Standard vs. Symmetric Derivative of Sin(x)Date: 02/07/2006 at 06:41:29 From: Greg Subject: Derivative of sin x. I am a high school mathematics teacher and was wondering if this idea for finding the derivative of f(x) = sin x was valid. (sin(x+h)-sin(x-h)) f'(x) = lim ------------------- h->0 2h 2cos(x)sin(h) = lim ------------- h->0 2h = cos x since lim h->0 sin(h)/h = 1. Thanks for your consideration. Date: 02/07/2006 at 12:12:42 From: Doctor Douglas Subject: Re: Derivative of sin x. Hi Greg, The distinction between the two limits lim f(x+h)-f(x) lim f(x+h)-f(x-h) h->0 ----------- and h->0 ------------- h 2h normally doesn't affect things too much, at least in the standard first-year calculus course, but there is an important distinction to make. The first limit is the usual definition of the derivative of f, while the second one is called the "symmetric derivative". When the derivative of f (the left expression above) exists, then either of the above expressions will be equal to it and of course equal to each other. However, the converse is not true. That is, the symmetric derivative may exist but the actual derivative may not. A simple example of this is the function f(x) = |x|, i.e., the absolute value of x. This function has a symmetric derivative equal to zero, but of course is not differentiable at x=0 because the limit of [f(x+h)-f(x)]/h does not exist as h->0. This function is an example that you can show to your students to help sharpen their understanding of what exactly the derivative of a function is. If you know in advance that a function is differentiable at a point, then you are allowed to use either of the above expressions to evaluate the derivative there. Either expression will do, but sometimes one is easier than the other. In particular, when doing numerical differentiation on computers the symmetric form is often easier to deal with and has better convergence properties. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 02/08/2006 at 03:07:02 From: Greg Subject: Thank you (Derivative of sin x.) Dr. Douglas, I appreciate your time for answering my question. The answer was so thorough and also favoured my idea (which is always nice). Thank you so much. |
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