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

##### 1 Answer

There are none.

#### Explanation:

A point on

So, we first need to find

#f(x)=x^(e^x)#

We will use logarithmic differentiation, which is helpful when finding a derivative such as this one, where we have a function whose power is another function. To use this method, begin by taking the natural logarithm of both sides of the equation.

#ln(f(x))=ln(x^(e^x))#

The right side can be simplified by bringing the power of

#ln(f(x))=e^xln(x)#

Now take the derivative of both sides of the equation. The left side will need the chain rule, since we're dealing with a composite function (a function within a function). On the right side, we'll use the product rule.

#1/f(x)f'(x)=(d/dxe^x)ln(x)+e^x(d/dxln(x))#

#1/f(x)f'(x)=e^xln(x)+e^x(1/x)#

#1/f(x)f'(x)=(xe^xln(x)+e^x)/x#

#1/f(x)f'(x)=(e^x(xln(x)+1))/x#

Now solving for the derivative by multiplying both sides by

#f'(x)=(x^(e^x)e^x(xln(x)+1))/x#

Rewriting the following:

#f'(x)=x^(e^x-1)e^x(xln(x)+1)#

We can try to solve for critical points. But first, realize that

Letting the derivative equal

#0=x^(e^x-1)e^x(xln(x)+1)#

We have three parts:

#x^(e^x-1)=0" "" or "" "e^x=0" " or """ "xln(x)+1=0#

The first is only true if

The second is never true.

The third can be graphed to show that

So, we have found no critical points.

A graph of

graph{x^(e^x) [-.2, 3, -20, 100]}