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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Aicharya · Accepted Answer

$\begin{aligned} & u=\cos ^3 x, v=\sin ^3 x \ & 	herefore \frac{d u}{d x}=3 \cos ^2 x(-\sin x) 	ext { and } \frac{d v}{d x}=3 \sin ^2 x(\cos x) \ & 	herefore \frac{d v}{d u}=\frac{3 \sin ^2 x \cos x}{-3 \sin x \cos ^2 x}=-	an x \ & 	herefore\left(\frac{d v}{d u}ight)_{x=\frac{\pi}{4}}=-	an \left(\frac{\pi}{4}=-1ight)\end{aligned}$

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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