Q. If $f^{'}(x) = \sin{(\log{x)\ and\ y = f\left( \frac{2x + 3}{3 - 2x} \right)}}$, then $\frac{dy}{dx}$ at $x = 1$ is equal to

A) $6\sin{\log{(5)}}$
B) $5\sin{\log{(6)}}$
C) $12\sin{\log{(5)}}$
D) $5\sin{\log{(12)}}$
Explanation

(c)
$\frac{dy}{dx} = f'\left( \frac{2x + 3}{3 - 2x} \right)\frac{d}{dx}\left( \frac{2x + 3}{3 - 2x} \right)$
$= \sin\left\lbrack \log\ \left( \frac{2x + 3}{3 - 2x} \right) \right\rbrack$
$\left( \frac{(3 - 2x)(2) - (2x + 3)( - 2)}{(3 - 2x)^{2}} \right)$
$= \frac{12}{(3 - 2x)^{2}}\sin\left\lbrack \ {\log\left( \frac{2x + 3}{3 - 2x} \right)} \right\rbrack$
$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 1} = \frac{12}{(3 - 2)^{2}}\sin{\log(5)}$
$= 12\sin{\log 5}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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