Product rule:
[tex]\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)[/tex]
[tex]=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}[/tex]
Power rule:
[tex]\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x[/tex]
The exponential function is its own derivative:
[tex]\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x[/tex]
Assuming the base of [tex]\log x[/tex] is [tex]e[/tex], its derivative is
[tex]\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x[/tex]
But if you mean a logarithm of arbitrary base [tex]b[/tex], we have
[tex]y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}[/tex]
[tex]\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}[/tex]
So we end up with
[tex]2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x[/tex]
[tex]=xe^x(2\log x+x\log x+1)[/tex]
