Download - 2.4 The Chain Rule Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002
2.4 The Chain Rule
Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002
Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002
U.S.S. AlabamaMobile, Alabama
HWQ
Let f(x) and g(x) be 2 differentiable functions such that:
x F(x) G(x) F’(x) G’(x)
4 1 7 8 -8
3 -5 -3 -4 6
-5 2 -10 9 -1
Find the derivative of f(x)g(x) at x = -5. -92
Calculus Warm-Up
3cosd
xdx
23cos sinx x
Calculus Warm-Up
2 1d
xdx
We will come back to this problem later.
The Chain Rule
Copyright © Cengage Learning. All rights reserved.
2.4
Find the derivative of a composite function using the Chain Rule.
Objective:
We now have a pretty good list of “shortcuts” to find derivatives of simple functions.
Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions.
We do this with the chain rule.
Consider a simple composite function:
6 10y x
2 3 5y x
If 3 5u x
then 2y u
6 10y x 2y u 3 5u x
6dy
dx 2
dy
du 3
du
dx
dy dy du
dx du dx
6 2 3
and another:
5 2y u
where 3u t
then 5 3 2y t
3u t
15dy
dt 5
dy
du 3
du
dt
dy dy du
dt du dt
15 5 3
5 3 2y t
15 2y t
5 2y u
and another:29 6 1y x x
23 1y x
If 3 1u x
3 1u x
18 6dy
xdx
2dy
udu
3du
dx
dy dy du
dx du dx
2y u
2then y u
29 6 1y x x
2 3 1dy
xdu
6 2dy
xdu
18 6 6 2 3x x This pattern is called the chain rule.
dy dy du
dx du dx Chain Rule:
Example: sinf x x 2 4g x x Find: at 2f g x
( ( )) '( ( )) '( )df g x f g x g x
dx
2sin 4f g x x
or:
2sin 4y f g x x
2 2cos 4 4dy d
x xdx dx
2cos 4 2dy
x xdx
Differentiate the outside function...
…then the inside function
at 2, 4x y
( ( )) '( ( )) '( )df g x f g x g x
dxChain Rule:
2cos 2 4 2 2dy
dx
cos 0 4dy
dx
4dy
dx
22 cos 4dy
x xdx
Use the chain rule to differentiate:
3cosd
xdx
23cos sinx x
Use the chain rule to differentiate:
2 1d
xdx
2 1
x
x
Another example:
2cos 3d
xdx
2cos 3
dx
dx
2 cos 3 cos 3d
x xdx
derivative of theoutside function
derivative of theinside function
It looks like we need to use the chain rule again!
con’t:
2cos 3d
xdx
2cos 3
dx
dx
2 cos 3 cos 3d
x xdx
2cos 3 sin 3 3d
x x xdx
2cos 3 sin 3 3x x
6cos 3 sin 3x x
The chain rule can be used more than once.
(That’s what makes the “chain” in the “chain rule”!)
The most common mistake on derivative tests is to forget to use the chain rule.
Every derivative problem could be thought of as a chain-rule problem:
2dx
dx2d
x xdx
2 1x 2x
derivative of outside function
derivative of inside function
The derivative of x is one.
Practice:
Differentiate: 32 2f x x
226 2f x x x
Practice:
Differentiate: 2
7
2 3g t
t
3
28
2 3g t
t
sin 2 ' ?
' cos 2 2
' 2cos 2y x
y x y
y x
2
2
tan 3 ' ?
' sec 3 3
' 3sec 3
y x y
y x
y x
' sin 1
cos 1 ' ?
' sin 1 1
y x y
y x
y x
2cos 3 ' ?
' 2cos 3
' 2 cos 3
y x y
y x
y x
2
2
2
cos 3 ' ?
' sin 3
' 6 sin 3
6
y x y
y x x
y x x
2
2
2
cos 3 ' ?
' sin 3 2 3
' 18 si
3
n 9
y x y
x
x x
x
y
y
BC Homework
• 2.4 Day 1 p. 137: 7-31 odd, 41-57 odd,
67-71 odd, 81,83
• 2.4 Day 2: MMM pgs. 44-46
• 2.4 Day 3: MMM pg. 50
2.4 The Chain Rule – Day 2
Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002
HWQ
Differentiate:
3sin 4f t t
212sin 4 cos 4f t t t
2.4 Warm-up
223
1/
2
322 / 32
3 2
1 Where does ' 0?
Where does ' not exist?
2' 1 2
34
' 0@ 0
' when 1
1
3 1
0 1
f x x f x
f x
f x x x
x
x
f
f x x
x x
f x DNE x x
223 1
f x x
2 2
1/ 2 1/ 22 2 2
1/ 2 1/ 22 3 2
1/ 2 1/ 22 2 3
1/ 2 1/ 22 2
2 3
2
2
3
1/ 2
2
2
1 ' ?
1' 2 1 1 2
2
2
2 3
1 1
2 1 1
1 1
2 1
1
1
1
f x x x f x
f x x x x x x
x x x x
x x x x
x x
x x xx
f x x
x
x
x
x
Common Denominator
3 2
1/ 3 2 / 32 2
2 / 32
21/ 32
2 / 32
2 / 32
1/ 3 2 / 32 2 2
2 / 3 2 / 32 2
2 / 32
2 2
2
1/ 32
2
4
/ 32 2 2
2 / 3 4 /32 32/2
' ?4
11 4 4 2
3'4
24
3 4
4
3 4 4 2
3 4 3 4
4
3 4 2
3 4 3 4 12
3
2
4 3 4 4
4
xf x f x
x
x x x xf x
x
xx
x
x
x x x
x x
x
x x
x x x
x x
x
x
xf
x
x
HWQ (no calculator)
Determine the point(s) at which the graph of
has a horizontal tangent. 2 1
xf x
x
1,1
2
2
2
2
2 22
2 2
2 22 32
3 1 ' ?
3
3 3 3 1 23 1' 2
3 3
3 1 3 2 3 19 6 22
3
3 2
33
9
xy y
x
x x xxy
x x
x x x x x
x
x x
xx
AB Homework
• 2.4 Day 1 p. 137: 1-31 odd, 41-57 odd
• 2.4 Day 2: p. 137: 59-73 odd, 79-89 odd
• 2.4 Day 3: MMM pgs. 44-46
• 2.4 Day 4: Chain Rule W/S