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

Results 1 to 2 of 2

- May 26th 2010, 04:00 AM #1

- Joined
- Apr 2010
- Posts
- 135

- May 26th 2010, 04:15 AM #2

- Joined
- May 2010
- Posts
- 251

Hi there. You must think about logaritms.

This is the way you can solve it:

$\displaystyle y=x^x$

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

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

Now the derivatives:

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

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