Derivative of x^(2x): Solved with ln

[3songs]

I only know that it is solved with ln. Something like lnx^2x...

 

[Dick]

x^(2x)=e^(ln(x)*(2x)). Use the chain rule.

 

[3songs]

x^(2x)=e^(ln(x)*(2x)). Use the chain rule.

I got:

(lnx 2x)'=1/x2x+lnx*2=x+2ln x

e^^x+2lnx
=e^x+e²e^lnx

?

 

[Dick]

Well, the derivative of 2*x*ln(x) isn't x+2*ln(x). You might want to work on that for a start.

 

[3songs]

Well, the derivative of 2*x*ln(x) isn't x+2*ln(x). You might want to work on that for a start.

I am sorry, that was a typo.
I meant
2+2lnx NOT x+2lnx

 

[Dick]

I am sorry, that was a typo.
I meant
2+2lnx NOT x+2lnx

That's better. Now work on the rest of it. The chain rule says (e^f(x))'=(e^f(x))*f'(x) doesn't it? Put f(x)=ln(x)*(2x).