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If $u=\cos ^3 x, v=\sin ^3 x$, then $\left(\frac{d v}{d u}\right)_{x=\frac{\pi}{4}}$ is equal to
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-1
$\begin{aligned} & u=\cos ^3 x, v=\sin ^3 x \\ & \therefore \frac{d u}{d x}=3 \cos ^2 x(-\sin x) \text { and } \frac{d v}{d x}=3 \sin ^2 x(\cos x) \\ & \therefore \frac{d v}{d u}=\frac{3 \sin ^2 x \cos x}{-3 \sin x \cos ^2 x}=-\tan x \\ & \therefore\left(\frac{d v}{d u}\right)_{x=\frac{\pi}{4}}=-\tan \left(\frac{\pi}{4}=-1\right)\end{aligned}$
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