Transcript Document
1.7 β The Chain Rule
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:
πβπ π₯ =π π π₯
The composition of functions is a function inside another function.
1.7 β The Chain Rule
πβπ π₯ =π π π₯
Given: π π₯ = 2π₯ + 3 πππ π π₯ = π₯ 2 + 5, find π β π π₯ .
πβπ π₯ =π π π₯
= 2 π₯2 + 5 + 3
= 2π₯ 2 + 10 + 3
πβπ π₯ =π π π₯
= 2π₯ 2 + 13
Find π β π π₯ .
πβπ π₯ =π π π₯
=
2π₯ + 3
2
+5
4π₯ 2 + 6π₯ + 6π₯ + 9 + 5
πβπ π₯ =π π π₯
= 4π₯ 2 + 12π₯ + 14
1.7 β The Chain Rule
πβπ π₯ =π π π₯
Given: π π₯ = π₯ 3 + π₯ β 6 πππ π π₯ = π₯ 2 + 2, find π β π π₯ .
πβπ π₯ =π π π₯
=
πβπ π₯ =π π π₯
=
π₯2 + 2
π₯2 + 2
3
+ π₯2 + 2 β 6
3
+ π₯2 β 4
Find π β π π₯ .
πβπ π₯ =π π π₯
=
π₯3 + π₯ β 6
2
+2
1.7 β The Chain Rule
Review of the Product Rule:
π¦ = 3π₯ 3 + 2π₯ 2
2
= 3π₯ 3 + 2π₯ 2 3π₯ 3 + 2π₯ 2
π¦β² = 3π₯ 3 + 2π₯ 2 9π₯ 2 + 4π₯ + 9π₯ 2 + 4π₯ 3π₯ 3 + 2π₯ 2
π¦β² = 2 3π₯ 3 + 2π₯ 2 9π₯ 2 + 4π₯
π¦ = 6π₯ 2 + π₯
3
= 6π₯ 2 + π₯ 6π₯ 2 + π₯ 6π₯ 2 + π₯
π¦ β² = 6π₯ 2 + π₯ 6π₯ 2 + π₯ 12π₯ + 1 + 6π₯ 2 + π₯ 12π₯ + 1 6π₯ 2 + π₯ + 12π₯ + 1 6π₯ 2 + π₯ 6π₯ 2 + π₯
π¦ β² = 6π₯ 2 + π₯
π¦β² = 3 6π₯ 2 + π₯
2
12π₯ + 1 + 6π₯ 2 + π₯
2
2
12π₯ + 1 + 6π₯ 2 + π₯
2
12π₯ + 1
12π₯ + 1
π¦ = 3π₯ 3 + 2π₯ 2
2
and π¦ = 6π₯ 2 + π₯
3
are composite functions.
1.7 β The Chain Rule
π¦ = 3π₯ 3 + 2π₯ 2
π¦ = 6π₯ 2 + π₯
2
π¦β² = 2 3π₯ 3 + 2π₯ 2 9π₯ 2 + 4π₯
3
π¦β² = 3 6π₯ 2 + π₯
2
12π₯ + 1
Additional Problems:
π¦ = π₯ 3 + 2π₯
9
π¦β² = 9 π₯ 3 + 2π₯
8
3π₯ 2 + 2
π¦ = 5π₯ 2 + 1
4
π¦β² = 4 5π₯ 2 + 1
3
10π₯
π¦ = 2π₯ 5 β 3π₯ 4 + π₯ β 3
13
π¦β² = 13 2π₯ 5 β 3π₯ 4 + π₯ β 3
12
10π₯ 4 β 12π₯ 3 + 1
1.7 β The Chain Rule
ππ¦ ππ¦ ππ’
=
β
ππ₯ ππ’ ππ₯
ππ¦
Find ππ₯ .
π¦ = π’3 β 7π’2
π’ = π₯2 + 3
ππ’
ππ¦
2
= 2π₯
= 3π’ β 14π’
ππ₯
ππ’
ππ¦
= 3π’2 β 14π’ β 2π₯
ππ₯
ππ¦
= 3 π₯ 2 + 3 2 β 14 π₯ 2 + 3 2π₯
ππ₯
ππ¦
= 2π₯ π₯ 2 + 3 3 π₯ 2 + 3 β 14
ππ₯
ππ¦
= 2π₯ π₯ 2 + 3 3π₯ 2 + 9 β 14
ππ₯
ππ¦
= 2π₯ π₯ 2 + 3 3π₯ 2 β 5
ππ₯
π’ = π₯2 + 3
π¦ = π’3 β 7π’2
π¦ = π₯2 + 3
3
β 7 π₯2 + 3
ππ¦
= 3 π₯2 + 3
ππ₯
2
2
2π₯ β 14 π₯ 2 + 3 2π₯
ππ¦
= 2π₯ π₯ 2 + 3 3 π₯ 2 + 3 β 14
ππ’
ππ¦
= 2π₯ π₯ 2 + 3 3π₯ 2 + 9 β 14
ππ’
ππ¦
= 2π₯ π₯ 2 + 3 3π₯ 2 β 5
ππ’
1.7 β The Chain Rule
Find the equation of the tangent line at π₯ = 1 for the previous problem.
π¦ = π₯2 + 3
3
β 7 π₯2 + 3
ππ¦
= 2π₯ π₯ 2 + 3 3π₯ 2 β 5
ππ₯
π₯=1
ππ‘ππ =
π¦ = β48
ππ¦
= β16
ππ₯
π¦ β π¦1 = π π₯ β π₯1
π¦ β β48 = β16 π₯ β 1
π¦ + 48 = β16π₯ + 16
π¦ = β16π₯ β 32
2
1.7 β The Chain Rule
The position of a particle moving along a coordinate line is, π π‘ = 12 + 4π‘, with s in
meters and t in seconds. Find the rate of change of the particle's position at π‘ = 6
seconds.
π π‘ = 12 + 4π‘
π π‘ = 12 + 4π‘
1
2
ππ
1
β²
= π π‘ = 12 + 4π‘
ππ‘
2
ππ
2
= π β² π‘ =
ππ‘
12 + 4π‘
ππ
ππ‘ π‘ = 6,
= π β² 6 =
ππ‘
1
β1 2
4
2
2
12 + 4 6
ππ
1
β²
= π 6 = πππ‘πππ /π ππππππ
ππ‘
3
1
2
1.7 β The Chain Rule
The total outstanding consumer credit of a certain country can be modeled by πΆ π₯ =
0.21π₯ 4 β 5.98π₯ 3 + 50.11π₯ 2 β 18.29π₯ + 1106.47 , where C is billion dollars and x is
the number of years since 2000.
ππΆ
a) Find .
ππ₯
b) Using this model, predict how quickly outstanding consumer credit will be rising in
2010.
a) πΆ π₯ = 0.21π₯ 4 β 5.98π₯ 3 + 50.11π₯ 2 β 18.29π₯ + 1106.47
ππΆ
ππ₯
= 0.84π₯ 3 β 17.94π₯ 2 + 100.22π₯ β 18.29
b) π₯ = 2010 β 2000 = 10 π¦ππππ
ππΆ
ππ‘ π₯ = 10, ππ₯ = 0.84 10
3
β 17.94 10
ππΆ
= 29.91 πππππππ πππππππ /π¦πππ
ππ₯
2
+ 100.22 10 β 18.29
1.8 βHigher-Order Derivatives
Higher-order derivatives provide a method to examine how a rate-of-change
changes.
Notations
1.8 βHigher-Order Derivatives
Find the requested higher-order derivatives.
Find
π β²β²β²
π₯ ,
π3 π¦
,
ππ₯ 3
π¦ β²β²β² .
π π₯ = 3π₯ 4 β 5π₯ 3 + 8π₯ + 12
Find π
4
π₯ ,
π4π¦
,
ππ₯ 4
π¦
4
π π₯ = 2π₯ 3 + 6π₯ 2 β 57π₯
π β² π₯ = 12π₯ 3 β 15π₯ 2 + 8
π β² π₯ = 6π₯ 2 + 12π₯ β 57
π β²β² π₯ = 36π₯ 2 β 30π₯
π β²β² π₯ = 12π₯ + 12
π β²β²β² π₯ = 72π₯ β 30
π β²β²β² π₯ = 12
π
4
π₯ =0
.
1.8 βHigher-Order Derivatives
Position, Velocity, and Acceleration
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
The velocity function, π£ π‘ , is obtain by differentiating the position function with
respect to time.
π£ π‘ =
π β²
ππ
π‘ =
ππ‘
π π‘ = 4π‘ 2 + π‘
π π‘ = 5π‘ 3 β 6π‘ 2 + 6
π£ π‘ = π β²(π‘) = 8π‘ + 1
π£ π‘ = π β²(π‘) = 15π‘ 2 β 12π‘
1.8 βHigher-Order Derivatives
Position, Velocity, and Acceleration
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
The velocity function, π£ π‘ , is obtain by differentiating the position function with
respect to time.
π£ π‘ =
π β²
ππ
π‘ =
ππ‘
π π‘ = 4π‘ 2 + π‘
π π‘ = 5π‘ 3 β 6π‘ 2 + 6
π£ π‘ = π β²(π‘) = 8π‘ + 1
π£ π‘ = π β²(π‘) = 15π‘ 2 β 12π‘
1.8 βHigher-Order Derivatives
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, π π‘ , is obtain by differentiating the velocity function
with respect to time. It is also the 2nd derivative of the position function.
ππ£
π2π
β²β²
π π‘ =π£ π‘ =
=π π‘ = 2
ππ‘
ππ‘
β²
π π‘ = 4π‘ 2 + π‘
π π‘ = 5π‘ 3 β 6π‘ 2 + 6
π£ π‘ = π β²(π‘) = 8π‘ + 1
π£ π‘ = π β²(π‘) = 15π‘ 2 β 12π‘
π π‘ = π£ β² π‘ = π β²β² π‘ = 8
π π‘ = π£β² π‘ = π β²β²(π‘) = 30π‘ β 12
1.8 βHigher-Order Derivatives
The position of an object is given by π π‘ = 2π‘ 2 + 8π‘ , where s is measured in feet and
t is measured in seconds.
ππ
ππ£
a) Find the velocity
and acceleration
functions.
ππ‘
ππ‘
b) What are the position, velocity, and acceleration of the object at 5 seconds?
a) π£ π‘ =
π π‘ =
ππ
= 4π‘ + 8
ππ‘
ππ£
=4
ππ‘
b) π 5 = 2 5
2
+8 5
= 90 ππππ‘
π£ 5 = 4 5 + 8 = 28 ππππ‘/π ππ
π 5 = 4 feet/sec/sec or ππππ‘/π ππ 2
1.8 βHigher-Order Derivatives
The 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?
a) π£ π‘ =
ππ
= 4π‘ 3 β 6π‘ 2 β 8π‘ + 12
ππ‘
b) π£ 1 = 2 πππβππ /π ππ
π£ 2 = 4 πππβππ /π ππ
π£ 4 = 140 πππβππ /π ππ
ππ£
c) π π‘ =
= 12π‘ 2 β 12π‘ β 8
ππ‘
π 4 = 136 ππππ‘/π ππ 2
d) π£ π‘ = 0 ππ‘ πππ π‘
0 = 4π‘ 3 β 6π‘ 2 β 8π‘ + 12
0 = 2π‘ 2 2π‘ β 3 β 4 2π‘ β 3
0 = 2π‘ β 3
2π‘ 2 β 4
3
π‘ = , 1.414 π ππ.
2