245
SECTION 2.5
.
.
The Chain Rule
189
2.5
THE CHAIN RULE
Suppose that the function
P
(
t
)
=
√
100
+
8
t
models the population of a city after
t
years. Then the rate of growth of that population after 2 years is given by
P
±
(2). At
present, we need the limit deﬁnition to compute this derivative. However, observe that
P
(
t
)i
s the composition of the two functions
f
(
t
)
=
√
t
and
g
(
t
)
=
100
+
8
t
,so that
P
(
t
)
=
f
(
g
(
t
)). Also, notice that both
f
±
(
t
) and
g
±
(
t
) are easily computed using existing
derivative rules. We now develop a general rule for the derivative of a composition of two
functions.
The following simple examples will help us to identify the form of the chain rule.
Notice that from the product rule
d
dx
[(
x
2
+
1)
2
]
=
d
[(
x
2
+
1)(
x
2
+
1)]
=
2
x
(
x
2
+
1)
+
(
x
2
+
1)2
x
=
2(
x
2
+
1)2
x
.
Of course, we can write this as 4
x
(
x
2
+
1), but the unsimpliﬁed form helps us to understand
the form of the chain rule. Using this result and the product rule, notice that
d
[(
x
2
+
1)
3
]
=
d
[(
x
2
+
1)(
x
2
+
1)
2
]
=
2
x
(
x
2
+
1)
2
+
(
x
2
+
1)2(
x
2
+
1)2
x
=
3(
x
2
+
1)
2
2
x
.
We leave it as a straightforward exercise to extend this result to
d
[(
x
2
+
1)
4
]
=
4(
x
2
+
1)
3
2
x
.
You should observe that, in each case, we have brought the exponent down, lowered
the power by one and then multiplied by 2
x
, the derivative of
x
2
+
1
.
Notice that we can
write (
x
2
+
1)
4
as the composite function
f
(
g
(
x
))
=
(
x
2
+
1)
4
, where
g
(
x
)
=
x
2
+
1 and
f
(
x
)
=
x
4
. Finally, observe that the derivative of the composite function is
d
[
f
(
g
(
x
))]
=
d
[(
x
2
+
1)
4
]
=
4(
x
2
+
1)
3
2
x
=
f
±
(
g
(
x
))
g
±
(
x
)
.
This is an example of the
chain rule,
which has the following general form.
THEOREM 5.1
(Chain Rule)
If
g
is differentiable at
x
and
f
is differentiable at
g
(
x
), then
d
[
f
(
g
(
x
))]
=
f
±
(
g
(
x
))
g
±
(
x
)
.
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View Full Document190
CHAPTER 2
.
.
Differentiation
246
PROOF
At this point, we can prove only the special case where
g
±
(
x
)
²=
0. Let
F
(
x
)
=
f
(
g
(
x
)).
Then,
d
dx
[
f
(
g
(
x
))]
=
F
±
(
x
)
=
lim
h
→
0
F
(
x
+
h
)
−
F
(
x
)
h
=
lim
h
→
0
f
(
g
(
x
+
h
))
−
f
(
g
(
x
))
h
Since
F
(
x
)
=
f
(
g
(
x
)).
=
lim
h
→
0
f
(
g
(
x
+
h
))
−
f
(
g
(
x
))
h
g
(
x
+
h
)
−
g
(
x
)
g
(
x
+
h
)
−
g
(
x
)
Multiply numerator
and denominator by
g
(
x
+
h
)
−
g
(
x
).
=
lim
h
→
0
f
(
g
(
x
+
h
))
−
f
(
g
(
x
))
g
(
x
+
h
)
−
g
(
x
)
lim
h
→
0
g
(
x
+
h
)
−
g
(
x
)
h
Regroup terms.
=
lim
g
(
x
+
h
)
→
g
(
x
)
f
(
g
(
x
+
h
))
−
f
(
g
(
x
))
g
(
x
+
h
)
−
g
(
x
)
lim
h
→
0
g
(
x
+
h
)
−
g
(
x
)
h
=
f
±
(
g
(
x
))
g
±
(
x
)
,
where the next to the last line is valid since as
h
→
0,
g
(
x
+
h
)
→
g
(
x
), by the continuity
of
g
. (Recall that since
g
is differentiable, it is also continuous.) You will be asked in the
exercises to ﬁll in some of the gaps in this argument. In particular, you should identify why
we need
g
±
(
x
)
²=
0in this proof.
It is often helpful to think of the chain rule in Leibniz notation. If
y
=
f
(
u
) and
u
=
g
(
x
), then
y
=
f
(
g
(
x
)) and the chain rule says that
dy
=
du
.
(5.1)
REMARK 5.1
The chain rule should make
sense intuitively as follows.
We think of
as the
(instantaneous) rate of change
of
y
with respect to
x
,
as the
(instantaneous) rate of change
of
y
with respect to
u
and
as
the (instantaneous) rate of
change of
u
with respect to
x
.
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 Spring '10
 Lagerstrom
 Chain Rule, Derivative, lim

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