Q. If $f(x) = \sqrt{1 + \cos^{2}{(x^{2})}},then\ f^{'}\left( \frac{\sqrt{\pi}}{2} \right)$ is
A) $\frac{\sqrt{\pi}}{6}$
B) $- \sqrt{\frac{\pi}{6}}$
C) $\frac{1}{\sqrt{6}}$
D) $\frac{\pi}{\sqrt{6}}$
Explanation
(b)
Given, $f(x) = \sqrt{1 + \cos^{2}{(x^{2})}}$
Now, $f^{'}(x) = \frac{- 2\sin{x^{2}\cos x^{2}}}{\sqrt{1 + \cos^{2}x^{2}}}(x) = \frac{- \sin{2x^{2}}}{\sqrt{1 + \cos^{2}x^{2}}}(x)$
$\Rightarrow f^{'}\left( \frac{\pi}{2} \right) = \frac{\sqrt{\pi}}{2}\frac{\sin{2\left( \frac{\pi}{4} \right)}}{\sqrt{1 + \left( \cos\frac{\pi}{4} \right)^{2}}} = - \sqrt{\frac{\pi}{6}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.