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Alternate Forms of the Derivative Formula


Date: 10/06/2001 at 23:37:55
From: Dina 
Subject: Equivalent Forms of the Derivative Formula

I am a first-year AP Calculus student and I understand the notion of 
the derivative and how/why the derivative of f at x is the limit as 
h approaches 0 of [f(x + h)- f(x)]/h, but I am not as clear about the 
alternate forms of the derivative and WHY they work. 

Specifically, why does f'(a) = limit as x approaches a of 
[f(x)-f(a)]/(x-a), and how do you use this formula to find the 
derivative? I always get confused about when to plug in x for a or a 
for x. 

Also, why does the symmetric difference quotient work (f'(a) = limit 
as h approaches 0 of [f(a+h)-f(a-h)]/2h)? Why do you even divide by 
2h? it looks as if you're taking the average, but I'm still not 
completely convinced. 

Any help you could provide to clarify and explain these issues would 
be greatly appreciated! Thanks a lot.


Date: 10/08/2001 at 11:28:54
From: Doctor Jubal
Subject: Re: Equivalent Forms of the Derivative Formula

Hi Dina,

Thanks for writing to Dr. Math.

You say you understand the usual definition of the derivative:

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

What I'm going to do is start with this form of the derivative 
definition and derive the alternate forms from it. Let me know whether 
I succeed in enlightening you in the process.

Here's a diagram showing the terms used in the traditional definition 
of the derivative. x is some point on the x axis. (x+h) is some point 
to its right. (x+h) - x = h is the distance between them.

  |
  |        ______
  |       /|
  | __   / |
  |   \_/| |
  |      | |
  |      | |
  |      | |
  +------------------
      x->|-|<- (x+h)
          h

The original form is a limit that focuses on the distance h between 
two points where you evaluate the function. Let's say instead that we 
want to focus on the position of the points themselves. We'll call the 
left point a and the right point x. Then the distance between them is 
x-a.

  |
  |        ______
  |       /|
  | __   / |
  |   \_/| |
  |      | |
  |      | |
  |      | |
  +----------------
      a->|-|<-x
         x-a

The only potentially confusing thing here is that the x we're using 
now is not the same thing as the x we were using before. It's an 
unfortunate choice of letters, but you can just look at the two 
diagrams and see which variables correspond to which other variables 
in our new naming scheme.

Making these substitutions (a for x, x for x+h, x-a for h) gives us:

  f'(a) = lim((x-a)->0) [f(x) - f(a)]/(x-a)

Well, if x-a is approaching zero, then x is approaching a, and we can
rewrite this as the second definition you asked about:

  f'(a) = lim(x->a) [f(x) - f(a)]/(x-a)

To get the symmetric formula, we have to try a third naming scheme: 
Let's call the point on the left a-h, and the point on the right a+h, 
and then the distance between them has to be (a+h) - (a-h) = 2h. 
This is where the 2h comes from. It is the distance bewteen (a+h) and 
(a-h). Here's a diagram to help you keep track of things.

  |
  |        ______
  |       /|
  | __   / |
  |   \_/| |
  |      | |
  |      | |
  |      | |
  +----------------
 (a-h) ->|-|<- (a+h)
         2h

Then, plugging these new names for the same old quantities back into 
the derivative definition (a-h for x, a+h for x+h, 2h for h), we get

  f'(a-h) = lim(2h->0) [f(a+h) - f(a-h)]/2h 

If 2h is approaching 0, h must be approaching zero, and if h is
approaching zero, the a-h on the left-hand side must be approaching a, 
so we can rewrite:

  f'(a) = lim(h->0) [f(a+h) - f(a-h)]/2h

As I hope you can see, all these definitions of the derivative are 
really the same definition, just with different names for the 
variables.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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