frac{d}{dx} left[ log left{ e^x left( frac{x | vedclass

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(C) Let $y = \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\}$.Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $

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

Vedclass · Answer

(C) Let $y = \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\}$.Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:$y = \log(e^x) + \log \left( \frac{x - 2}{x + 2} \right)^{3/4}$$y = x + \frac{3}{4} [\log(x - 2) - \log(x + 2)]$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx}(\log(x - 2)) - \frac{d}{dx}(\log(x + 2)) \right]$$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x - 2} - \frac{1}{x + 2} \right]$$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{(x + 2) - (x - 2)}{(x - 2)(x + 2)} \right]$$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{4}{x^2 - 4} \right]$$\frac{dy}{dx} = 1 + \frac{3}{x^2 - 4} = \frac{x^2 - 4 + 3}{x^2 - 4} = \frac{x^2 - 1}{x^2 - 4}$.

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

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