Question

(2x² − 3) sin x

Answer 1

$$\begin{gathered} \mathrm{Let}u=2x^2-3;v=\sin x \\ \mathrm{Then},u\text{'}=4x;v\text{'}=\cos x \\ \mathrm{Using}\mathrm{the}\text{product rule:} \\ \frac{d}{dx}\left(uv\right)=uv\text{'}+vu\text{'} \\ \frac{d}{dx}\left[\left(2x^2-3\right)\sin x\right]=\left(2x^2-3\right)\cos x+4x\sin x \end{gathered}$$