Differentiate ^3 x ^3 x w.r.t x - Vidyadip

Question

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

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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]
$\begin{array}{l}=\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cdot \cos x \\ =-3 \sin ^4 x \cos ^2 x+3 \cos ^4 x \sin ^2 x \\ =3 \sin ^2 x \cos ^2 x\left(-\sin ^2 x+\cos ^2 x\right) \\ =3 \sin ^2 x \cos ^2 x \cdot \cos 2 x \\ =\frac{3}{4} \cdot 4 \sin ^2 x \cos ^2 x \cdot \cos 2 x=\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x \\ =\frac{3}{4} \sin ^2 2 x \cdot \cos 2 x\end{array}$

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