Question
Differentiate of the following w.r.t. x:
$\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}$

Answer

Let $\text{y}=\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\sin\text{x})^\text{m}\cdot(\cos\text{x})^\text{n}\big]$
$=(\sin\text{x})^\text{m}\cdot\frac{\text{d}}{\text{dx}}(\cos\text{x})^\text{n}+(\cos\text{x})^\text{n}\cdot\frac{\text{d}}{\text{dx}}(\sin\text{x})^\text{m}$
$=(\sin\text{x})^\text{m}\cdot\text{n}(\cos\text{x})^{\text{n}-1}\cdot\frac{\text{d}}{\text{dx}}\cos\text{x}+(\cos\text{x})^\text{n}\text{m}(\sin\text{x})^{\text{m}-1}\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}$
$=(\sin\text{x})^\text{m}\cdot\text{n}(\cos\text{x})^{\text{n}-1}(-\sin\text{x})+(\cos\text{x})^\text{n}\cdot\text{m}(\sin\text{x})^{\text{m}-1}\cos\text{x}$
$=-\text{n}\cdot\sin^\text{m}\text{x}\cdot\sin\text{x}\cdot\cos^\text{n}\text{x}\cdot\frac{1}{\cos\text{x}}+\text{m}\cdot\sin^\text{m}\text{x}\cdot\frac{1}{\sin\text{x}}\cdot\cos^\text{n}\text{x}\cdot\cos\text{x}$
$=-\text{n}\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\cdot\tan\text{x}+\text{m}\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\cdot\cot\text{x}$
$=\sin^\text{m}\text{x}\cdot\cos^\text{n}\text{x}\big[\text{m}\cot\text{x}-\text{n}\tan\text{x}\big]$

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