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Q.
If $f \left(x\right)=\sqrt{1+cos^{2}\left(x^{2}\right)}$, then the value of $f '\left(\frac{\sqrt{\pi}}{2}\right)$
Continuity and Differentiability
Solution:
We have, $f \left(x\right)=\sqrt{1+cos^{2}\left(x^{2}\right)}$
$\Rightarrow f '\left(x\right)=\frac{1}{2}\times\frac{-2\,sin\,x^{2}\,cos\,x^{2}}{\sqrt{1+cos^{2}\,x^{2}}}\left(2x\right)$
$=\frac{-sin\,2x^{2}}{\sqrt{1+cos^{2}\,x^{2}}} (x)$
$\therefore f '\left(\frac{\sqrt{\pi}}{2}\right)$
$=-\frac{\sqrt{\pi}}{2}\cdot\frac{sin\,2\left(\frac{\pi}{4}\right)}{\sqrt{1+cos^{2}\left(\frac{\pi}{4}\right)}}$
$=-\sqrt{\frac{\pi}{6}}$