T
he
C
hain
R
ule
So far we have considered partial derivative and directional derivatives. The goal of
this section is to address the following question: What is the derivative of a function
f
(
x
,
y
)
where both
x
and
y
are functions of other variables?
We begin with the simple case where
x
and
y
are both functions of just a single
variable,
t
. That is,
x
=
g
(
t
) and
y
=
h
(
t
). This means that
z
=
f
(
x
,
y
) =
f
(
g
(
t
),
h
(
t
)). Thus, at
the end of the day,
z
is actually just a function of
t
. It might help to see a tangible example.
Suppose
z
=
f
(
x
,
y
) =
x
2
+
xy
3
,
x
=
g
(
t
) =
t
+ 1 and
y
=
h
(
t
) =
t
2
. If we substitute in for
x
and
y
, then we have that
z
=
f
(
x
,
y
) =
f
(
g
(
t
),
h
(
t
)) = (
t
+ 1)
2
+ (
t
+ 1)(
t
2
)
3
= (
t
+ 1)(
t
6
+
t
+ 1).
We can take the derivative of this function with respect to
t
just like we did when we
first learned about derivatives, since this is now just a function of only one-variable,
t
.
Thus, we see how we can take the derivative. However, this approach does seem a bit
cumbersome. Let us try to develop a more concise way to express
dz
/
dt
in terms of
derivatives of
x
’s and
y
’s.
Using the notation from the section on local linearity, if
x
=
x
(
t
) and
y
=
y
(
t
), then we
have that
dx
x
t
dt
and
dy
y
t
dt
. And if
z
=
f
(
x
,
y
), then
zz
zxy
x
y
.
Substituting in
D
x
and
D
y
, we have
zdx
zdy
ztt
x dt
y dt
x dt
y dt
t
.
Dividing both sides by
D
t
, we have
d
xz
d
tx
d
ty
d
y
t
. But as
t
ö
0, we have that
dz
z dx
z dy
dt
x dt
y dt
. We record this as the following:
The Chain Rule for
z
=
f
(
x
,
y
),
x
=
g
(
t
),
y
=
h
(
t
)
If
z
=
f
(
x
,
y
),
x
=
g
(
t
), and
y
=
h
(
t
) are differentiable functions, then
dz
z dx
z dy
dt
x dt
y dt
1