Q:

Proving that the derivative of an odd function is even.

For an assignment I had, I had to prove that the derivative of an odd function is even. In the assignment we also had to prove that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even, which I did do. Using that fact I stated the following:
Let us define $F(x)=\int_0^xf(t)dt$ such that $F(x)$ is odd.
\begin{equation}
F'(x)=\frac{d}{dx}\int_0^xf(t)dt=f(x) \nonumber
\end{equation}
Using 1.1 (the section where I proved that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even) we know that $f(x)$ is even.
However, the teacher felt that the answer was not rigorous enough and that I was simply going backwards. Am I indeed simply moving backwards and not proving anything in which case: could someone point out places where I could make it more succinct and rigorous or alternatively supply better proof altogether.

A:

The proof is quite simple from the definition of the derivative: if $f$ is odd then
$$
f'(-x) = \lim\limits_{h\to 0}\frac{f(-x+h)-f(-x)}{h} =  -\lim\limits_{h\to 0}\frac{f(x-h)-f(x)}{h} = -f'(x).
$$
W.r.t. your proof. You have showed that if $f$ is even, then $F = \int f$ is odd. You proved it - but you didn't prove that any odd function is an anti-derivative of the even function. That would be a reverse statement, as Alex has already told you. 
Generally, you have $A\Rightarrow B$ where $A = \{f\text{ is even}\}$ and $B = \{F\text{ is odd}\}$ but to prove that the derivative of the odd function is even you need $B\Rightarrow A$ which you don't know at the moment.

