Q. If $f(x) = x\tan^{- 1}{x,\ then\ f^{'}(1)}\$is equal to
A) $\frac{1}{2} + \frac{\pi}{4}$
B) $- \frac{1}{2} + \frac{\pi}{4}$
C) $- \frac{1}{2} - \frac{\pi}{4}$
D) $\frac{1}{2} - \frac{\pi}{4}$
Explanation
(a)
Given, $f(x) = x\tan^{- 1}x$
$\therefore\ f^{'}(x) = \frac{x}{1 + x^{2}} + \tan^{- 1}x$
$\Rightarrow f^{'}(1) = \frac{1}{1 + 1^{2}} + \tan^{- 1}{\ \ 1 = \frac{1}{2} + \frac{\pi}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.