Proving two functions are equal with their derivatives I just post a question to check if my proof to the following is true : 
Let $f$ and $g$ be two differentiable functions on $E=[a,b],$ then
if $f'(t)=g'(t)$ and $f(a)=g(a)$ then $f=g, \forall t \in E$  
(1)we admit that $f$ and $g$ are differentiable, thus they are continuous thus integrable.
(2) we also admit that $f'(t)=g'(t) \implies \int_a^b f'(t) dt =\int^b_a g'(t) dt $
(2) We'll admit the fundamental theorem of calculus
then 
let $x \in E$
$$\int_a^x f'(t) dt =f(x)-f(a)$$
$$\int_a^x g'(t) dt=g(x)-g(a)$$
so 
$$f(x)-f(a)=g(x)-f(a)$$
i.e.
$$f(x)=g(x), \forall x \in E$$
QED
Is there any way to prove this using only the derivate (and no integral) ? 
Thanks 
Tom
 A: Let $h(x)=f(x)-g(x)$ for $x\in[a,b]$.
Let $x\in (a,b]$. By Mean Value Theorem, there exists $\zeta\in(a,x)$ such that
$$\frac{h(x)-h(a)}{x-a}=f'(\zeta)=0$$
So, $h(x)=h(a)=0$ and hence $f(x)=g(x)$.
A: Hint: Fix $x\in(a,b]$. Applying the Mean Value Theorem to $f-g$ on $[a,x]$ you can conclude $(f-g)(x)=0$.
You can even relax a little bit the hypotheses and ask for differentiability on $(a,b)$ and continuity at the endpoints instead of differentiability on $[a,b]$.
A: You could use the fact that if a function has derivative zero, then it is a constant function on each connected component of its domain, in particular in your situation on $[a,b]$. More specifically, $h=f-g$ satisfies $h^{\prime}=0$, hence $h=c$ for some $c\in \mathbb R$ and necessarily $c=0$ since $h$ has a zero by assumption.
The above fact can be proved using the mean value theorem.
A: For fun, proof by contradiction .
Let $x \in (a,b].$
$h(x):=f(x)-g(x)$; $h(a)=0$; and $h'(x) = 0$, $x \in (a,b)$.
Assume $h(y) \not =0$, for a $y \in (a,b].$
MVT:
$\dfrac{h(y)-h(a)}{y}= h'(t)$, $a<t<y.$
By assumption the LHS $\not =0.$
Contradiction, hence
$h(x)=0$ for $x \in [a,b].$
