2010 – Chain Rule
AP CALCULUS
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Activity: Teacher-Directed Instruction
COMPOSITE FUNCTIONS
Know: Need:
Know: Need: REM:
2( 2)y x 8( 2)y x 2( 2)y x 4 3 2 2(4 3 2 7 5)y x x x x
2010
10
t(x)=1.0825(x)
d(x)=.5(x)10.83
𝑓 (𝑥 )=𝑡(𝑑 (𝑥 ))
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𝑑 (𝑥 ) h𝑤 𝑎𝑡 𝑖𝑠 𝑑𝑜𝑛𝑒 𝑓𝑖𝑟𝑠𝑡
COMPOSITE FUNCTIONS
( ( ))y f g x Let and
y = outside u = inside
function function
( )u g x( )y f u
g(x) - what is done first.
f(x)
xg(x)
g(x)
f(g(x))
COMPOSITE FUNCTIONS
DECOMPOSE y = (outside) u = (inside).
2 3( 1)y x
23 1y x x
1
1y
x
tan 2y x2tany x
2tany x
𝑦=𝑢3 𝑢=𝑥2+1
𝑦=𝑢12 𝑢=3 𝑥2 −𝑥+1
𝑦=𝑢− 1 𝑢=𝑥+1
𝑦=tan(𝑢) 𝑢=2𝑥
𝑦=tan(𝑢) 𝑢=𝑥2
𝑦=𝑢2 𝑢=tan(𝑥)
tan𝑥2
¿
( ( ( ))d
f g xdx
In Words:
__________________________________________________________
Derivative of __________________________________________________
a composite function2
func
tions
2 pa
rts ¿ 𝑓 ′ (𝑔 (𝑥 ) )∗𝑔 ′ (𝑥)
d(outside)leave the inside alone*d(inside)
Texas two step
Chain Rule
( )u g x
Leibniz Notation:
In Words:
__________________________________________________________
( ( ))y f g x ( )y f u
𝑑𝑦𝑑𝑥
=𝑑𝑦𝑑𝑢
∗𝑑𝑢𝑑𝑥
d(outside)leave the inside alone*d(inside)
Example 1:
Example:
Let y = u =
_______
dy
du
du
dx
dy
dx
2 3(3 2 1)y x x
𝑢3 3 𝑥2+2𝑥+1
3𝑢2 (6 𝑥+2)
3𝑢2∗(6 𝑥+2)
𝑑𝑦𝑑𝑥
=3(3𝑥¿¿2+2𝑥+1)2∗(6 𝑥+2)¿
Example 2:2 3(3 2 )y x x
Two steps: 𝑦=𝑢3 𝑢=3 𝑥−2 𝑥2
𝑦 ′=3 (3 𝑥− 2𝑥2 ) 2∗(3 − 4 𝑥)
𝑦 ′=(9−12 𝑥 )∗(3 𝑥− 2𝑥2)2
Example 3:
23 ( 2)y x 𝑦=(𝑥2+2)13
𝑦=𝑢13 𝑢=(𝑥2+2)2
𝑦 ′=13(𝑥2+2)
− 23 ∗(2 𝑥)
𝑦 ′=(23𝑥)¿
Example 4:
Example:1
(2 3)y
t
2
7
(3 5)y
x
Note: 2
1
4(3 5)y
x
¿ (2 𝑡−3)−1
𝑦=𝑢− 1 𝑢=(2𝑡− 3)
𝑦 ′=−1 (2 𝑡−3 )− 2∗(2)
𝑦 ′=−2(2𝑡− 3)−2
𝑦 ′=− 2
(2 𝑡−3)2
2
7
(3 5)y
x
𝑦=−7 ¿
𝑦 ′=14 (3 𝑥−5)− 3∗(3)
𝑦 ′=42 (3 𝑥−5)− 3
𝑦 ′=42
(3 𝑥− 5)2
𝑦=−7𝑢−2 𝑢=3 𝑥−5
2
1
4(3 5)y
x
𝑦=
14
(3 𝑥−5)− 2
𝑦=14𝑢− 2 𝑢=3 𝑥−5
𝑦 ′=− 24
(3 𝑥−5)− 3∗(3)
𝑦 ′=− 64
(3 𝑥− 5)−3
𝑦 ′=−3
2(3 𝑥−5)3
Example 6:
cos(3 )y x
2cosy x
𝑦 ′=−sin (3 𝑥 )∗3
𝑦 ′=−3 sin (3 𝑥)
𝑦=cos𝑢 𝑢=3 𝑥
𝑦=𝑢2 𝑢=cos 𝑥𝑦 ′=2¿
𝑦 ′=−2 sin (𝑥 )∗ cos (𝑥)
Example 7:Extended Chain:
Ex: OR
WORDS:
Extended Chain: ___________
d(outside)*d(middle)*d(inside)
¿¿𝑦 ′=3¿¿
𝑦 ′=6 𝑥 𝑠𝑖𝑛2(𝑥¿¿2)cos (𝑥¿¿ 2)¿¿
Number of functions = number of parts
𝑦=𝑠𝑖𝑛3 (𝑥¿¿ 2)¿
Derivative of the Absolute Value Function
REM:2x x
2d dx x
dx dx Do not simplify.
Use the Chain Rule.¿ (𝑥¿¿2)
12 ¿
𝑦 ′=12(𝑥2)
12∗(2 𝑥)
𝑦 ′=𝑥√𝑥
=𝑥
|𝑥|
𝑦=|𝑢|
𝑦 ′=𝑢|𝑢|
∗𝑢 ′
General Rules: Working with number values
Find the derivative.
1) f(x) + g(x) at x = 3 2) 2f(x) – 3g(x) at x =2
3) f(x)*g(x) at x = 2 4) f(x) / g(x) at x = 3
5) f(g(x)) at x = 2 6) (f(x))3 at x = 3
x f(x) g(x) f / (x) g / (x)
2 8 3 -4
3 2 -6 -512
23
𝑓 ′ (𝑥 )+𝑔 ′ (𝑥) = 2 𝑓 ′ (𝑥 )− 3𝑔 ′ (𝑥)¿2 (− 4 )− 3( 23 )=− 10
𝑓 𝑔′+𝑔𝑓 ′ 8 ( 23 )+3(− 4 ) 𝑔 𝑓 ′ − 𝑓𝑔′
𝑔2
(−6 )( 12 )−(2)(−5)
(−6)2
𝑓 ′ (𝑔 (𝑥 ) )∗𝑔 ′ (𝑥) 𝑓 ′ (𝑔 (2 ) )∗𝑔 ′ (2)
𝑓 ′ (3 )∗( 23 )=( 1
2 )( 23 )=1
3
𝑑𝑦𝑑𝑥
3
𝑓 (𝑥 )∗ 𝑓 ′(𝑥 )
3¿3∗22∗
12=3∗4∗
12=6
Last Update
• 10/13/07
• Assignment p. 153 # 13 – 31 odd, 56
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