sinm x . cosn x
Let $y=\sin ^{m} x \cdot \cos ^{n} x$
Differentiating both sides w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{m} x \cdot \cos ^{n} x\right)$
$=\sin ^{m} x \cdot \frac{d}{d x}\left(\cos ^{n} x\right)+\cos ^{n} x \cdot \frac{d}{d x} \sin ^{m} x$
$=\sin ^{m} x \cdot n \cdot \cos ^{n-1} x \frac{d}{d x}(\cos x)+\cos ^{n} x \cdot m \cdot \sin ^{m-1} x$
$\frac{d}{d x}(\sin x)$
$=n \cdot \sin ^{m} x \cdot \cos ^{n-1} x \cdot(-\sin x)+m \cdot \cos ^{n} x \cdot \sin ^{m-1} x \cdot \cos x$
$=-n \cdot \sin ^{m+1} x \cdot \cos ^{n-1} x+m \cos ^{n+1} x \cdot \sin ^{m-1} x$
$=\sin ^{m} x \cdot \cos ^{n} x\left[-n \frac{\sin x}{\cos x}+m \cdot \frac{\cos x}{\sin x}\right]$
Thus, $\frac{d y}{d x}=\sin ^{m} x \cdot \cos ^{n} x[-n \tan x+m \cdot \cot x]$