- #1

mcastillo356

Gold Member

- 278

- 146

- Homework Statement:
- Calculate the critical point of ##y=x^x##

- Relevant Equations:
- Chain Rule, Product Rule

Hi everybody

We can't differentiate ##x^x## neither like a power function nor an exponential function. But ##x^x=e^{x\mbox{ln}x}##. So

##\dfrac{d}{dx}x^x=\dfrac{d}{dx}e^{x\mbox{ln}x}=x^x(\mbox{ln}x+1)##

And here comes the doubt: prove the domain of ##x^x## is ##(0, +\infty)##

Why is only positive?; why it will never be 0?

So ##(\mbox{ln}x+1)=0\Rightarrow{x=1/e}##

It's a solved exercise at Calculus, by R.A.A. It's the only question not solved (the domain, I mean).

We can't differentiate ##x^x## neither like a power function nor an exponential function. But ##x^x=e^{x\mbox{ln}x}##. So

##\dfrac{d}{dx}x^x=\dfrac{d}{dx}e^{x\mbox{ln}x}=x^x(\mbox{ln}x+1)##

And here comes the doubt: prove the domain of ##x^x## is ##(0, +\infty)##

Why is only positive?; why it will never be 0?

So ##(\mbox{ln}x+1)=0\Rightarrow{x=1/e}##

It's a solved exercise at Calculus, by R.A.A. It's the only question not solved (the domain, I mean).