TITLE: Equivalence Relation between Derivative Being Odd and Function Being Even
QUESTION [6 upvotes]: In the exercise, I am required to prove that
$f'$ is odd $\iff$ $f$ is even
Moving from right to left was pretty trivial, however, I couldn't move from left to right. Note that we can only use very basic things like the definition of the derivative and rules of differentiation (No mean value theorem or any other theorems.)
I cannot even see the left implying the right, but I hope I will see it after you guys explain how. Thanks in advance.

REPLY [8 votes]: Consider the function $g(x)=f(x)-f(-x)$. If $f'(-x)=-f'(x)$, then
$$
\begin{align}
g'(x)
&=f'(x)+f'(-x)\tag{1}\\
&=f'(x)-f'(x)\tag{2}\\
&=0\tag{3}
\end{align}
$$
Explanation:
$(1)$: chain rule and the definition of $g$
$(2)$: $f'$ is odd
$(3)$: $a-a=0$
Thus, $g(x)$ is constant. Since $g(0)=f(0)-f(0)=0$, $g(x)=0$ for all $x$. Thus,
$$
f(x)=f(-x)
$$
Therefore, $f$ is even.