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### Alternate Forms of the Derivative Formula

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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

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) ->|-|<- (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.

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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