Question

If $f\left(x\right)=\sqrt{1+{\cos }^2\left(x^2\right)}$, then $f^{\prime }\left(\frac{\sqrt{\pi }}{2}\right)$ equal to

• A. $\frac{\sqrt{\pi }}{6}$
• B. $-\sqrt{\frac{\pi }{6}}$
• C. $\frac{1}{\sqrt{6}}$
• D. $\frac{\pi }{\sqrt{6}}$

Answer 1

Answer: (B)

$-\sqrt{\frac{\pi }{6}}$

Given,
$f\left(x\right)=\sqrt{1+{\cos }^2\left(x^2\right)}$ ......(i)

On differentiating equation (i) with respect to $x$, we get

$f^{\prime }\left(x\right)=\frac{-2\sin x^2\cos x^2}{2\sqrt{1+{\cos }^2\left(x^2\right)}}\left(2x\right).........({\cos }^{2}x=\frac{\cos 2x+1}{2})$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-\sin 2x^2}{\sqrt{1+{\cos }^2\left(x^2\right)}}\left(x\right)$

$\therefore f^{\prime }\left(\frac{\sqrt{\pi }}{2}\right)=-\frac{\sqrt{\pi }}{2}\cdot \frac{\sin 2\left(\frac{\pi }{4}\right)}{\sqrt{1+\frac{1}{2}}}=-\sqrt{\frac{\pi }{6}}$