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

1 Answer
Jim H
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.

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