Derivative of an Odd Function

Date: 02/27/2002 at 00:22:13
From: Sheri
Subject: Derivative of an odd function is even

How do you show that the derivative of an odd function is even?  

That is, if f(-x) = -f(x), then f'(-x) = f'(x).

I have been working on this problem for two days and still can't 
figure out the answer.

Date: 03/01/2002 at 01:27:42
From: Doctor Douglas
Subject: Re: Derivative of an odd function is even

Hi, Sheri,

Thanks for submitting your question to the Math Forum.  

We start from the definition of the derivative:

  f'(x) = lim{h->0} [f(x+h) - f(x)]/h

Note that this must be defined for all h sufficiently close to zero.  
In particular, the limit must exist whether h approaches zero from the 
right or from the left (and these limits must be equal to one another, 
of course).

Now let us consider f'(-x) for a function that is odd.

  f'(-x) = lim{h->0} [f(-x+h) - f(-x)]/h      now let g = -h
         = lim{g->0} [f(-x-g) - f(-x)]/(-g)   now use "f is odd" twice
         = lim{g->0} [-f(x+g) - (-f(x))]/-g   the rest is just 
                                                manipulation...
         = lim{g->0} [f(x+g) - f(x)]/g
         = f'(x)                              by def'n of derivative

Thus, if an odd function possesses a derivative, it is even. 

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/