1 §4.2 The Chain Rule (Pages 251~261) Composite Functions: The Chain Rule: Lectures 9 & 10.
Chapter 14 – Partial Derivatives 14.5 The Chain Rule 1 Objectives: How to use the Chain Rule and...
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Transcript of Chapter 14 – Partial Derivatives 14.5 The Chain Rule 1 Objectives: How to use the Chain Rule and...
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Chapter 14 – Partial Derivatives14.5 The Chain Rule
14.5 The Chain Rule
Objectives: How to use the Chain Rule and
applying it to applications
How to use the Chain Rule for Implicit Differentiation
Dr. Erickson
14.5 The Chain Rule 2
Chain Rule: Single Variable Functions
Recall that the Chain Rule for functions of a single variable gives the following rule for differentiating a composite function.
If y = f (x) and x = g (t), where f and g are differentiable functions, then y is indirectly a differentiable function of t, and
dy dy dx
dt dx dt
Dr. Erickson
14.5 The Chain Rule 3
Chain Rule: Multivariable Functions
For functions of more than one variable, the Chain Rule has several versions.◦ Each gives a rule for differentiating
a composite function.The first version (Theorem 2) deals with
the case where z = f (x, y) and each of the variables x and y is, in turn, a function of a variable t. ◦ This means that z is indirectly a function of t,
z = f (g(t), h(t)), and the Chain Rule gives a formula for differentiating z as a function of t.
Dr. Erickson
14.5 The Chain Rule 4
Chain Rule: Case 1
Since we often write ∂z/∂x in place of ∂f/∂x, we can rewrite the Chain Rule in the form
dz z dx z dy
dt x dt y dt
Dr. Erickson
14.5 The Chain Rule 5
Example 1 – pg. 930 # 2Use the chain rule to find dz/dt or dw/dt.
tytxyxz /1,5,4cos 4
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14.5 The Chain Rule 6
Example 2 – pg. 930 # 6Use the chain rule to find dz/dt or dw/dt.
2 2 2ln , sin , cos , tanw x y z x t y t z t
Dr. Erickson
14.5 The Chain Rule 7
Chain Rule: Case 2
Case 2 of the Chain Rule contains three types of variables: ◦ s and t are independent variables.◦ x and y are called intermediate variables.◦ z is the dependent variable.
Dr. Erickson
14.5 The Chain Rule 8
Using a Tree Diagram with Chain Rule
We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y.
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14.5 The Chain Rule 9
Tree DiagramThen, we draw branches from x and y to the independent
variables s and t.
◦On each branch, we write the corresponding partial derivative.
Dr. Erickson
14.5 The Chain Rule 10
Tree DiagramTo find ∂z/∂s, we find the product of the partial
derivatives along each path from z to s and then add these products:
z z x z y
s x s y s
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14.5 The Chain Rule 11
Example 3 – pg. 930 # 12Use the Chain rule to find ∂z/∂s and ∂z/∂t.
tsvtsuvuz 23,32),/tan(
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14.5 The Chain Rule 12
Chain Rule: General Version
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14.5 The Chain Rule 13
Example 4Use the Chain Rule to find the indicated
partial derivatives.
1 when ,,
;2,2,2,ln 222
yxy
R
x
R
xywyxvyxuwvuR
Dr. Erickson
14.5 The Chain Rule 14
Example 5Use the Chain Rule to find the indicated
partial derivatives.
0,1 when ,,
;,,,tan 1
strt
Y
s
Y
r
Y
rtwtsvsruuvwY
Dr. Erickson
14.5 The Chain Rule 15
Implicit DifferentiationThe Chain Rule can be used to give
a more complete description of the process of implicit differentiation that was introduced in Sections 3.5 and 14.3
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14.5 The Chain Rule 16
Implicit Differentiation If F is differentiable, we can apply Case 1 of the Chain
Rule to differentiate both sides of the equation F(x, y) = 0 with respect to x.◦ Since both x and y are functions of x,
we obtain:
0F dx F dy
x dx y dx
Dr. Erickson
14.5 The Chain Rule 17
Implicit DifferentiationHowever, dx/dx = 1.So, if ∂F/∂y ≠ 0, we solve for dy/dx
and obtain:
x
y
FFdy x
Fdx Fy
Dr. Erickson
14.5 The Chain Rule 18
Implicit DifferentiationNow, we suppose that z is given implicitly as a function
z = f(x, y) by an equation of the form F(x, y, z) = 0. ◦ This means that F(x, y, f(x, y)) = 0
for all (x, y) in the domain of f. If F and f are differentiable, then we can use the Chain
Rule to differentiate the equation F(x, y, z) = 0 as follows:
0F x F y F z
x x y x z x
Dr. Erickson
14.5 The Chain Rule 19
Implicit DifferentiationHowever,
So, that equation becomes:
( ) 1 and ( ) 0x yx x
0F F z
x z x
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14.5 The Chain Rule 20
Implicit DifferentiationIf ∂F/∂z ≠ 0, we solve for ∂z/∂x and
obtain the first formula in these equations.
◦The formula for ∂z/∂y is obtained in a similar manner.
FFz z yx
F Fx yz z
Dr. Erickson
14.5 The Chain Rule 21
Example 6Use equation 7 to find ∂z/∂x and ∂z/∂y .
zxyz ln
Dr. Erickson
14.5 The Chain Rule 22
More Examples
The video examples below are from section 14.5 in your textbook. Please watch them on your own time for extra instruction. Each video is about 2 minutes in length. ◦Example 2◦Example 4◦Example 5
Dr. Erickson