If y = cos(sin x^2),then frac{dy}{dx} at x = | vedclass

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(B) Given $y = \cos(\sin x^2)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x

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(B) Given $y = \cos(\sin x^2)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x^2)$$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \cos(x^2) \cdot 2x$Now,substitute $x = \sqrt{\frac{\pi}{2}}$ into the derivative:$\frac{dy}{dx} = -\sin(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) \cdot 2\sqrt{\frac{\pi}{2}}$Since $\cos(\frac{\pi}{2}) = 0$,the entire expression becomes:$\frac{dy}{dx} = -\sin(1) \cdot 0 \cdot 2\sqrt{\frac{\pi}{2}} = 0$.

If $y = \cos(\sin x^2)$,then $\frac{dy}{dx}$ at $x = \sqrt{\frac{\pi}{2}}$ is

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