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

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

Finding $\frac{dy}{dx}$ at $x = \frac{\sqrt{\pi}}{2}$

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Applying the chain rule to differentiate $y = \cos(\sin x^2)$:

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

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

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

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

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

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Evaluating at $x = \frac{\sqrt{\pi}}{2}$:

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

$= \frac{\pi}{4}$

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

$= \frac{1}{\sqrt{2}}$

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

$= \frac{1}{\sqrt{2}}$

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Substituting into the derivative:

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

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

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

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

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

Therefore, $\frac{dy}{dx}\bigg|_{x=\frac{\sqrt{\pi}}{2}} = -\sqrt{\frac{\pi}{2}} \sin\left(\frac{1}{\sqrt{2}}\right)$