Question

Differentiate x^2 e^x logx

Answer 1

Product rule:

`(\mathrm d)/(\mathrm dx)(x^2e^x\log x)`

`=(\mathrm d(x^2))/(\mathrm dx)e^x\log x+x^2(\mathrm d(e^x))/(\mathrm dx)\log x+x^2e^x(\mathrm d(\log x))/(\mathrm dx)`

Power rule:

`(\mathrm d(x^2))/(\mathrm dx)=2x`

The exponential function is its own derivative:

`(\mathrm d(e^x))/(\mathrm dx)=e^x`

Assuming the base of
`\log x` is
`e`, its derivative is

`(\mathrm d(\log x))/(\mathrm dx)=\frac1x`

But if you mean a logarithm of arbitrary base
`b`, we have

`y=\log_bx\implies x=b^y=e^(y\ln b)\implies1=e^(y\ln b)\ln b(\mathrm dy)/(\mathrm dx)`

`\implies(\mathrm dy)/(\mathrm dx)=(e^(-y\ln b))/(\ln b)=\frac1{b^y\ln b}`

`\implies(\mathrm d(\log_bx))/(\mathrm dx)=\frac1{x\ln b}`

So we end up with

`2xe^x\log x+x^2e^x\log x+\frac{x^2e^x}x`

`=xe^x(2\log x+x\log x+1)`