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(A) Given the function $f(x) = \log \left[e^x \left(\frac{x-2}{x+2}\right)^{3/4}\right]$.Using the properties of logarithms, $\log(ab) = \log a + \log

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$\frac{1}{4}$

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(A) Given the function $f(x) = \log \left[e^x \left(\frac{x-2}{x+2}\right)^{3/4}\right]$.Using the properties of logarithms, $\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$, we can simplify the expression:$f(x) = \log(e^x) + \log \left(\left(\frac{x-2}{x+2}\right)^{3/4}\right)$$f(x) = x + \frac{3}{4} [\log(x-2) - \log(x+2)]$.Now, differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx}(\log(x-2)) - \frac{d}{dx}(\log(x+2)) \right]$$f'(x) = 1 + \frac{3}{4} \left( \frac{1}{x-2} - \frac{1}{x+2} \right)$.Simplify the term inside the parenthesis:$\frac{1}{x-2} - \frac{1}{x+2} = \frac{(x+2) - (x-2)}{(x-2)(x+2)} = \frac{4}{x^2-4}$.Substituting this back into the derivative:$f'(x) = 1 + \frac{3}{4} \left( \frac{4}{x^2-4} \right) = 1 + \frac{3}{x^2-4}$.Finally, evaluate at $x = 0$:$f'(0) = 1 + \frac{3}{0^2-4} = 1 - \frac{3}{4} = \frac{1}{4}$.

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

Let $f: R \rightarrow R$ be defined by $f(x) = \log \left[e^x \left(\frac{x-2}{x+2}\right)^{3/4}\right]$. Find the value of $f'(0)$.

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