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.
A: I think your teacher is right that the argument is not quite rigorous. You've proved that the integral of an even function is odd. However, you haven't proved (or don't say specifically that you've proved) that the integral of a function that is not even is not odd. (Every dog has four legs, but it is not true that everything that is not a dog does not have four legs.)
Logically, there could be (a) even functions whose integrals are odd, (b) odd functions whose integrals are odd, and (c) functions that are neither even nor odd, whose integrals are odd. To complete your proof, you would have to show that cases b and c don't exist.
A: Comment:
The closest answer given was the direct proof from the limit definition.
One major correction, however:
Technically, the last limit is NOT the claimed -1 times (the definition of) f '(x), owing to the negative h instead of the usual h. In fact, to make this work, the negative sign needs to be in the denominator so that h everywhere is replaced by -h. As a result, the right side is and should be just f '(x). So, now we get f '(-x) = f '(x) and NOT -f '(x)!
Besides, you want f '(-x) = f '(x) for the function f ' to be even! If you really had the negative one multiplier, then your derivative of your odd f would have also been odd!
(P.S. Sorry but I wrote those derivatives f ' because if I wrote f' it may look like just f. Please read very closely, as I am writing only about f prime, not original function f.)
Dr. Michael W. Ecker/
Associate Professor of Mathematics/
Pennsylvania State University/
Wilkes-Barre Campus/
Lehman, PA 18627
