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### Assigning a Piecewise Function in a Proof

```
Date: 01/11/2001 at 14:24:25
From: Mahesh Bhat
Subject: Chain rule proof

Hi,

As I was trying to teach my niece the chain rule, I came across a step
which I have trouble understanding. I'll narrate the proof until the
step with which I have the problem.

Proof:

lim  f o g(x) - f o g(c)   lim  f(g(x)) -  f(g(c))
(f o g)'(c) = x->c ------------------- = x->c ------------------
x - c                     x - c

We must show that the limit on the right hand side has the value
f'(g(c)).g'(x).

Note that if we knew that g(x) - g(c) is not equal to zero for all x
near c, x not equal to c we could write:

f(g(x)) -  f(g(c))   f(g(x)) - f(g(c))   g(x) - g(c)
------------------ = ----------------- . -----------
x - c             g(x) - g(c)         x - c

and show that the first factor on the right has limit f'(g(c)) and the
other has g'(c). The problem is that g(x) - g(c) could be zero even if
x is not equal to c.

(Now the step I do not understand.)
===================================

We introduce a function:

/
| f(u) - f(g(c))
| -------------- - f'(g(c)),   when u <> g(c)
e(u) = {     u - g(c)
|
| 0,                           when u = g(c)
\

How can we say that e(u) is 0 when u = g(c)?

The only way I thought that could happen is if the numerator of the
first term is (u - g(c)).f'(g(c)) = f(u) - f(g(c)). Even so how can I
be sure that the f(u) - f(g(c)) can have (u - g(c)) and f'(g(c)) as
its factors? Can we guarantee that this manipulation will always be
true?

Even though we have introduced e(u), how can we have a rule that
conveniently says e(u) = 0 when u = g(c)?

I have seen other proofs, but I would like to understand this step.

When I was in school I probably just accepted this step and went on.

Thank you,
Mahesh
```

```
Date: 01/11/2001 at 14:32:25
From: Doctor Schwa
Subject: Re: Chain rule proof

>How can we say that e(u) is 0 when u = g(c)?

This is simply the DEFINITION we have given e(u). Depending on the
value of u, it has different definitions.

It's like saying: f(x) = {x^2 when x is positive, x when x is
negative}. There's no *reason* f has to be defined that way; that's
just the definition.

>The only way I thought that could happen is if the numerator of the
>first term is (u - g(c)).f'(g(c)) = f(u) - f(g(c)). Even so how can I
>be sure that the f(u) - f(g(c)) can have (u - g(c)) and f'(g(c)) as
>its factors? Can we guarantee that this manipulation will always be
>true?

So you see there's no manipulation to be done. It's just the
definition of a new function. We choose to define it any way we want
to. We define it this particular way to patch up the hole at g(x) =
g(c).

detail. That's an important part of mathematics! You're right that on
seeing it for the first time most people probably just accept this
step on faith, memorize it, or ignore it... but for future
mathematicians, this kind of thinking is important.

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

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