Q. $\frac{d}{dx}\left\lbrack \log\left\{ e^{x}\left( \frac{x - 2}{x + 2} \right)^{\frac{3}{4}} \right\} \right\rbrack$is equal to

A) 1
B) $\frac{x^{2} + 1}{x^{2} - 4}$
C) $\frac{x^{2} - 1}{x^{2} - 4}$
D) $e^{x}.\frac{x^{2} - 1}{x^{2} - 4}$
Explanation

(c)
Let $y = \left\lbrack \log\left\{ e^{x}\left( \frac{x - 2}{x + 2} \right)^{3/4} \right\} \right\rbrack$
$= \log{e^{x} + {\log\left( \frac{x - 2}{x + 2} \right)}^{3/4}}$
$\Rightarrow y = x + \frac{3}{4}\lbrack\log{(x - 2) - \log{(x + 2)\rbrack}}$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx} = \frac{d}{dx}\left\lbrack x + \frac{3}{4}\{\log{(x - 2) - \log{(x + 2)\}}} \right\rbrack$
$= 1 + \frac{3}{4}\left\lbrack \frac{1}{x - 2} - \frac{1}{x + 2} \right\rbrack = 1 + \frac{3}{(x^{2} - 4)}$
$\Rightarrow \frac{dy}{dx} = \frac{x^{2} - 1}{x^{2} - 4}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.