Proving $f'$=$g'$ has some c such that $g=f+c$ Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.

So far, I started with this:

Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$

After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!
Fixing your proof a little bit, I can extend your proof.

Suppose that your domain of function is (a,b). And let's choose any $\beta\in (a,b)$.

So for any $x\in (a,b)$ $\exists \alpha\in (\beta,x)$ (or $\alpha \in(x,\beta)$ if $x&lt;\beta$) such that $h'(\alpha)=\frac{h(x)-h(\beta)}{x-\beta}=0$.

Thus,

$$\frac{h(x)-h(\beta)}{x-\beta}=0 \implies h(x)-h(\beta)=0\\ \implies f(x)-g(x)=f(\beta)-g(\beta)\\ \implies f(x)=g(x)+f(\beta)-g(\beta)$$

Thus, the $c$, we were finding is $f(\beta)-g(\beta)$.
Assume f-g is non-constant. Then f'-g' is not identically zero. But f'=g' by hypothesis. Contradiction. Hence f-g is constant.