Derivative of an even function is odd and vice versa

Official, shmofficial: I think the following might prove to be easier to grasp for some: suppose $\,f\,$ is odd, then

$$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x)+f(x_0)}{-x+x_0}=$$

$$=\lim_{x\to x_0}\frac{f(-x)-f(-x_0)}{(-x)-(-x_0)}\stackrel{-x\to y}=\lim_{y\to -x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=:f'(-x_0)$$

The above remains, mutatis mutandis, in case $\,f\,$ is even.

Following the official solution, we have $f(-x) = -f(x)$ by assumption. Thus, by considering the function $g(x) = -x = (-1) \cdot x$, we have $f(g(x)) = (-1)\cdot f(x)$. Differentiating on both sides gives $$\frac{d}{dx} f(g(x)) - \frac{d}{dx} (-1)\cdot f(x) = -1 \cdot \frac{d}{dx} f(x).$$

Now, applying the chain rule, we get $$\frac{d}{dx} f(g(x)) = \frac{df}{dx} g(x) \cdot \frac{d}{dx} g(x) = \frac{df}{dx} (-x) \cdot \frac{d}{dx} (-1 \cdot x) = \frac{df}{dx} (-x) \cdot (-1).$$

Equating both sides and simplifying gives $$\frac{d}{dx} f(-x) = \frac{d}{dx} f(x).$$

If $f(x)$ is an even function,

Then: $f(−x)=f(x)$

Now, differentiate above equation both side: $f'(−x)(−1)=f'(x)$

$⇒f'(−x)=−f' (x)$

Hence $f'(x)$ is an odd function.

Rest part can be proved similarly.