If $f(x)$ is even, is $f'(0)=0$ always true Let $f(x)$ be differentiable function from $\mathbb R$ to $\mathbb R$, If $f(x)$ is even, then $f'(0)=0$. Is it always true?
 A: Given: $f(x)=f(-x)$
then we obtain: $f'(x)=-f'(-x) \implies f'(0)=-f'(0) \iff 2f'(0)=0$ , hence $f'(0)=0$
A: Hint: If a function $f$ is diffentiable at $x$, then
$$
f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}
$$
A: Let $f$ be an even function defined on a (symmetric) neighborhood of $0$ and differentiable at $0$. From the definition of differentiability:
$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}h.$$
By composition, we also have:
$$f'(0)=\lim_{h\to0}\frac{f(-h)-f(0)}{-h}$$
and using the fact that $f$ is even, this equality can be written as:
$$f'(0)=\lim_{h\to0}-\frac{f(h)-f(0)}h=-f'(0).$$
Hence $f'(0)=0$.
A: Trick question. Remember that "continuous" does not imply "differentiable".
If the function is differentiable at $0$, refer to the answer by @Brandon.
Update: The question has been edited so that it says "differentiable" rather than "continuous". The answer above applied to the originally posted question.
A: Let f (x) = square root of absolute value of x. Is it an even function? Is it continuous? What is f' (0)? 
