Question

x² e^x log x

Answer 1

$$\begin{gathered} \mathrm{Let}u=x^2;v=e^x;w=\log x \\ \mathrm{Then},u\text{'}=2x;v\text{'}=e^x,w=\frac{1}{x} \\ \text{Using the product rule:} \\ \frac{d}{dx}\left(uvw\right)=u\text{'}vw++uv\text{'}w+uvw\text{'} \\ \frac{d}{dx}\left(x^2{e}^x\log x\right)=2x{e}^x\log x+x^2{e}^x\log x+x^2{e}^x\frac{1}{x} \\ =2x{e}^x\log x+x^2{e}^x\log x+x{e}^x \\ =x{e}^x\left(2\log x+x\log x+1\right) \end{gathered}$$