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

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

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

Given: $y = \cos(\sin(x^2))$

Differentiate using chain rule:

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

Now, evaluate at $x = \frac{\sqrt{\pi}}{2}$

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

So,

$\frac{dy}{dx} = -\sin(\sin(\frac{\pi}{4})) \cdot \cos(\frac{\pi}{4}) \cdot 2 \cdot \frac{\sqrt{\pi}}{2}$

Now, $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Thus,

$\frac{dy}{dx} = -\sin\left(\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{\pi}$