Question

Differentiate the function given below w.r.t. $x:$

$x^2\sin x+\cos 2x$

Answer 1

Let $y=x^2\sin x+\cos 2x$

Differentiating with respect to $x,$ we get

$\frac{dy}{dx}=\frac{d}{dx}(x^2\sin x+\cos 2x)$

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^2\sin x\right)+\frac{d}{dx}\cos 2x$

$\Rightarrow \frac{dy}{dx}=\sin x\frac{d}{dx}\left(x^2\right)+x^2\frac{d}{dx}\left(\sin x\right)+\frac{d}{dx}\cos 2x$

$\Rightarrow \frac{dy}{dx}=(2x\sin x+x^2\cos x)-2\sin 2x$

$[\because \frac{d{x}^n}{dx}=n{x}^{n-1},\frac{d\left(\cos \theta \right)}{dx}=-\sin \theta \&\frac{d\left(\sin \theta \right)}{dx}=\cos \theta ]$

$\Rightarrow \frac{dy}{dx}=2x\sin x+x^2\cos x-2\sin 2x$