How do you differentiate the following: v=x^x?

My steps:
1. v=x^x
2. x^x = e^ln(x^x) = e^(x lnx)
3. (e^(x lnx))' = (e^(x lnx))lnx
4. (dv)/dx = (x^x)lnx

Correct Answer: (dv)/dx = x^x(1+lnx)

1 Answer
Jul 28, 2017

(dv)/dx = x^x(1 + lnx)

Explanation:

My approach would be to use implicit logarithm differentiation:

We have:

v=x^x

Taking Natural logarithms we have:

lnv = ln {x^x}
" " = xlnx

Implicitly differentiating and applying the product rule we get:

d/dx lnv = (x)(d/dx ln x) + (d/dx x)(lnx)
:. 1/v (dv)/dx = x 1/x + 1 lnx
" " = 1 + lnx

:. (dv)/dx = v(1 + lnx)
" " = x^x(1 + lnx)