# What are the critical points of #f(x) =x^(x^2)#?

##### 1 Answer

Apr 5, 2016

The only critical number is

#### Explanation:

The domain of

The negative integers are discrete, so we cannot find a (real) derivative of

For the calculus, we can restrict the domain to the positive real numbers, which is the domain of

# = e^(x^2lnx)[(2x)lnx+x^2 (1/x)]#

# = e^(x^2lnx)(2xlnx+x)#

# = x^(x^2)x(2lnx+1)# .

Since

So,

And,