Derivative of a function is odd prove the function is even. $f:\mathbb{R} \rightarrow \mathbb{R}$ is such that $f'(x)$ exists $\forall x.$
And $f'(-x)=-f'(x)$
I would like to show $f(-x)=f(x)$
In other words a function with odd derivative is even.
If I could apply the fundamental theorem of calculus
$\int_{-x}^{x}f'(t)dt = f(x)-f(-x)$ but since the integrand is odd we have $f(x)-f(-x)=0 \Rightarrow f(x)=f(-x)$
but unfortunately I don't know that f' is integrable. 

Let $g(x)=f(-x)$.  Then $g'(x)=-f'(-x)=f'(x)$.
Since $g(0)=f(0)$ and $g'=f'$, it follows from the mean value theorem that $g=f$.
