Differentiate the function $\sin^n(ax^2 + bx + c)$ with respect to $x$.

$n(2ax + b) \sin^{n-1}(ax^2 + bx + c) \cos(ax^2 + bx + c)$

The correct answer is Option (2) → $n(2ax + b) \sin^{n-1}(ax^2 + bx + c) \cos(ax^2 + bx + c)$ ##

Let $y = \sin^n(ax^2 + bx + c)$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx} [\sin(ax^2 + bx + c)]^n$

$= n \cdot [\sin(ax^2 + bx + c)]^{n-1} \cdot \frac{d}{dx} \sin(ax^2 + bx + c)$

$= n \cdot \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot \frac{d}{dx} (ax^2 + bx + c)$

$= n \cdot \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b)$

$= n \cdot (2ax + b) \cdot \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c)$