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

The function is $y = \sin(\cos x^2)$. This is a composite function with three layers: $\sin(...)$, $\cos(...)$, and $x^2$.

Using the chain rule:

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

The derivative of the outer layer:

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

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The derivative of $\cos(x^2)$ requires the chain rule again:

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

$= -\sin(x^2) \times 2x$

$= -2x\sin(x^2)$

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Combining both parts:

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

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

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

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

Therefore:

$\sin(x^2) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

$\cos(x^2) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

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Substituting into the derivative:

$\frac{dy}{dx}\bigg|_{x=\frac{\sqrt{\pi}}{2}} = -2 \times \frac{\sqrt{\pi}}{2} \times \sin\left(\frac{\pi}{4}\right) \times \cos\left(\cos\left(\frac{\pi}{4}\right)\right)$

$= -\sqrt{\pi} \times \frac{1}{\sqrt{2}} \times \cos\left(\frac{1}{\sqrt{2}}\right)$

$= -\frac{\sqrt{\pi}}{\sqrt{2}} \times \cos\left(\frac{1}{\sqrt{2}}\right)$

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

Therefore, the value is $-\sqrt{\frac{\pi}{2}} \cos\left(\frac{1}{\sqrt{2}}\right)$.