Higher Order Deriavatives and Examples

1. 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
1.7 – The Chain Rule
The composition of functions is a function inside another function.

2. 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥
1.7 – The Chain Rule
Given: 𝑓 𝑥 = 2𝑥 + 3 𝑎𝑛𝑑 𝑔 𝑥 = 𝑥2
+ 5, find 𝑓 ∘ 𝑔 𝑥 .
𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 = 2 𝑥2
+ 5 + 3
= 2𝑥2
+ 10 + 3
2𝑥2
+ 13𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 =
Find 𝑔 ∘ 𝑓 𝑥 .
𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 = 2𝑥 + 3 2
+ 5
4𝑥2
+ 6𝑥 + 6𝑥 + 9 + 5
4𝑥2 + 12𝑥 + 14𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 =

3. 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥
1.7 – The Chain Rule
Given: 𝑓 𝑥 = 𝑥3
+ 𝑥 − 6 𝑎𝑛𝑑 𝑔 𝑥 = 𝑥2
+ 2, find 𝑓 ∘ 𝑔 𝑥 .
𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 = 𝑥2
+ 2 3
+ 𝑥2
+ 2 − 6
Find 𝑔 ∘ 𝑓 𝑥 .
𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 = 𝑥3 + 𝑥 − 6 2 + 2
𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 = 𝑥2 + 2 3 + 𝑥2 − 4

4. 1.7 – The Chain Rule
Review of the Product Rule:
𝑦 = 3𝑥3 + 2𝑥2 2 = 3𝑥3 + 2𝑥2 3𝑥3 + 2𝑥2
𝑦′ = 3𝑥3
+ 2𝑥2
9𝑥2
+ 4𝑥 + 9𝑥2
+ 4𝑥 3𝑥3
+ 2𝑥2
𝑦′ = 2 3𝑥3 + 2𝑥2 9𝑥2 + 4𝑥
𝑦 = 6𝑥2 + 𝑥 3 = 6𝑥2 + 𝑥 6𝑥2 + 𝑥 6𝑥2 + 𝑥
𝑦′
= 6𝑥2
+ 𝑥 6𝑥2
+ 𝑥 12𝑥 + 1 + 6𝑥2
+ 𝑥 12𝑥 + 1 6𝑥2
+ 𝑥 + 12𝑥 + 1 6𝑥2
+ 𝑥 6𝑥2
+ 𝑥
𝑦′ = 3 6𝑥2
+ 𝑥 2
12𝑥 + 1
𝑦′
= 6𝑥2
+ 𝑥 2
12𝑥 + 1 + 6𝑥2
+ 𝑥 2
12𝑥 + 1 + 6𝑥2
+ 𝑥 2
12𝑥 + 1
𝑦 = 3𝑥3
+ 2𝑥2 2 𝑦 = 6𝑥2 + 𝑥 3
and are composite functions.

5. Additional Problems:
𝑦 = 3𝑥3
+ 2𝑥2 2
𝑦′ = 2 3𝑥3
+ 2𝑥2
9𝑥2
+ 4𝑥
𝑦 = 6𝑥2
+ 𝑥 3
𝑦′ = 3 6𝑥2 + 𝑥 2 12𝑥 + 1
𝑦 = 𝑥3
+ 2𝑥 9
𝑥3
+ 2𝑥 89 3𝑥2
+ 2
𝑦 = 5𝑥2 + 1 4 5𝑥2 + 1 34 10𝑥
𝑦′ =
𝑦′ =
𝑦 = 2𝑥5
− 3𝑥4
+ 𝑥 − 3 13 2𝑥5
− 3𝑥4
+ 𝑥 − 3 1213 10𝑥4
− 12𝑥3
+ 1𝑦′ =
1.7 – The Chain Rule

6. Find
𝑑𝑦
𝑑𝑥
.
𝑦 = 𝑢3 − 7𝑢2 𝑢 = 𝑥2
+ 3
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
∙
𝑑𝑢
𝑑𝑥
1.7 – The Chain Rule
𝑑𝑦
𝑑𝑢
= 3𝑢2
− 14𝑢
𝑑𝑢
𝑑𝑥
= 2𝑥
𝑑𝑦
𝑑𝑥
= 3𝑢2
− 14𝑢 ∙ 2𝑥
𝑑𝑦
𝑑𝑥
= 3 𝑥2 + 3 2 − 14 𝑥2 + 3 2𝑥
𝑑𝑦
𝑑𝑥
= 2𝑥 𝑥2
+ 3 3 𝑥2
+ 3 − 14
𝑑𝑦
𝑑𝑥
= 2𝑥 𝑥2
+ 3 3𝑥2
+ 9 − 14
𝑑𝑦
𝑑𝑥
= 2𝑥 𝑥2 + 3 3𝑥2 − 5
𝑦 = 𝑢3
− 7𝑢2 𝑢 = 𝑥2 + 3
𝑦 = 𝑥2
+ 3 3
− 7 𝑥2
+ 3 2
𝑑𝑦
𝑑𝑥
= 3 𝑥2
+ 3 2
2𝑥 − 14 𝑥2
+ 3 2𝑥
𝑑𝑦
𝑑𝑢
= 2𝑥 𝑥2
+ 3 3 𝑥2
+ 3 − 14
𝑑𝑦
𝑑𝑢
= 2𝑥 𝑥2
+ 3 3𝑥2
+ 9 − 14
𝑑𝑦
𝑑𝑢
= 2𝑥 𝑥2 + 3 3𝑥2 − 5

7. Find the equation of the tangent line at 𝑥 = 1 for the previous problem.
1.7 – The Chain Rule
𝑦 = −48𝑥 = 1
𝑦 = 𝑥2
+ 3 3
− 7 𝑥2
+ 3 2
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝑚 𝑡𝑎𝑛 =
𝑑𝑦
𝑑𝑥
= −16
𝑑𝑦
𝑑𝑥
= 2𝑥 𝑥2
+ 3 3𝑥2
− 5
𝑦 − −48 = −16 𝑥 − 1
𝑦 + 48 = −16𝑥 + 16
𝑦 = −16𝑥 − 32

8. 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
𝑑𝑡
= 𝑠′ 𝑡 =
1
2
12 + 4𝑡 −1
2 4
𝑑𝑠
𝑑𝑡
= 𝑠′ 𝑡 =
2
12 + 4𝑡
1
2
𝑎𝑡 𝑡 = 6,
𝑑𝑠
𝑑𝑡
= 𝑠′ 6 =
2
12 + 4 6
1
2
𝑑𝑠
𝑑𝑡
= 𝑠′ 6 =
1
3
𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐𝑜𝑛𝑑𝑠

9. 1.7 – The Chain Rule
The total outstanding consumer credit of a certain country can be modeled by 𝐶 𝑥 =
0.21𝑥4 − 5.98𝑥3 + 50.11𝑥2 − 18.29𝑥 + 1106.47 , where C is billion dollars and x is
the number of years since 2000.
a) Find
𝑑𝐶
𝑑𝑥
.
b) Using this model, predict how quickly outstanding consumer credit will be rising in
2010.
𝐶 𝑥 = 0.21𝑥4
− 5.98𝑥3
+ 50.11𝑥2
− 18.29𝑥 + 1106.47
𝑑𝐶
𝑑𝑥
= 0.84𝑥3 − 17.94𝑥2 + 100.22𝑥 − 18.29
a)
b) 𝑥 = 2010 − 2000 = 10 𝑦𝑒𝑎𝑟𝑠
𝑎𝑡 𝑥 = 10,
𝑑𝐶
𝑑𝑥
= 0.84 10 3
− 17.94 10 2
+ 100.22 10 − 18.29
𝑑𝐶
𝑑𝑥
= 29.91 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑑𝑜𝑙𝑙𝑎𝑟𝑠/𝑦𝑒𝑎𝑟

10. 1.8 –Higher-Order Derivatives
Higher-order derivatives provide a method to examine how a rate-of-change
changes.
Notations

11. 1.8 –Higher-Order Derivatives
Find the requested higher-order derivatives.
Find 𝑓′′′ 𝑥 ,
𝑑3 𝑦
𝑑𝑥3 , 𝑦′′′.
𝑓 𝑥 = 3𝑥4 − 5𝑥3 + 8𝑥 + 12
𝑓′
𝑥 = 12𝑥3
− 15𝑥2
+ 8
𝑓′′ 𝑥 = 36𝑥2 − 30𝑥
𝑓′′′
𝑥 = 72𝑥 − 30
𝑓 𝑥 = 2𝑥3
+ 6𝑥2
− 57𝑥
𝑓′
𝑥 = 6𝑥2
+ 12𝑥 − 57
𝑓′′ 𝑥 = 12𝑥 + 12
𝑓′′′
𝑥 = 12
𝑓 4 𝑥 = 0
Find 𝑓 4 𝑥 ,
𝑑4 𝑦
𝑑𝑥4 , 𝑦 4 .

12. 1.8 –Higher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
𝑣 𝑡 = 𝑠′ 𝑡 =
𝑑𝑠
𝑑𝑡
The velocity function, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑠 𝑡 = 4𝑡2
+ 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
Position, Velocity, and Acceleration

13. 1.8 –Higher-Order Derivatives
Velocity: the change in position with respect to a change in time. It is a rate of
change with direction.
𝑣 𝑡 = 𝑠′ 𝑡 =
𝑑𝑠
𝑑𝑡
The velocity function, 𝑣 𝑡 , is obtain by differentiating the position function with
respect to time.
𝑠 𝑡 = 4𝑡2
+ 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
Position, Velocity, and Acceleration

14. 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.
𝑎 𝑡 = 𝑣′
𝑡 =
𝑑𝑣
𝑑𝑡
= 𝑠′′
𝑡 =
𝑑2 𝑠
𝑑𝑡2
𝑠 𝑡 = 4𝑡2 + 𝑡
𝑣 𝑡 = 𝑠′(𝑡) = 8𝑡 + 1
𝑠 𝑡 = 5𝑡3
− 6𝑡2
+ 6
𝑣 𝑡 = 𝑠′(𝑡) = 15𝑡2 − 12𝑡
𝑎 𝑡 = 𝑣′
𝑡 = 𝑠′′
𝑡 = 8 𝑎 𝑡 = 𝑣′ 𝑡 = 𝑠′′(𝑡) = 30𝑡 − 12
1.8 –Higher-Order Derivatives

15. The position of an object is given by 𝑠 𝑡 = 2𝑡2 + 8𝑡 , where s is measured in feet and
t is measured in seconds.
a) Find the velocity
𝑑𝑠
𝑑𝑡
and acceleration
𝑑𝑣
𝑑𝑡
functions.
b) What are the position, velocity, and acceleration of the object at 5 seconds?
𝑣 𝑡 =
𝑑𝑠
𝑑𝑡
= 4𝑡 + 8a)
b)
1.8 –Higher-Order Derivatives
𝑎 𝑡 =
𝑑𝑣
𝑑𝑡
= 4
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

16. 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?
𝑣 𝑡 =
𝑑𝑠
𝑑𝑡
= 4𝑡3 − 6𝑡2 − 8𝑡 + 12a)
b)
c)
d)
𝑣 1 = 2 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 2 = 4 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑣 4 = 140 𝑖𝑛𝑐ℎ𝑒𝑠/𝑠𝑒𝑐
𝑎 𝑡 =
𝑑𝑣
𝑑𝑡
= 12𝑡2 − 12𝑡 − 8
𝑎 4 = 136 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐2
𝑣 𝑡 = 0 𝑎𝑡 𝑟𝑒𝑠𝑡
0 = 4𝑡3
− 6𝑡2
− 8𝑡 + 12
0 = 2𝑡2 2𝑡 − 3 − 4 2𝑡 − 3
0 = 2𝑡 − 3 2𝑡2 − 4
𝑡 =
3
2
, 1.414 𝑠𝑒𝑐.