Derivative of an Odd FunctionDate: 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/ |
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