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Let $f(x)=\int_{\sin x}^{\cos x} e^{-t^2} d t$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals
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The correct answer is:
$-\sqrt{2 / e}$
$f^{\prime}(x)=-e^{-\cos ^2 x} \sin x-e^{-\sin ^2 x} \cos x$
$\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}$
$\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}$
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