TITLE: Derivative of an even function is odd and vice versa
QUESTION [11 upvotes]: 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. The following block shows the official solution, but I didn't quite understand it. (Particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$.)

(1F-6). Following the hint, let $z=-x$. If $f$ is even, then $f(x)=f(z).$ Differentiating and using the chain rule:
$$f'(x)=f'(z) \frac{dz}{dx}=-f'(z),$$ because $\frac{dz}{dx}=-1.$ But this means that $f'$ is odd.
Similarly, if $g$ is odd, then $g$($x=-g(z)$. Differentiating and using the chain rule: $$g'(x)=-g'(z) \frac{dz}{dx}=g'(z),$$ because $\frac{dz}{dx}=-1.$

Thanks in advance =]

REPLY [8 votes]: 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.