A differentiable function on $(a,b)$ with nonzero derivative over $(a,b)$ where $f'(c)>0$ for some $c\in(a,b)$ means $f'(x)>0$ for all $x\in(a,b)$. It certainly seems correct that if you have at least one point $c\in(a,b)$ such that $f'(c)>0$, where $f'(x)\ne0$ for all $x\in(a,b)$, then $f'(x)>0$ for all $x\in(a,b)$. At least from a calculus standpoint, if $f$ is well-defined (not necessarily continuous), it should be the case that this is true. But how does one show this formally?
The exact statement I am attempting to prove is this:

Suppose $f:(a,b)\to\mathbb{R}$ is a differentiable function such that $f'(x)\ne0$ for all $x\in(a,b)$. Suppose there exists a point $c\in(a,b)$ with $f'(c)>0$, then $f'(x)>0$ for all $x\in(a,b)$.

I thought it might be easiest to prove by contradiction, and then use the intermediate value property to derive a contradiction. Any pointers would be helpful on this one!
EDIT: My original direct proof went something like this:

Let $f:(a,b)\to\mathbb{R}$ be a differentiable function such that $f'(x)\ne0$ for all $x\in(a,b)$. Take a point $c\in(a,b)$ such that $f'(c)>0$. Since $f'(x)\ne0$ for any $x\in(a,b)$ there can be no extreme values on $(a,b)$. Then take some $y\in\mathbb{R}$ such that $f'(c)=y$ and by the intermediate value property, we are guaranteed that $f'(a)<y<f'(b)$, that is $f$ is strictly increasing on $(a,b)$. Hence $f'(x)>0$ for all $x\in(a,b)$.

I'm a little worried about some of the logic in this. It feels incomplete to me.

According to Darboux's Thereom, if $f$  is a differentiable function in an interval $I$, then the range of its derivative function $f'$ must still be an interval. Using the given condition, we can easily get the conclusion.
