sinm x . cosn x
Question:

sinm x . cosn x

Solution:

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, $\frac{d y}{d x}=\sin ^{m} x \cdot \cos ^{n} x[-n \tan x+m \cdot \cot x]$

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