If $f$ is an even function, and $f$ is differentiable at $x=0$, prove that $f'(0)=0$? I don't really know where to begin - intuitively I understand that the y-axis intersection must be an extrema, so the derivative is obviously 0, but I'm having difficulty writing the proof..
We haven't learned yet that the derivative of an extrema is 0, so it's not sufficient to simply prove that $f(0)$ an extrema.
Any help would be appreciated.
I'd also appreciate a general explanation of how to approach problems like these. 
 A: We have $f(-x)=f(x)$ and so differentiating, $$f^{\prime}(-x)(-x)^{\prime}=f^{\prime}(x)\Rightarrow -f^{\prime}(-x)=f^{\prime}(x)$$
for the $x$ that $f$ is differentiable on $x$ and $-x$ (one such $x$ is $0$). Plugging $x=0$ yields the result.
A: You know that 
$$f'(0)= \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}$$
In particular
$$f'(0)= \lim_{x \to 0^+} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0} $$
Now we use the fact that $f$ is even
$$f'(0)= \lim_{x \to 0^-} \frac{f(x)-f(0)}{x}= \lim_{x \to 0^-} \frac{f(-x)-f(0)}{x}$$
Subbing in $y=-x$ you get
$$f'(0)=\lim_{y \to 0^+} \frac{f(y)-f(0)}{-y}=-\lim_{y \to 0^+} \frac{f(y)-f(0)}{y}=-f'(0)$$
A: Suppose that $f\,'(0)=\alpha>0$. Then
$$\lim_{h\to 0}\frac{f(h)-f(0)}h=\alpha\;,$$
so there is a $\delta>0$ such that $$\frac{f(h)-f(0)}h>\frac{\alpha}2\quad\text{whenever}\quad |h|<\delta\;.$$
Let $x=\dfrac{\delta}2$. Then $$\frac{f(x)-f(0)}x>\frac{\alpha}2\quad\text{and}\quad\frac{f(-x)-f(0)}{-x}>\frac{\alpha}2\;,$$ so 
$$f(x)>\frac{\alpha}2x+f(0)\quad\text{and}\quad f(-x)<-\frac{\alpha}2+f(0)\;.$$ 
But $f$ is even, so $f(-x)=f(x)$, and we have
$$f(x)<-\frac{\alpha}2+f(0)<\frac{\alpha}2+f(0)<f(x)\;,$$
which is absurd. A similar argument shows that $f\,'(0)$ cannot be negative, and we conclude that it must be $0$.
A: If you know that the derivative of an even function is odd, and that every odd function vanishes at $x=0$, then the result is immediate. The first can be shown by playing with negatives in the limit of the difference quotient, and the latter by noticing $f(0)=-f(0)$ when $f$ is odd.
