Q. If $y = f\left( \frac{2x + 3}{3 - 2x} \right)$ and $f(x) = \sin{(\log{x),\ then\frac{dy}{dx}\ is\ equal\ to\ }}$
A) $\frac{12}{9 - 4x^{2}}\ \cos\left\{ \log\left( \frac{2x + 3}{3 - 2x} \right) \right\}$
B) $\frac{12}{4x^{2} - 9}\cos\left\{ \log\left( \frac{2x + 3}{3 - 2x} \right) \right\}$
C) $\frac{12}{9 - 4x^{2}}\cos\left\{ \log\left( \frac{3 - 2x}{2x + 3} \right) \right\}$
D) $\frac{12}{9 - 4x^{2}}\cos\left\{ \log\left( \frac{2x + 3}{2x - 3} \right) \right\}$
Explanation
(a)
Here, $y = \sin\left\lbrack \log\left( \frac{2x + 3}{3 - 2x} \right) \right\rbrack$
$\Rightarrow \frac{dy}{dx} = \cos\left\lbrack \log\left( \frac{2x + 3}{3 - 2x} \right) \right\rbrack.\frac{3 - 2x}{2x + 3}$.$\frac{12}{(3 - 2x)^{2}}$
$= \frac{12}{9 - 4x^{2}}.\cos\left\lbrack \log\left( \frac{2x + 3}{3 - 2x} \right) \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.