$f'$ odd implies that $f$ is even I can prove that if $f$ is even then $f'$ is odd (and $f$ odd implying $f'$ is even).
However, I'm not sure how to prove it the other way around by using limits.
To prove that $f'$ being odd implies that $f$ is even, I tried $f'(a) = -f'(-a).$
So, 
$$f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} = -\lim_{h \rightarrow 0} \frac{f(-a+h) - f(-a)}{h} = -f'(-a) $$
Is this correct, and if so, where do I go from here?
 A: Alternative solution:
Any function $g$ can be written as the sum of its even and odd parts. That is, for an arbitrary function $g : \mathbb{R} \to \mathbb{R}$ is a function, we can write
$$g_{\text{even}}(x) = \dfrac{g(x)+g(-x)}{2} \qquad \text{and} \qquad g_{\text{odd}}(x) = \dfrac{g(x)-g(-x)}{2}$$
Then $g_{\text{even}}$ is even, $g_{\text{odd}}$ is odd, and $g = g_{\text{even}} + g_{\text{odd}}$. It's then easy to see that $g$ is even if and only if $g_{\text{odd}}=0$ and $g$ is odd if and only if $g_{\text{even}}=0$.
Now suppose $f : \mathbb{R} \to \mathbb{R}$ is differentiable and $f'$ is odd. Then
$$(f_{\text{odd}})'(x) = \dfrac{f'(x)+f'(-x)}{2} = (f')_{\text{even}}(x) = 0$$
So $f_{\text{odd}}$ is constant. And the only odd constant function is zero, so $f_{\text{odd}}=0$ and $f=f_{\text{even}}$.
A: Under the assumption that $f'$ is Riemann-integrable, we can do the following. Since $f'$ is odd, then $$\int_{-x}^xf'(t)\,dt=0$$ for all $x$. What's another way we can write that integral, in terms of $f$?
A: $\rm f'$ odd $\rm\:\Rightarrow\: 0 = f'(x)\!+f'(-x) = (f(x)\!-f(-x))'\Rightarrow\:f(x)\!-\! f(-x) =c.\ $ Eval at $\rm\:x=0\:\Rightarrow\: c=0.$
A: Let $f(x)=g(x)+g(-x)$
So, $f(-x)=g(-x)+g(x)=f(x)$  i.e., even
$\implies f'(x)=g'(x)-g'(-x)$
So, $f'(-x)=g'(-x)-g'(x)=-f'(x)$  i.e., odd
Now, start from $f'(x)=g'(x)-g'(-x)$ and integrate
