Let $y=\sin(\cos x^2)$, then the value of $\frac{dy}{dx}$ at $x=\frac{\sqrt{\pi}}{2}$ is equal to

$-\sqrt{\frac{\pi}{2}}\cos(\frac{1}{\sqrt{2}})$

The correct answer is Option (2) → $-\sqrt{\frac{\pi}{2}}\cos(\frac{1}{\sqrt{2}})$

$y=\sin(\cos x^{2})$

$\frac{dy}{dx}=\cos(\cos x^{2})\cdot \frac{d}{dx}(\cos x^{2})$

$\frac{d}{dx}(\cos x^{2})=-\sin(x^{2})\cdot 2x$

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

At $x=\frac{\sqrt{\pi}}{2}$:

$x^{2}=\frac{\pi}{4}$

$\sin(x^{2})=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$

$\cos(\cos x^{2})=\cos\left(\frac{\sqrt{2}}{2}\right)$

$\frac{dy}{dx}=-2\left(\frac{\sqrt{\pi}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\cos\left(\frac{\sqrt{2}}{2}\right)$

$\frac{dy}{dx}=-\frac{\sqrt{2\pi}}{2}\cos\left(\frac{\sqrt{2}}{2}\right)$

Final answer: $-\frac{\sqrt{2\pi}}{2}\cos\left(\frac{\sqrt{2}}{2}\right)$