Question

Find $\frac{d}{dx}{\sin }^{3}4x$.

Answer 1

Using chain rule for differentiation

$\frac{d}{dx}\left({\sin }^{3}4x\right)=\frac{d}{dx}\left\{\right(\sin 4x{)}^3\}$

$=\frac{d}{d\left(\sin 4x\right)}\left\{\right(\sin 4x{)}^3\left\}\frac{d}{dx}\right(\sin 4x)$

$=3\left(\sin 4x{)}^{3-1}\frac{d}{d\left(4x\right)}\right\{\sin \left(4x\right)\}\frac{d\left(4x\right)}{dx}$

$=3{\sin }^{2}4x\cdot \cos \left(4x\right)\cdot 4\frac{dx}{dx}$ $[\because \frac{dx}{dx}=1]$ and $[\frac{d\left(\sin x\right)}{dx}=\cos x]$

$=12{\sin }^{2}4x\cos 5x$.