Derivative of Logarithmic Function problems

You need the derivatives of these functions? If that's the case then it's not very complicated.

The answers are: h'(x)=e^x+x*e^x, g'(x)=2(x+1) and f'(x)=-1/(x^2) (if I didn't do any mistake)

Generally the rule is: [f(g(x))]'=f '(g(x))*g'(x). (the ' represents the derivative).
Let's start with the first function. h(x)=e^(lnx+x). I make these auxiliary functions: H(x)=e^x and G(x)=lnx+x.
So apparently we have: h(x)=H(G(x)). In order to find out what the derivative is we apply the rule I said earlier.
We know that (e^x)'=e^x and (lnx+x)'=(1/x)+1
So we have: h'(x)= [H(G(x))]'=H'(G(x))*G'(x)=e^(lnx+x)*((1/x)+1)= e^lnx*e^x*((1/x)+1)=x*e^x*((1/x)+1)=...=e^x+x*e^x

The other are a bit simpler because you have: g(x)=e^(2ln(x+1))=e^(ln(x+1)^2)=(x+1)^2 and g'(x)=[(x+1)^2]'=2(x+1)
And the last one: f(x)=e^(-lnx)=e^(ln(x^-1))=x^-1=1/x and f'(x)= (1/x)'=-1/(x^2)

hope I helped, though I have no idea why you need these derivatives..

Thank you very much! It really helped! And I don't know why I needed the derivatives either, my professor never went over these with us. It was just a problem from my homework that I was trying to understand.

Thank you!!!