Question

If y = cos (sin x²), then $\frac{dy}{dx}\mathrm{at}x=\sqrt{\frac{\mathrm{\pi }}{2}}$ is equal to ______________________.

Answer 1

$y=\cos \left(\sin x^2\right)$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\cos \left(\sin x^2\right)$

$\Rightarrow \frac{dy}{dx}=-\sin \left(\sin x^2\right)\times \frac{d}{dx}\sin x^2$

$\Rightarrow \frac{dy}{dx}=-\sin \left(\sin x^2\right)\times \cos x^2\times \frac{d}{dx}{x}^2$

$\Rightarrow \frac{dy}{dx}=-\sin \left(\sin x^2\right)\times \cos x^2\times 2x$

$\Rightarrow \frac{dy}{dx}=-2x\cos x^2\sin \left(\sin x^2\right)$

Putting $x=\sqrt{\frac{\mathrm{\pi }}{2}}$, we get

${\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2\times \sqrt{\frac{\mathrm{\pi }}{2}}\times \cos {\left(\sqrt{\frac{\mathrm{\pi }}{2}}\right)}^2\sin \left[\sin {\left(\sqrt{\frac{\mathrm{\pi }}{2}}\right)}^2\right]$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2\times \sqrt{\frac{\mathrm{\pi }}{2}}\times \cos \frac{\mathrm{\pi }}{2}\times \sin \left(\sin \frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2\times \sqrt{\frac{\mathrm{\pi }}{2}}\times 0\times \sin 1$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=0$

Thus, $\frac{dy}{dx}$ at $x=\sqrt{\frac{\mathrm{\pi }}{2}}$ is 0.


If y = cos (sin x²), then $\frac{dy}{dx}\mathrm{at}x=\sqrt{\frac{\mathrm{\pi }}{2}}$ is equal to ___0___.