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If $\phi(x)=\int_{1 / x}^{\sqrt{x}} \sin \left(t^2\right) d t$, then $\phi^{\prime}(1)=$
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The correct answer is:
$\frac{3}{2} \sin 1$
$\begin{aligned} \phi^{\prime}(x) & =\sin x \frac{d}{d x} \sqrt{x}-\sin \frac{1}{x^2} \frac{d}{d x}\left(\frac{1}{x}\right) \\ & =\sin x \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{x^2} \sin \frac{1}{x^2} \\ \Rightarrow \phi^{\prime}(1) & =\frac{1}{2} \sin 1+\sin 1=\frac{3}{2} \sin 1\end{aligned}$
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