# Math Help - derevatitive

1. ## derevatitive

hello how would i find dy/dx of $y = x^x$

2. Hi there. You must think about logaritms.

This is the way you can solve it:

$y=x^x$

$\ln(y)=\ln(x^x)$

$\ln(y)=x\ln(x)$

Now the derivatives:

$\frac{1}{y}y'=\ln(x)+x\frac{1}{x}$

$y'= y(\ln(x)+1)$