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 =]
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 $$\frac{d}{dx} f(-x) = \frac{d}{dx} f(x).$$