Chapter 16
Section 16.3
The Mean-Value Theorem; The Chain
Rule
Chain Rule for a 1 Variable Function
The derivative of a complicated function that can be formed by substitutions (compositions) of
simpler functions can be found by apply the chain rule. That is taking the derivatives of each
simpler function and multiplying them together.
Example
The variables y, u, v, and x
are related by the equations
below. Find 𝑑𝑦
.
𝑑𝑥
𝑦 = 𝑒 𝑢 + 5𝑢
𝑢 = tan 𝑣
𝑣= 𝑥
Dependence tree
y
y depends on u
u
u depends on v
v
v depends on x
Chain Rule for a 2 Variable Function
If 𝑧 = 𝑓 𝑥, 𝑦 and 𝑥 = 𝑔 𝑠, 𝑡 and 𝑦 =
ℎ 𝑠, 𝑡 then ultimately the variable z is a
function of the variables s and t. The
variable z has 2 partial derivatives one
with respect to s and the other t.
Partial Derivative Formula (respect to s):
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
+
𝜕𝑠 𝜕𝑥 𝜕𝑠 𝜕𝑦 𝜕𝑠
Formula with derivatives:
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑣
=
𝑑𝑥 𝑑𝑢 𝑑𝑣 𝑑𝑥
𝑑𝑦
𝑑𝑢
𝑑𝑢
𝑑𝑣
Formula with Variables:
𝑑𝑦
1
𝑢
2
= 𝑒 + 5 sec 𝑣
𝑑𝑥
2 𝑥
𝑑𝑣
𝑑𝑥
x
Dependence tree
z
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑠
s
x
𝜕𝑥
𝜕𝑡
t
𝜕𝑦
𝜕𝑦
𝜕𝑠
s
y
𝜕𝑦
𝜕𝑡
t
Partial Derivative Formula (respect to t):
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
+
𝜕𝑡 𝜕𝑥 𝜕𝑡 𝜕𝑦 𝜕𝑡
Example
𝑤 = 𝑥𝑦 3 + 𝑒 𝑧
The variables u, v, w, x, y, and z all depend on each other as
𝑥 = 𝑢 ln 𝑣
given to the right. Draw the dependence tree and give the
𝑦 = 𝑢2 + 2𝑣
partial derivatives using partial derivative and using the
𝑧= 𝑣
variables.
Partial Derivative Formula:
Partial Derivative Formula:
Dependence tree
𝜕𝑤 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦
𝜕𝑤 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 𝜕𝑤 𝑑𝑧
w
=
+
=
+
+
𝜕𝑢
𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢
𝜕𝑣
𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 𝜕𝑧 𝑑𝑣
z
x
y
With the variables:
With the variables:
𝜕𝑤
𝑢
1
𝜕𝑤
v
u v
u v
= 𝑦 3 + 3𝑥𝑦 2 2 + 𝑒 𝑧
= 𝑦 3 ln 𝑣 + 3𝑥𝑦 2 2𝑢
𝜕𝑣
𝑣
𝜕𝑢
2 𝑣
Notice each position in the tree that ends with the independent variable that you are taking
the derivative with respect to represents a term in the partial derivative. Each tier (level) of the
tree represents a factor in that term.
Example
Find the derivative of the function 𝑧 = 𝑓 𝑥, 𝑦 along the
curve r 𝑡 that are given to the right.
Dependence tree
z
x
y
t
t
𝑧 = 𝑥 4 − 𝑒 𝑥 𝑦 2 + ln 𝑦
r 𝑡 = cos 𝑡 , 𝑡
Derivative Formula using partial derivatives:
𝑑𝑧
𝑑𝑡
=
𝜕𝑧 𝑑𝑥
𝜕𝑥 𝑑𝑡
+
Derivative Formula using the variables:
𝑑𝑧
1 1
3
𝑥 2
𝑥
= 4𝑥 − 𝑒 𝑦 − sin 𝑡 + −2𝑒 𝑦 +
𝑑𝑡
𝑦 2 𝑡
𝜕𝑧 𝑑𝑦
𝜕𝑦 𝑑𝑡
Example
The variables u, v, w, x, y, z, r, and 𝜃 all
depend on each other as given to the
𝜕𝑤
right. Find formulas for 𝜕𝑤
and
using
𝜕𝑟
𝜕𝜃
both partial derivatives and the
variables.
Partial derivative formula for
𝜕𝑤 𝜕𝑤
:
𝜕𝑟 𝜕𝑟
Dependence tree
w
𝑤 = 𝑥𝑒 𝑦 + cosh 𝑧
𝑥= 𝑢
𝑦 = 𝑢5 +𝑣 3
𝑧 = 𝑟2
𝑢 = sin 𝜃
𝑣 = 𝑟 cos 𝜃
=
𝜕𝑤 𝜕𝑦 𝜕𝑣
𝜕𝑦 𝜕𝑣 𝜕𝑟
+
𝜕𝑤 𝑑𝑧
𝜕𝑧 𝑑𝑟
Partial derivative with the variables:
𝜕𝑤
= 𝑥𝑒 𝑦 3𝑣 2 cos 𝜃 + sinh 𝑧 2𝑟
𝜕𝑟
𝜕𝑤 𝜕𝑤
Partial derivative formula for 𝜕𝜃 : 𝜕𝜃 =
𝜕𝑤 𝑑𝑥 𝑑𝑢
𝜕𝑥 𝑑𝑢 𝑑𝜃
𝜕𝑤 𝜕𝑦 𝑑𝑢
𝜕𝑤 𝜕𝑦 𝜕𝑣
+ 𝜕𝑦 𝜕𝑢 𝑑𝜃 + 𝜕𝑦 𝜕𝑣 𝜕𝜃
Partial derivative with variables:
𝜕𝑤
1
𝑦
=𝑒
cos 𝜃 + 𝑥𝑒 𝑦 5𝑢4 cos 𝜃 + 𝑥𝑒 𝑦 3𝑣 2 −𝑟 sin 𝜃
𝜕𝜃
2 𝑢
z
y
x
u
u
𝜃
𝜃 𝜃 r
v
r