Question

Question #7e03a

Answer 1

`(xe^xlogx)^'=(e^x+x*e^x)*logx+e^x/(ln10)`

Explanation:

`y=xe^xlogx`

Let:
`f=xe^x`
`g=logx`
Then: `(f*g)^'=f^'*g+f*g^'`

`y^'=(xe^x)'*logx+xe^x(logx)^'`

`(x*e^x)^'=x^'*e^x+x(e^x)^'`
`(e^x)^'=e^x(x)^'=e^x*1`
`(x*e^x)^'=1*e^x+x*e^x*1=e^x+xe^x`

`y^'=(e^x+x*e^x)*logx+xe^x(logx)^'`

I guess we are talking about logarithm with the base equal to 10
`(logx)^'=1/(x*ln10)`

`y^'=(e^x+x*e^x)*logx+cancelxe^x*1/(cancelx*ln10)`

`y^'=(e^x+x*e^x)*logx+e^x/(ln10)`