Question
Differentiate following w.r.t. x:
$\sin^\text{n}\big(\text{ax}^2+\text{bx}+\text{c}\big)$

Answer

Let $\text{y}=\sin^\text{n}\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)\Big]^\text{n}$
$=\text{n}\cdot\Big[\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)\Big]^{\text{n}-1}\cdot\frac{\text{d}}{\text{dx}}\sin\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$=\text{n}\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\frac{\text{d}}{\text{dx}}\big(\text{ax}^2+\text{bx}+\text{c}\big)$
$=\text{n}\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot(2\text{ax + b})$
$=\text{n}\cdot(2\text{ax + b})\cdot\sin^{\text{n}-1}\big(\text{ax}^2+\text{bx}+\text{c}\big)\cdot\cos\big(\text{ax}^2+\text{bx}+\text{c}\big)$

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