Question
Differentiate $\sin ^3 x \cos ^3 x\ \ w.r.t\ x$

Answer

We have, $\frac{d}{d x}\left(\sin ^3 x \cos ^3 x\right)=\sin ^3 x \cdot \frac{d}{d x} \cos ^3 x+\cos ^3 x \cdot \frac{d}{d x}\left(\sin ^3 x\right) [$Using Product Rule of differentiation$]$
$=\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cdot \cos x$
$=-3 \sin ^4 x \cos ^2 x+3 \cos ^4 x \sin ^2 x$
$=3 \sin ^2 x \cos ^2 x\left(-\sin ^2 x+\cos ^2 x\right)$
$=3 \sin ^2 x \cos ^2 x \cdot \cos 2 x$
$=\frac{3}{4} \cdot 4 \sin ^2 x \cos ^2 x \cdot \cos 2 x$
$=\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x$
$=\frac{3}{4} \sin ^2 2 x \cdot \cos 2 x$

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