Question

How do you use logarithmic differentiation to find the derivative of the function `y=x^x`?

Answer 1

`y=x^x`

`lny=lnx^x`

`lny=xlnx`

Now differentiate implicitly:

`1/y (dy)/(dx)=(1)(lnx)+(x)(1/x)`

`1/y (dy)/(dx)=1+lnx`

Solve for `(dy)/(dx)`

`(dy)/(dx)=y(1+lnx)`

Recall that `y=x^x`, so

`(dy)/(dx)=x^x (1+lnx)`

Answer 2

It's easy if you remember the logarithmic rule:

`[f(x)]^g(x)=e^ln([f(x)]^g(x))=e^(g(x)*lnf(x)`,

so:

`y=x^x=e^(xlnx)rArr`

`y'=e^(xlnx)(1*lnx+x*1/x)=e^(xlnx)(lnx+1)`,

(or, if you want: `y'=x^x(lnx+1)`).