If function f:RR->RR have one parity (even or odd) then hers derivate have the opposite parity?

1 Answer
Apr 21, 2017

Yes, it is true (assuming that f is differentiable on RR). There is a proof below.

Explanation:

If f(-x) = f(x), then the derivatives must be equal. By the chain rule d/dx(f(-x)) = -f'(-x).

So -f'(-x) = f'(x). That is f'(-x) = -f'(x) so f' is odd.

Similarly, if f(-x) = -f(x), then the derivatives must be equal, so -f'(-x) = -f'(x). That is f'(-x) =f'(x) so f' is even.