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

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Question

(A) Let $y = \log \left\{ e^x \left( \frac{x + 2}{x - 2} \right)^{3/4} \right\}$.Using the property $\log(ab) = \log a + \log b$,we get $y = \log(e^x)

Vedclass · Accepted Answer

$\frac{x^2 - 7}{x^2 - 4}$

Vedclass · Answer

(A) Let $y = \log \left\{ e^x \left( \frac{x + 2}{x - 2} \right)^{3/4} \right\}$.Using the property $\log(ab) = \log a + \log b$,we get $y = \log(e^x) + \log \left( \frac{x + 2}{x - 2} \right)^{3/4}$.Since $\log(e^x) = x$ and $\log(a^n) = n \log a$,we have $y = x + \frac{3}{4} \log \left( \frac{x + 2}{x - 2} \right)$.Using $\log \left( \frac{a}{b} \right) = \log a - \log b$,we get $y = x + \frac{3}{4} [ \log(x + 2) - \log(x - 2) ]$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{3}{4} \left[ \frac{d}{dx} \log(x + 2) - \frac{d}{dx} \log(x - 2) \right]$.$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{1}{x + 2} - \frac{1}{x - 2} \right]$.$\frac{dy}{dx} = 1 + \frac{3}{4} \left[ \frac{(x - 2) - (x + 2)}{(x + 2)(x - 2)} \right] = 1 + \frac{3}{4} \left[ \frac{-4}{x^2 - 4} \right]$.$\frac{dy}{dx} = 1 - \frac{3}{x^2 - 4} = \frac{x^2 - 4 - 3}{x^2 - 4} = \frac{x^2 - 7}{x^2 - 4}$.

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

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