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.