Q. If $\frac{d}{dx}\{ f(x) = \frac{1}{1 + x^{2}}$, then $\frac{d}{dx}\{ f\left( x^{3} \right)\}$ is
A) $\frac{3x}{1 + x^{3}}$
B) $\frac{3x^{2}}{1 + x^{6}}$
C) $\frac{- 6x^{5}}{\left( 1 + x^{6} \right)^{2}}$
D) $\frac{- 6x^{5}}{1 + x^{6}}$
Explanation
(b)
Given , $\frac{d}{dx}\{ f(x)\} = \frac{1}{1 + x^{2}}$
On integrating both sides, we get
$f(x) = \tan^{- 1}x$
$\therefore\ \frac{d}{dx}\ f\left( x^{3} \right) = \frac{d}{dx}\ (\tan^{- 1}{x^{3})}$
$= \frac{1}{1 + \left( x^{3} \right)^{2}}.3x^{2}$
$= \frac{3x^{2}}{1 + x^{6}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.