Question

Find $\frac{dy}{dx}$, if $y={\sin }^{3}x{\cos }^{5}x$.

Answer 1

$y={\sin }^{3}x{\cos }^{5}x$

$\frac{dy}{dx}={\sin }^{3}x\frac{d}{dx}{\cos }^{5}x+{\cos }^{5}x\frac{d}{dx}{\sin }^{3}x$

$\frac{dy}{dx}={\sin }^{3}x5{\cos }^{4}x(-\sin x)+{\cos }^{5}x3{\sin }^{2}x\cos x$

$\frac{dy}{dx}=-5{\sin }^{4}x{\cos }^{4}x+3{\cos }^{6}x{\sin }^{2}x$

$\frac{dy}{dx}={\sin }^{2}x{\cos }^{4}x(-5{\sin }^{2}x+3{\cos }^{2}x)$