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11.26
Definition (.)
Let
be sets and let
be functions.
The composition of
and
is the function
defined by:
i.e.,
is the set of all points
such that
is defined. The rule for
is
11.27
Example.
If
and
, then
and
Thus
So in this case
. Thus composition is not a commutative
operation.
If and , then
and
11.28
Exercise.
For each of the functions
below,
find functions
and
such that
. Then find a formula
for
.
- a)
-
.
- b)
-
.
- c)
-
.
11.29
Exercise.
A
Let
Calculate formulas for
,
,
,
,
, and
.
11.30
Entertainment (Composition problem.)
From the previous exercise you
should be able to find a subset
of
, and a function
such that
for all
. You should also be able to find a subset
of
and a function
such that
for all
. Can you find a subset
of
, and a function
such that
for
all
? One obvious example is the function
from the previous
example. To make the
problem more interesting, also add the condition that
for some
in
.
Before we prove the theorem we will give a few examples of how it is used:
11.32
Example.
Let
. Then
where
Hence
Let
. Then where
Hence
Usually I will not write out all of the details of a calculation like this.
I will just write:
Let
. Then
.
Proof of chain rule: Suppose is differentiable at and is differentiable at
. Then
|
(11.33) |
Since is differentiable at , we know that
Hence the theorem will follow from (11.33), the definition of
derivative, and the
product rule for limits of functions, if we can show that
Since is differentiable at , it follows from lemma 11.17
that
|
(11.34) |
Let be a generic sequence in
, such
that
. Then by (11.34), we have
|
(11.35) |
Since
is differentiable at , we have
From this and (11.35) it follows that
Since this holds for a generic sequence in
, we have
which is what we wanted to prove.
To complete the proof, I should show that is an interior point of
. This turns out to be rather tricky, so I will omit the proof.
Remark: Our proof of the chain rule is not valid in all cases,
but it
is valid in all cases where you are likely to use it.
The proof fails in the case where every interval
contains a point for which
. (You should check the proof to see where this
assumption was made.) Constant functions satisfy this condition, but if is
constant then is also constant so the chain rule holds trivially in
this case. Since the proof in the general case is more technical than
illuminating, I am going to omit it. Can you find a non-constant function
for which the proof fails?
11.36
Example.
If
is differentiable at
, and
, then
Also
i.e.,
|
(11.37) |
I will use this relation frequently.
11.38
Example (Logarithmic differentiation.)
Let
|
(11.39) |
The derivative of
can be found by using the quotient rule and the product
rule
and the chain rule. I will use a trick here which is frequently useful. I
have
Now differentiate both sides of this equation using (
11.37) to get
Multiply both sides of the equation by
to get
This formula is not valid at points where
, because we took logarithms
in the calculation. Thus
is differentiable at
, but our formula for
is not defined when
.
The process of calculating by first taking the logarithm of the
absolute value of
and then differentiating the result, is called logarithmic
differentiation.
11.40
Exercise.
A
Let
be the function defined in (
11.39) Show that
is differentiable at
, and calculate
.
Next: 12. Extreme Values of
Up: 11. Calculation of Derivatives
Previous: 11.2 Some General Differentiation
  Index
Ray Mayer
2007-09-07