Derivative of an even function is odd and vice versa This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even.
(Write the equation that says f is even, and differentiate both sides, using the
chain rule.)"

I already read numerous solutions online. This was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).

Thanks in advance =]
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.
Well, geometrically, even function means reflection along y axis, so any direction will reflect, that mean, the derivative on the right is the same as the derivative on the left, but the direction change. It means the value is the same, but with different sign.

Odd function means rotational symmetric, if you rotate an arrow, I.e. direction, you will change by 180 degree, so it is the same slope, hence the derivative of odd function is even.