$\frac{d}{d x} [\log {\{e^{x} (\frac{x - 2}{x + 2})\}}^{3 / 4}]$ is

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$\frac{x^{2} + 1}{x^{2} - 4}$

$\frac{x^{2} - 1}{x^{2} - 4}$

$e^{x} \frac{x^{2} - 1}{x^{2} - 4}$

answer is C.

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Detailed Solution

let $y = [\log \{e^{x} {(\frac{x - 2}{x + 2})}^{3 / 4}\}]$
$\begin{array}{l} = \log e^{x} + \log {(\frac{x - 2}{x + 2})}^{3 / 4} \\ \Rightarrow y = x + \frac{3}{4} [\log (x - 2) - \log (x + 2)] \end{array}$
On differentiating w.r.t. x, we get
$\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} [x + \frac{3}{4} {\log (x - 2) - \log (x + 2)}] \\ = 1 + \frac{3}{4} [\frac{1}{x - 2} - \frac{1}{x + 2}] = 1 + \frac{3}{x^{2} - 4} \Rightarrow \frac{d y}{d x} = \frac{x^{2} - 1}{x^{2} - 4} \end{array}$