composition of functions: the process of combining two or more functions in order to create another...
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Composition of Functions:The process of combining two or more functions in order to create another function.One function is evaluated at a value of the independent variable and the result is substituted into the other function as the independent variable.The composition of functions f and g is written as:
( π βπ ) (π₯ )ΒΏ π (π (π₯ ) )
1.7 β The Chain Rule
The composition of functions is a function inside another function.
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( π βπ ) (π₯ )ΒΏ π (π (π₯ ) )1.7 β The Chain Rule
Given:, find .
( π βπ ) (π₯ )= π (π (π₯ ) )=ΒΏ2 (π₯2+5 )+3
ΒΏ2 π₯2+10+3
2 π₯2+1 3( π βπ ) (π₯ )= π (π (π₯ ) )=ΒΏ
Find .
(πβ π ) (π₯ )=π ( π (π₯ ) )=ΒΏ(2 π₯+3 )2+5
4 π₯2+6π₯+6 π₯+9+5
4 π₯2+12 π₯+14(πβ π ) (π₯ )=π ( π (π₯ ) )=ΒΏ
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( π βπ ) (π₯ )ΒΏ π (π (π₯ ) )1.7 β The Chain Rule
Given:, find .
( π βπ ) (π₯ )= π (π (π₯ ) )=ΒΏ(π₯2+2 )3+ (π₯2+2 )β6
Find .
(πβ π ) (π₯ )=π ( π (π₯ ) )=ΒΏ(π₯3+π₯β6 )2+2
( π βπ ) (π₯ )= π (π (π₯ ) )=ΒΏ(π₯2+2 )3+π₯2β4
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1.7 β The Chain RuleReview of the Product Rule:
π¦=( 3π₯3+2 π₯2 )2ΒΏ ( 3π₯3+2 π₯2 ) (3 π₯3+2π₯2 )
π¦ β²= (3 π₯3+2π₯2 ) (9 π₯2+4 π₯ )+( 9 π₯2+4 π₯ ) (3 π₯3+2π₯2 )
π¦ β²=2 ( 3π₯3+2 π₯2 ) (9 π₯2+4 π₯ )
π¦=( 6 π₯2+π₯ )3ΒΏ ( 6 π₯2+π₯ ) (6 π₯2+π₯ ) ( 6π₯2+π₯ )+
π¦ β²=3 (6 π₯2+π₯ )2 (12π₯+1 )
π¦ β²=(6 π₯2+π₯ )2 (12 π₯+1 )+ (6 π₯2+π₯ )2 (12π₯+1 )+( 6 π₯2+π₯ )2 (12π₯+1 )
π¦=( 3π₯3+2 π₯2 )2 π¦=( 6 π₯2+π₯ )3and are composite functions.
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Additional Problems:
π¦=( 3π₯3+2 π₯2 )2 π¦ β²=2 ( 3π₯3+2 π₯2 ) (9 π₯2+4 π₯ )
π¦=( 6 π₯2+π₯ )3 π¦ β²=3 (6 π₯2+π₯ )2 (12π₯+1 )
π¦=(π₯3+2 π₯ )9 (π₯3+2 π₯ )89 (3 π₯2+2 )
π¦=(5 π₯2+1 )4 (5 π₯2+1 )34 (10 π₯ )
π¦ β²=ΒΏπ¦ β²=ΒΏ
π¦=( 2π₯5β3π₯4 +π₯β3 )13 (2 π₯5β3π₯4+π₯β3 )1213 (10 π₯4β12 π₯3+1 )π¦ β²=ΒΏ
1.7 β The Chain Rule
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Find
π¦=π’3β7π’2 π’=π₯2+3
ππ¦ππ₯
=ππ¦ππ’βππ’ππ₯
1.7 β The Chain Rule
ππ¦ππ’
=3π’2β14π’ππ’ππ₯
=2π₯
ππ¦ππ₯
=ΒΏ(3π’2β14π’ )β2π₯ππ¦ππ₯
=ΒΏ(3 (π₯2+3 )2β14 (π₯2+3 ))2 π₯
ππ¦ππ₯
=2π₯ (π₯2+3 ) (3 (π₯2+3 )β14 )ππ¦ππ₯
=2π₯ (π₯2+3 ) (3 π₯2+9β14 )
ππ¦ππ₯
=2π₯ (π₯2+3 ) (3 π₯2β5 )
π¦=π’3β7π’2 π’=π₯2+3
π¦=(π₯2+3 )3β7 (π₯2+3 )2
ππ¦ππ₯
=3 (π₯2+3 )22 π₯β14 (π₯2+3 ) 2π₯
ππ¦ππ’
=2π₯ (π₯2+3 ) (3 (π₯2+3 )β14 )
ππ¦ππ’
=2π₯ (π₯2+3 ) (3 π₯2+9β14 )
ππ¦ππ’
=2π₯ (π₯2+3 ) (3 π₯2β5 )
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Find the equation of the tangent line at for the previous problem.
1.7 β The Chain Rule
π¦=β48π₯=1
π¦=(π₯2+3 )3β7 (π₯2+3 )2
π¦β π¦1=π (π₯βπ₯1 )
ππ‘ππ=ππ¦ππ₯
=β16
ππ¦ππ₯
=2π₯ (π₯2+3 ) (3 π₯2β5 )
π¦ββ48=β16 (π₯β1 )
π¦+48=β16π₯+16
π¦=β16 π₯β32
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1.7 β The Chain RuleThe position of a particle moving along a coordinate line is, , with s in meters and t in seconds. Find the rate of change of the particle's position at seconds.
π (π‘ )=β12+4 π‘
π (π‘ )=(12+4 π‘ )12
ππ ππ‘
=π β² (π‘ )=12
(12+4 π‘ )β 1
2 (4 )
ππ ππ‘
=π β² (π‘ )= 2
(12+4 π‘ )12
ππ‘ π‘=6 ,ππ ππ‘
=π β² (6 )= 2
(12+4 (6 ) )12
ππ ππ‘
=π β² (6 )=13πππ‘πππ /π ππππππ
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1.7 β The Chain RuleThe total outstanding consumer credit of a certain country can be modeled by , where C is billion dollars and x is the number of years since 2000. a) Find .b) Using this model, predict how quickly outstanding consumer credit will be rising in 2010.
a)
b) π₯=2010β2000=10 π¦ππππ
ππΆππ₯
=29.91ππππππππππππππ /π¦πππ
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1.8 βHigher-Order DerivativesHigher-order derivatives provide a method to examine how a rate-of-change changes.
Notations
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1.8 βHigher-Order DerivativesFind the requested higher-order derivatives.
Find
π β² (π₯ )=12 π₯3β15 π₯2+8
π β² β² (π₯ )=36 π₯2β30 π₯
π β² β² β² (π₯ )=72π₯β30
π β² (π₯ )=6π₯2+12 π₯β57
π β² β² (π₯ )=12π₯+12
π β² β² β² (π₯ )=12
π ( 4 ) (π₯ )=0
Find
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1.8 βHigher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of change with direction.
π£ (π‘ )=π β² (π‘ )=ππ ππ‘
The velocity function, , is obtain by differentiating the position function with respect to time.
π (π‘ )=4 π‘2+π‘π£ (π‘ )=π β² (π‘)=8 π‘+1
π (π‘ )=5 π‘3β6 π‘ 2+6π£ (π‘ )=π β² (π‘)=15 π‘2β12 π‘
Position, Velocity, and Acceleration
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1.8 βHigher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of change with direction.
π£ (π‘ )=π β² (π‘ )=ππ ππ‘
The velocity function, , is obtain by differentiating the position function with respect to time.
π (π‘ )=4 π‘2+π‘π£ (π‘ )=π β² (π‘)=8 π‘+1
π (π‘ )=5 π‘3β6 π‘ 2+6π£ (π‘ )=π β² (π‘)=15 π‘2β12 π‘
Position, Velocity, and Acceleration
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Position, Velocity, and Acceleration
Acceleration: the change in velocity with respect to a change in time. It is a rate of change with direction.
The acceleration function, , is obtain by differentiating the velocity function with respect to time. It is also the 2nd derivative of the position function.
π (π‘ )=π£ β² (π‘ )=ππ£ππ‘
=π β² β² (π‘ )= π2π π π‘2
π (π‘ )=4 π‘2+π‘
π£ (π‘ )=π β² (π‘)=8 π‘+1
π (π‘ )=5 π‘3β6 π‘ 2+6
π£ (π‘ )=π β² (π‘)=15 π‘2β12 π‘
π (π‘ )=π£ β² (π‘ )=π β² β² (π‘ )=8 π (π‘ )=π£ β² (π‘ )=π β² β² (π‘)=30 π‘β12
1.8 βHigher-Order Derivatives
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The position of an object is given by , where s is measured in feet and t is measured in seconds. a) Find the velocity and acceleration functions.b) What are the position, velocity, and acceleration of the object at 5 seconds?
π£ (π‘ )=ππ ππ‘
=4 π‘+8a)
b)
1.8 βHigher-Order Derivatives
π (π‘ )= ππ£ππ‘
=4
ππππ‘
π£ (5 )=4 (5 )+8 ππππ‘ / π ππ
π (5 )=4feet/sec/sec or
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1.8 βHigher-Order DerivativesThe position of a particle (in inches) moving along the x-axis after t seconds have elapsed is given by the following equation: s(t) = t4 β 2t3 β 4t2 + 12t.(a) Calculate the velocity of the particle at time t.(b) Compute the particle's velocity at t = 1, 2, and 4 seconds.(c) Calculate the acceleration of the particle after 4 seconds.(d) When is the particle at rest?
π£ (π‘ )=ππ ππ‘
=4 π‘3β6 π‘ 2β8 π‘+12a)
b)
c)
d)
π£ (1 )=2 hπππ ππ /π ππ
π£ (2 )=4 hπππ ππ /π ππ
π£ (4 )=140 hπππ ππ /π ππ
π (π‘ )= ππ£ππ‘
=12 π‘2β12 π‘β8
π (4 )=136 ππππ‘ /π ππ2
π£ (π‘ )=0ππ‘ πππ π‘
0=4 π‘ 3β6 π‘2β8 π‘+12
0=2 π‘ 2 (2 π‘β3 )β4 (2π‘β3 )
0=(2 π‘β3 ) (2 π‘ 2β4 )
π‘=32,1.414 π ππ .
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