For x^2-4 neq 0,the value of frac{d}{d x}left | vedclass

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(A) Let $y = \log \left\{e^x \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}$.Using logarithmic properties,we have:$y = \log e^x + \log \left(\frac

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$\frac{8}{5}$

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(A) Let $y = \log \left\{e^x \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}$.Using logarithmic properties,we have:$y = \log e^x + \log \left(\frac{x-2}{x+2}\right)^{\frac{3}{4}} = x \log e + \frac{3}{4} \log \left(\frac{x-2}{x+2}\right) = x + \frac{3}{4} [\log(x-2) - \log(x+2)]$.Differentiating with respect to $x$:$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x-2} - \frac{1}{x+2} \right] = 1 + \frac{3}{4} \left[ \frac{(x+2) - (x-2)}{(x-2)(x+2)} \right] = 1 + \frac{3}{4} \left[ \frac{4}{x^2-4} \right] = 1 + \frac{3}{x^2-4}$.At $x = 3$:$\left(\frac{dy}{dx}\right)_{x=3} = 1 + \frac{3}{3^2-4} = 1 + \frac{3}{9-4} = 1 + \frac{3}{5} = \frac{8}{5}$.

For $x^2-4 \neq 0$,the value of $\frac{d}{d x}\left[\log \left\{e^x\left(\frac{x-2}{x+2}\right)^{\frac{3}{4}}\right\}\right]$ at $x=3$ is

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