If $f$ is of given parity, what can be said of its derivative and primitive?

Question

If $f$ is of given parity, what can be said of it's derivative and primitive? Clearly for power functions or simple trigonometric functions it seems that the parity of the derivative and antiderivative are opposite to that of the function itself. But is this true in general? If so, how to prove/disprove it?

Answer 1

Using the chain rule: $$f'(x)=\frac{d}{dx}f(x) = \frac{d}{dx}h(f(g(x))) = h'(f(g(x)))f'(g(x))g'(x)$$ where $g,h$ are functions such that $f(x)=h(f(g(x)))$. In the case of even functions you can take $h(x)=x$ and $g(x)=-x$, giving you $f(x)=f(-x)$ and thus: $$f'(x)=-f'(-x)$$ In the case of odd functions we have $h(x)=g(x)=-x$, and thus: $$f'(x)=f'(-x)$$ Notice that the rule above is much more general than that. For example consider the condition that $f$ is periodic of period $1$. This can be translated in terms of functions $g$ and $h$ by $g(x)=x+1$, $h(x)=x$, giving: $$f'(x)=f'(x+1)$$ And thus the derivative is also periodic with period $1$.

Answer 2

Hint: $$f'(-x) = \lim_{h\to 0}\frac{f(-x-h)-f(-x)}{h}.$$