How do I go about proving that an even function about x=0 has a derivative of 0? I do not have mean value theorem and can't use it. All I have are IVT and that the derivative of an even function is odd and the definition of a derivative. With those tools, what can I do?
 A: Suppose you mean the even function is differentiable at $x=0$, then use the definition. The left derivative is:
$$\lim_{h\rightarrow 0^+}\frac{f(0)-f(-h)}{h}=\lim_{h\rightarrow 0^+}-\frac{f(-h)-f(0)}{h}$$
The right derivative is:
$$\lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}$$
Since $f$ is even, then $f(-h)=f(h)$. The left derivative must equal to right derivative, so they have to be $0$.
A: Hint: What do you know about an odd function evaluated at $x=0$?
A: I could not understand the accepted answer, so here is another proof of the following statement:

If $f$ is an even function that is differentiable at $x=0$, then $f'(0)=0$.


Proof
By assumption, we know that $\displaystyle \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=M$. WLOG, suppose $M \gt 0$. Now, consider an $0\lt\varepsilon \lt \frac{M}{2}$. Then there is a corresponding $\delta_{\varepsilon} \gt 0: \forall x \in (0-\delta_{\varepsilon}, 0+\delta_{\varepsilon})\setminus\{0\}:\left|\frac{f(x)-f(0)}{x-0} -M\right| \lt \varepsilon \quad (\dagger_1)$.
Consider a $z_L$ and a $z_R$ such that $z_R \in (0,\delta_{\varepsilon})$ and $z_L=-z_R$. By $(\dagger_1)$, we know the following:

*

*$\left|\frac{f(z_R)-f(0)}{z_R}-M \right| \lt \varepsilon \implies M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt M+\varepsilon \quad (\dagger_2)$


*$\left|\frac{f(z_L)-f(0)}{z_L}-M \right| \lt \varepsilon \implies M-\varepsilon \lt \frac{f(z_L)-f(0)}{z_L} \lt M+\varepsilon \quad (\dagger_3)$
Because $f$ is an even function, which means that $\forall x \in \text{dom}(f): f(x)=f(-x)$, and we designated that $z_L=-z_R$, we can rewrite $(\dagger_3)$ in the following way:
$(\dagger_3) \iff M-\varepsilon \lt \frac{f(z_R)-f(0)}{-z_R} \lt M +\varepsilon$.
Multiplying the inequality by $(-1)$ gives us:

*

*$-M+\varepsilon \gt \frac{f(z_R)-f(0)}{z_R} \gt -M -\varepsilon \quad (\dagger_4)$
Because $M \gt 0$ and $0 \lt \varepsilon \lt \frac{M}{2}$, we know that $-M + \varepsilon \lt 0$ and $M-\varepsilon \gt 0$. Under these circumstances, $(\dagger_2)$ and $(\dagger_4)$ can be combined to give:
$$-M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt -M+\varepsilon \lt 0 \lt M-\varepsilon \lt \frac{f(z_R)-f(0)}{z_R} \lt M+\varepsilon$$
which, yields the contradictory statement of: $\frac{f(z_R)-f(0)}{z_R} \lt \frac{f(z_R)-f(0)}{z_R}$.
A similar contradiction will arise if $M \lt 0$.
We therefore conclude that $M=0$.
