$\text{ If }y=x^2\sin x\text{ , then }\frac{dy}{dx}\text{ will be}$

Solution:

$\begin{array}{l}y=x^2\sin x\\ \frac{dy}{dx}=\frac{d}{dx}\left(x^2\sin x\right)\\ =x^2\frac{d\sin x}{dx}+\sin x\frac{d{x}^2}{dx}\left[\because \frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{dv}{dx}\right]\\ =x^2[\cos x]+\sin x\left[2x\right]\left(\because \frac{d}{dx}\sin x=\cos x,\frac{d{x}^n}{dx}=n{x}^{n-1}\right)\\ =x^2\cos x+2x\sin x\end{array}$