Question

(x sin x + cos x ) (e^x + x² log x)

Answer 1

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