ and $dx = e^u du$. We want to find the derivative of $f(x) = e^x (\ln x)^2$ with respect to $u = \ln x$. Using the chain rule,$\frac{df}{du} = \frac{df}{dx} \cdot \frac{dx}{du}$. First,calculate $\frac{df}{dx} = \frac{d}{dx}(e^x (\ln x)^2) = e^x (\ln x)^2 + e^x \cdot 2 \ln x \cdot \frac{1}{x} = e^x (\ln x)^2 + \frac{2 e^x \ln x}{x}$. Since $u = \ln x$,we have $\frac{dx}{du} = \frac{d}{du}(e^u) = e^u = x$. Thus,$\frac{df}{du} = (e^x (\ln x)^2 + \frac{2 e^x \ln x}{x}) \cdot x = x e^x (\ln x)^2 + 2 e^x \ln x$. At $x = e$,we have $\ln x = 1$. Substituting these values: $\frac{df}{du} = e \cdot e^e \cdot (1)^2 + 2 \cdot e^e \cdot 1 = e^{e+1} + 2 e^e = e^e (e + 2)$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

The value of $\frac{d}{{d(\ln x)}}({e^x}{\ln ^2}x)$ at $x=e$ is:

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