Question

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

Answer 1

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.