MAT 473
9
4.
Chain Rule. Partial Derivatives.
Proposition 4.1
(Chain Rule)
.
Suppose
f
:
U
⊆
R
n
→
R
m
is di
ff
erentiable at
x
∈
U
and
g
:
V
⊆
R
m
→
R
is di
ff
erentiable at
f
(
x
)
∈
V
. Then
g
◦
f
is di
ff
erentiable at
x
, with
(
g
◦
f
)
(
x
) =
g
(
f
(
x
))
◦
f
(
x
)
.
Proof.
Put
y
=
f
(
x
) and choose
s, W
s
and
t, W
t
as in Lemma 3.6 such that
f
(
x
+
h
)
=
f
(
x
) +
f
(
x
)(
h
) +
s
(
h
)
h
for all
h
∈
W
s
,
g
(
y
+
k
)
=
g
(
y
) +
g
(
y
)(
k
) +
t
(
k
)
k
for all
k
∈
W
t
,
lim
h
→
0
s
(
h
) =
0
, and lim
k
→
0
t
(
k
) =
0
.
Now let
k
(
h
) =
f
(
x
)(
h
) +
s
(
h
)
h
; certainly
lim
h
→
0
k
(
h
) =
0
, so we can choose
>
0 such that
k
(
h
)
∈
W
t
whenever
h
<
. Thus for
all
h
∈
W
=
W
s
∩
B
(
0
), we have
(
g
◦
f
)(
x
+
h
)
=
g
(
f
(
x
) +
f
(
x
)(
h
) +
s
(
h
)
h
)
=
g
(
y
+
k
(
h
))
=
g
(
y
) +
g
(
y
)(
k
(
h
)) +
t
(
k
(
h
))
k
(
h
)
=
g
(
f
(
x
)) +
g
(
f
(
x
))(
f
(
x
)(
h
) +
s
(
h
)
h
) +
t
(
k
(
h
))
k
(
h
)
=
(
g
◦
f
)(
x
) +
g
(
f
(
x
))(
f
(
x
)(
h
)) +
g
(
f
(
x
))(
s
(
h
))
h
+
t
(
k
(
h
))
k
(
h
)
h
h
.
Now clearly lim
h
→
0
g
(
f
(
x
))(
s
(
h
)) =
0
. Moreover,
0
≤
t
(
k
(
h
))
k
(
h
)
h
=
t
(
k
(
h
))
f
(
x
)(
h
) +
s
(
h
)
h
h
≤
t
(
k
(
h
))
f
(
x
)
h
+
s
(
h
)
h
h
=
t
(
k
(
h
))
(
f
(
x
)
+
s
(
h
)
)
→
0
as
h
→
0
, so lim
h
→
0
t
(
k
(
h
))
k
(
h
)
h
=
0
. Since
h
→
g
(
f
(
x
))(
f
(
x
)(
h
)) = (
g
(
f
(
x
))
◦
f
(
x
)) (
h
)
is linear, another application of Lemma 3.6 completes the proof.
Definition 4.2.
For
f
:
U
⊆
R
n
→
R
m
, and for 1
≤
i
≤
m
, define
f
i
:
U
→
R
by
f
i
(
x
) =
e
i
, f
(
x
)
=
f
(
x
)
i
for all
x
∈
R
n
. We call
f
i
the
i
th
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 Spring '09
 Calculus, Chain Rule, Derivative, Rn Rm

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