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If $\mathrm{f}(\mathrm{x})=\sqrt{1+\cos ^2\left(\mathrm{x}^2\right)}$, then the value of $f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)$
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The correct answer is:
$-\sqrt{\frac{\pi}{6}}$
On differentiating (i) w.r.t.x, we get
$f^{\prime}(x)=\frac{-2 \sin x^2 \cos x^2}{\sqrt{1+\cos ^2 x^2}}(x)$
Put, $x=\frac{\sqrt{\pi}}{2}$ in (ii), we get
$\begin{aligned}
& \mathrm{f}^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin 2\left(\frac{\pi}{4}\right)}{\sqrt{1+\frac{1}{2}}} \\
& =-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin \frac{\pi}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{\pi}{6}}
\end{aligned}$
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