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$y=\cos \left(\operatorname{sir} x^2\right)$, then at $x=\sqrt{\frac{\pi}{2}}, \frac{d y}{d x}=$
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Since $\frac{d y}{d x}=-\sin \left(\sin x^2\right) \cdot \cos x^2 \cdot 2 x$
Therefore, at $x=\sqrt{\frac{\pi}{2}}, \quad \cos x^2=\cos \frac{\pi}{2}=0$
$\frac{d y}{d x}=0$.
Therefore, at $x=\sqrt{\frac{\pi}{2}}, \quad \cos x^2=\cos \frac{\pi}{2}=0$
$\frac{d y}{d x}=0$.
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