Q. $\frac{d}{dx}\left\lbrack \tan^{- 1}\left( \frac{a - x}{1 + ax} \right) \right\rbrack$is equal to
A) $- \frac{1}{1 + x^{2}}$
B) $\frac{1}{1 + a^{2}} - \frac{1}{1 + x^{2}}$
C) $\frac{1}{1 + \left( \frac{a - x}{1 + ax} \right)^{2}}$
D) $\frac{- 1}{\sqrt{1 - \left( \frac{a - x}{1 + ax} \right)^{2}}}$
Explanation
(a)
$\frac{d}{dx}\left\lbrack \tan^{- 1}\left( \frac{a - x}{1 + ax} \right) \right\rbrack = \frac{d}{dx}\lbrack\tan^{- 1}{a - \tan^{- 1}{x\rbrack}}$
$= 0 - \frac{1}{1 + x^{2}} = - \frac{1}{1 + x^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.