Transcript Document

1.7 – 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:
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
The composition of functions is a function inside another function.
1.7 – The Chain Rule
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
Given: 𝑓 𝑥 = 2𝑥 + 3 𝑎𝑛𝑑 𝑔 𝑥 = 𝑥 2 + 5, find 𝑓 ∘ 𝑔 𝑥 .
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
= 2 𝑥2 + 5 + 3
= 2𝑥 2 + 10 + 3
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
= 2𝑥 2 + 13
Find 𝑔 ∘ 𝑓 𝑥 .
𝑔∘𝑓 𝑥 =𝑔 𝑓 𝑥
=
2𝑥 + 3
2
+5
4𝑥 2 + 6𝑥 + 6𝑥 + 9 + 5
𝑔∘𝑓 𝑥 =𝑔 𝑓 𝑥
= 4𝑥 2 + 12𝑥 + 14
1.7 – The Chain Rule
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
Given: 𝑓 𝑥 = 𝑥 3 + 𝑥 − 6 𝑎𝑛𝑑 𝑔 𝑥 = 𝑥 2 + 2, find 𝑓 ∘ 𝑔 𝑥 .
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
=
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
=
𝑥2 + 2
𝑥2 + 2
3
+ 𝑥2 + 2 − 6
3
+ 𝑥2 − 4
Find 𝑔 ∘ 𝑓 𝑥 .
𝑔∘𝑓 𝑥 =𝑔 𝑓 𝑥
=
𝑥3 + 𝑥 − 6
2
+2
1.7 – The Chain Rule
Review of the Product Rule:
𝑦 = 3𝑥 3 + 2𝑥 2
2
= 3𝑥 3 + 2𝑥 2 3𝑥 3 + 2𝑥 2
𝑦′ = 3𝑥 3 + 2𝑥 2 9𝑥 2 + 4𝑥 + 9𝑥 2 + 4𝑥 3𝑥 3 + 2𝑥 2
𝑦′ = 2 3𝑥 3 + 2𝑥 2 9𝑥 2 + 4𝑥
𝑦 = 6𝑥 2 + 𝑥
3
= 6𝑥 2 + 𝑥 6𝑥 2 + 𝑥 6𝑥 2 + 𝑥
𝑦 ′ = 6𝑥 2 + 𝑥 6𝑥 2 + 𝑥 12𝑥 + 1 + 6𝑥 2 + 𝑥 12𝑥 + 1 6𝑥 2 + 𝑥 + 12𝑥 + 1 6𝑥 2 + 𝑥 6𝑥 2 + 𝑥
𝑦 ′ = 6𝑥 2 + 𝑥
𝑦′ = 3 6𝑥 2 + 𝑥
2
12𝑥 + 1 + 6𝑥 2 + 𝑥
2
2
12𝑥 + 1 + 6𝑥 2 + 𝑥
2
12𝑥 + 1
12𝑥 + 1
𝑦 = 3𝑥 3 + 2𝑥 2
2
and 𝑦 = 6𝑥 2 + 𝑥
3
are composite functions.
1.7 – The Chain Rule
𝑦 = 3𝑥 3 + 2𝑥 2
𝑦 = 6𝑥 2 + 𝑥
2
𝑦′ = 2 3𝑥 3 + 2𝑥 2 9𝑥 2 + 4𝑥
3
𝑦′ = 3 6𝑥 2 + 𝑥
2
12𝑥 + 1
Additional Problems:
𝑦 = 𝑥 3 + 2𝑥
9
𝑦′ = 9 𝑥 3 + 2𝑥
8
3𝑥 2 + 2
𝑦 = 5𝑥 2 + 1
4
𝑦′ = 4 5𝑥 2 + 1
3
10𝑥
𝑦 = 2𝑥 5 − 3𝑥 4 + 𝑥 − 3
13
𝑦′ = 13 2𝑥 5 − 3𝑥 4 + 𝑥 − 3
12
10𝑥 4 − 12𝑥 3 + 1
1.7 – The Chain Rule
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
∙
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑𝑦
Find 𝑑𝑥 .
𝑦 = 𝑢3 − 7𝑢2
𝑢 = 𝑥2 + 3
𝑑𝑢
𝑑𝑦
2
= 2𝑥
= 3𝑢 − 14𝑢
𝑑𝑥
𝑑𝑢
𝑑𝑦
= 3𝑢2 − 14𝑢 ∙ 2𝑥
𝑑𝑥
𝑑𝑦
= 3 𝑥 2 + 3 2 − 14 𝑥 2 + 3 2𝑥
𝑑𝑥
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3 𝑥 2 + 3 − 14
𝑑𝑥
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3𝑥 2 + 9 − 14
𝑑𝑥
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3𝑥 2 − 5
𝑑𝑥
𝑢 = 𝑥2 + 3
𝑦 = 𝑢3 − 7𝑢2
𝑦 = 𝑥2 + 3
3
− 7 𝑥2 + 3
𝑑𝑦
= 3 𝑥2 + 3
𝑑𝑥
2
2
2𝑥 − 14 𝑥 2 + 3 2𝑥
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3 𝑥 2 + 3 − 14
𝑑𝑢
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3𝑥 2 + 9 − 14
𝑑𝑢
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3𝑥 2 − 5
𝑑𝑢
1.7 – The Chain Rule
Find the equation of the tangent line at 𝑥 = 1 for the previous problem.
𝑦 = 𝑥2 + 3
3
− 7 𝑥2 + 3
𝑑𝑦
= 2𝑥 𝑥 2 + 3 3𝑥 2 − 5
𝑑𝑥
𝑥=1
𝑚𝑡𝑎𝑛 =
𝑦 = −48
𝑑𝑦
= −16
𝑑𝑥
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝑦 − −48 = −16 𝑥 − 1
𝑦 + 48 = −16𝑥 + 16
𝑦 = −16𝑥 − 32
2
1.7 – The Chain Rule
The position of a particle moving along a coordinate line is, 𝑠 𝑡 = 12 + 4𝑡, with s in
meters and t in seconds. Find the rate of change of the particle's position at 𝑡 = 6
seconds.
𝑠 𝑡 = 12 + 4𝑡
𝑠 𝑡 = 12 + 4𝑡
1
2
𝑑𝑠
1
′
= 𝑠 𝑡 = 12 + 4𝑡
𝑑𝑡
2
𝑑𝑠
2
= 𝑠′ 𝑡 =
𝑑𝑡
12 + 4𝑡
𝑑𝑠
𝑎𝑡 𝑡 = 6,
= 𝑠′ 6 =
𝑑𝑡
1
−1 2
4
2
2
12 + 4 6
𝑑𝑠
1
′
= 𝑠 6 = 𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑑𝑡
3
1
2
1.7 – The Chain Rule
The total outstanding consumer credit of a certain country can be modeled by 𝐶 𝑥 =
0.21𝑥 4 − 5.98𝑥 3 + 50.11𝑥 2 − 18.29𝑥 + 1106.47 , where C is billion dollars and x is
the number of years since 2000.
𝑑𝐶
a) Find .
𝑑𝑥
b) Using this model, predict how quickly outstanding consumer credit will be rising in
2010.
a) 𝐶 𝑥 = 0.21𝑥 4 − 5.98𝑥 3 + 50.11𝑥 2 − 18.29𝑥 + 1106.47
𝑑𝐶
𝑑𝑥
= 0.84𝑥 3 − 17.94𝑥 2 + 100.22𝑥 − 18.29
b) 𝑥 = 2010 − 2000 = 10 𝑦𝑒𝑎𝑟𝑠
𝑑𝐶
𝑎𝑡 𝑥 = 10, 𝑑𝑥 = 0.84 10
3
− 17.94 10
𝑑𝐶
= 29.91 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑑𝑜𝑙𝑙𝑎𝑟𝑠/𝑦𝑒𝑎𝑟
𝑑𝑥
2
+ 100.22 10 − 18.29
1.8 –Higher-Order Derivatives
Higher-order derivatives provide a method to examine how a rate-of-change
changes.
Notations
1.8 –Higher-Order Derivatives
Find the requested higher-order derivatives.
Find
𝑓 ′′′
𝑥 ,
𝑑3 𝑦
,
𝑑𝑥 3
𝑦 ′′′ .
𝑓 𝑥 = 3𝑥 4 − 5𝑥 3 + 8𝑥 + 12
Find 𝑓
4
𝑥 ,
𝑑4𝑦
,
𝑑𝑥 4
𝑦
4
𝑓 𝑥 = 2𝑥 3 + 6𝑥 2 − 57𝑥
𝑓 ′ 𝑥 = 12𝑥 3 − 15𝑥 2 + 8
𝑓 ′ 𝑥 = 6𝑥 2 + 12𝑥 − 57
𝑓 ′′ 𝑥 = 36𝑥 2 − 30𝑥
𝑓 ′′ 𝑥 = 12𝑥 + 12
𝑓 ′′′ 𝑥 = 72𝑥 − 30
𝑓 ′′′ 𝑥 = 12
𝑓
4
𝑥 =0
.
1.8 –Higher-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, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑣 𝑡 =
𝑠′
𝑑𝑠
𝑡 =
𝑑𝑡
𝑠 𝑡 = 4𝑡 2 + 𝑡
𝑠 𝑡 = 5𝑡 3 − 6𝑡 2 + 6
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡 2 − 12𝑡
1.8 –Higher-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, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑣 𝑡 =
𝑠′
𝑑𝑠
𝑡 =
𝑑𝑡
𝑠 𝑡 = 4𝑡 2 + 𝑡
𝑠 𝑡 = 5𝑡 3 − 6𝑡 2 + 6
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡 2 − 12𝑡
1.8 –Higher-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, 𝑎 𝑡 , is obtain by differentiating the velocity function
with respect to time. It is also the 2nd derivative of the position function.
𝑑𝑣
𝑑2𝑠
′′
𝑎 𝑡 =𝑣 𝑡 =
=𝑠 𝑡 = 2
𝑑𝑡
𝑑𝑡
′
𝑠 𝑡 = 4𝑡 2 + 𝑡
𝑠 𝑡 = 5𝑡 3 − 6𝑡 2 + 6
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡 2 − 12𝑡
𝑎 𝑡 = 𝑣 ′ 𝑡 = 𝑠 ′′ 𝑡 = 8
𝑎 𝑡 = 𝑣′ 𝑡 = 𝑠′′(𝑡) = 30𝑡 − 12
1.8 –Higher-Order Derivatives
The position of an object is given by 𝑠 𝑡 = 2𝑡 2 + 8𝑡 , where s is measured in feet and
t is measured in seconds.
𝑑𝑠
𝑑𝑣
a) Find the velocity
and acceleration
functions.
𝑑𝑡
𝑑𝑡
b) What are the position, velocity, and acceleration of the object at 5 seconds?
a) 𝑣 𝑡 =
𝑎 𝑡 =
𝑑𝑠
= 4𝑡 + 8
𝑑𝑡
𝑑𝑣
=4
𝑑𝑡
b) 𝑠 5 = 2 5
2
+8 5
= 90 𝑓𝑒𝑒𝑡
𝑣 5 = 4 5 + 8 = 28 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐
𝑎 5 = 4 feet/sec/sec or 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐 2
1.8 –Higher-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 – 2t3 – 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) 𝑣 𝑡 =
𝑑𝑠
= 4𝑡 3 − 6𝑡 2 − 8𝑡 + 12
𝑑𝑡
b) 𝑣 1 = 2 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 2 = 4 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 4 = 140 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑑𝑣
c) 𝑎 𝑡 =
= 12𝑡 2 − 12𝑡 − 8
𝑑𝑡
𝑎 4 = 136 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐 2
d) 𝑣 𝑡 = 0 𝑎𝑡 𝑟𝑒𝑠𝑡
0 = 4𝑡 3 − 6𝑡 2 − 8𝑡 + 12
0 = 2𝑡 2 2𝑡 − 3 − 4 2𝑡 − 3
0 = 2𝑡 − 3
2𝑡 2 − 4
3
𝑡 = , 1.414 𝑠𝑒𝑐.
2