Question

Differentiate the function given below w.r.t. $x:$

${\sin }^{3}x.{\cos }^{3}x$

Answer 1

Let $y={\sin }^{3}x\cdot {\cos }^{3}x$

Differentiating with respect to $x,$ we get

$\frac{dy}{dx}={\cos }^{3}x\frac{d}{dx}\left({\sin }^{3}x\right)+{\sin }^{3}x\frac{d}{dx}\left({\cos }^{3}x\right)$

$\Rightarrow \frac{dy}{dx}={\cos }^{3}x\left(3{\sin }^{2}x\cos x\right)$

$+{\sin }^{3}x(3{\cos }^{2}x\cdot (-\sin x\left)\right)$

$\Rightarrow \frac{dy}{dx}=3{\sin }^{2}x.{\cos }^{4}x-3{\cos }^{2}x.{\sin }^{4}x$

$\Rightarrow \frac{dy}{dx}=3{\sin }^{2}x.{\cos }^{2}x({\cos }^{2}x-{\sin }^{2}x)$

$\text{divide and multiply by 4}$

$\Rightarrow \frac{dy}{dx}=\frac{3}{4}\times 4{\sin }^{2}x.{\cos }^{2}x.\cos 2x$

$(\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta )$

$\Rightarrow \frac{dy}{dx}=\frac{3}{4}(2\sin x\cos x{)}^2.\cos 2x$

$(\because \sin 2\theta =2\sin \theta .\cos \theta )$

$\Rightarrow \frac{dy}{dx}=\frac{3}{4}{\sin }^{2}2x.\cos 2x$