If $f(x)$ is odd, then prove that $f'(0)=0$

Question

If $f(x)$ is odd, then prove that $f'(0)=0$, assuming that $f'(0)$ exists.

We have that $f(x)$ is odd. Therefore, $f(-x)=-f(x)\implies -f'(-x)=-f'(x)$. What I do after this?

Answer 1

Hint: Consider $f(x) = \sin(x)$ and observe that $f'(x) = \cos(x)$.

Answer 2

As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $h\to 0$ though, since in that case $$ f'(0) = \lim_{h\to 0}\frac{f(h) - f(-h)}{2h} = \lim_{h\to 0}\frac{f(h)+f(h)}{2h} = \lim_{h\to 0}\frac{f(h)}{h} = 0. $$ This holds for $f(x) = x^{2n+1}$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$, $$ f'(0) = \lim_{h\to 0}\frac{f(h) - f(-h)}{2h} = \lim_{h\to 0}\frac{f(h)-f(h)}{2h} = \lim_{h\to 0}\frac{0}{h} = 0 $$ as desired.

Answer 3

Your way is nice for even $f$;

$f(x) = f(-x) \implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.