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.

( 𝑓 ∘𝑔 ) (𝑥 )¿ 𝑓 (𝑔 (𝑥 ) )1.7 – The Chain Rule

Given:, find .

( 𝑓 ∘𝑔 ) (𝑥 )= 𝑓 (𝑔 (𝑥 ) )=¿2 (𝑥2+5 )+3

¿2 𝑥2+10+3

2 𝑥2+1 3( 𝑓 ∘𝑔 ) (𝑥 )= 𝑓 (𝑔 (𝑥 ) )=¿

Find .

(𝑔∘ 𝑓 ) (𝑥 )=𝑔 ( 𝑓 (𝑥 ) )=¿(2 𝑥+3 )2+5

4 𝑥2+6𝑥+6 𝑥+9+5

4 𝑥2+12 𝑥+14(𝑔∘ 𝑓 ) (𝑥 )=𝑔 ( 𝑓 (𝑥 ) )=¿

( 𝑓 ∘𝑔 ) (𝑥 )¿ 𝑓 (𝑔 (𝑥 ) )1.7 – The Chain Rule

Given:, find .

( 𝑓 ∘𝑔 ) (𝑥 )= 𝑓 (𝑔 (𝑥 ) )=¿(𝑥2+2 )3+ (𝑥2+2 )−6

Find .

(𝑔∘ 𝑓 ) (𝑥 )=𝑔 ( 𝑓 (𝑥 ) )=¿(𝑥3+𝑥−6 )2+2

( 𝑓 ∘𝑔 ) (𝑥 )= 𝑓 (𝑔 (𝑥 ) )=¿(𝑥2+2 )3+𝑥2−4

1.7 – The Chain RuleReview of the Product Rule:

𝑦=( 3𝑥3+2 𝑥2 )2¿ ( 3𝑥3+2 𝑥2 ) (3 𝑥3+2𝑥2 )

𝑦 ′= (3 𝑥3+2𝑥2 ) (9 𝑥2+4 𝑥 )+( 9 𝑥2+4 𝑥 ) (3 𝑥3+2𝑥2 )

𝑦 ′=2 ( 3𝑥3+2 𝑥2 ) (9 𝑥2+4 𝑥 )

𝑦=( 6 𝑥2+𝑥 )3¿ ( 6 𝑥2+𝑥 ) (6 𝑥2+𝑥 ) ( 6𝑥2+𝑥 )+

𝑦 ′=3 (6 𝑥2+𝑥 )2 (12𝑥+1 )

𝑦 ′=(6 𝑥2+𝑥 )2 (12 𝑥+1 )+ (6 𝑥2+𝑥 )2 (12𝑥+1 )+( 6 𝑥2+𝑥 )2 (12𝑥+1 )

𝑦=( 3𝑥3+2 𝑥2 )2 𝑦=( 6 𝑥2+𝑥 )3and are composite functions.

Additional Problems:

𝑦=( 3𝑥3+2 𝑥2 )2 𝑦 ′=2 ( 3𝑥3+2 𝑥2 ) (9 𝑥2+4 𝑥 )

𝑦=( 6 𝑥2+𝑥 )3 𝑦 ′=3 (6 𝑥2+𝑥 )2 (12𝑥+1 )

𝑦=(𝑥3+2 𝑥 )9 (𝑥3+2 𝑥 )89 (3 𝑥2+2 )

𝑦=(5 𝑥2+1 )4 (5 𝑥2+1 )34 (10 𝑥 )

𝑦 ′=¿𝑦 ′=¿

𝑦=( 2𝑥5−3𝑥4 +𝑥−3 )13 (2 𝑥5−3𝑥4+𝑥−3 )1213 (10 𝑥4−12 𝑥3+1 )𝑦 ′=¿

1.7 – The Chain Rule

Find

𝑦=𝑢3−7𝑢2 𝑢=𝑥2+3

𝑑𝑦𝑑𝑥

=𝑑𝑦𝑑𝑢∙𝑑𝑢𝑑𝑥

1.7 – The Chain Rule

𝑑𝑦𝑑𝑢

=3𝑢2−14𝑢𝑑𝑢𝑑𝑥

=2𝑥

𝑑𝑦𝑑𝑥

=¿(3𝑢2−14𝑢   )∙2𝑥𝑑𝑦𝑑𝑥

=¿(3 (𝑥2+3 )2−14 (𝑥2+3 ))2 𝑥

𝑑𝑦𝑑𝑥

=2𝑥 (𝑥2+3 ) (3 (𝑥2+3 )−14 )𝑑𝑦𝑑𝑥

=2𝑥 (𝑥2+3 ) (3 𝑥2+9−14 )

𝑑𝑦𝑑𝑥

=2𝑥 (𝑥2+3 ) (3 𝑥2−5 )

𝑦=𝑢3−7𝑢2 𝑢=𝑥2+3

𝑦=(𝑥2+3 )3−7 (𝑥2+3 )2

𝑑𝑦𝑑𝑥

=3 (𝑥2+3 )22 𝑥−14 (𝑥2+3 ) 2𝑥

𝑑𝑦𝑑𝑢

=2𝑥 (𝑥2+3 ) (3 (𝑥2+3 )−14 )

𝑑𝑦𝑑𝑢

=2𝑥 (𝑥2+3 ) (3 𝑥2+9−14 )

𝑑𝑦𝑑𝑢

=2𝑥 (𝑥2+3 ) (3 𝑥2−5 )

Find the equation of the tangent line at 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 )

𝑚𝑡𝑎𝑛=𝑑𝑦𝑑𝑥

=−16

𝑑𝑦𝑑𝑥

=2𝑥 (𝑥2+3 ) (3 𝑥2−5 )

𝑦−−48=−16 (𝑥−1 )

𝑦+48=−16𝑥+16

𝑦=−16 𝑥−32

1.7 – 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.

𝑠 (𝑡 )=√12+4 𝑡

𝑠 (𝑡 )=(12+4 𝑡 )12

𝑑𝑠𝑑𝑡

=𝑠′ (𝑡 )=12

(12+4 𝑡 )− 1

2 (4 )

𝑑𝑠𝑑𝑡

=𝑠′ (𝑡 )= 2

(12+4 𝑡 )12

𝑎𝑡 𝑡=6 ,𝑑𝑠𝑑𝑡

=𝑠′ (6 )= 2

(12+4 (6 ) )12

𝑑𝑠𝑑𝑡

=𝑠′ (6 )=13𝑚𝑒𝑡𝑒𝑟𝑠 /𝑠𝑒𝑐𝑜𝑛𝑑𝑠

1.7 – 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) 𝑥=2010−2000=10 𝑦𝑒𝑎𝑟𝑠

𝑑𝐶𝑑𝑥

=29.91𝑏𝑖𝑙𝑙𝑖𝑜𝑛𝑑𝑜𝑙𝑙𝑎𝑟𝑠 /𝑦𝑒𝑎𝑟

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

Notations

1.8 –Higher-Order DerivativesFind the requested higher-order derivatives.

Find

𝑓 ′ (𝑥 )=12 𝑥3−15 𝑥2+8

𝑓 ′ ′ (𝑥 )=36 𝑥2−30 𝑥

𝑓 ′ ′ ′ (𝑥 )=72𝑥−30

𝑓 ′ (𝑥 )=6𝑥2+12 𝑥−57

𝑓 ′ ′ (𝑥 )=12𝑥+12

𝑓 ′ ′ ′ (𝑥 )=12

𝑓 ( 4 ) (𝑥 )=0

Find

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 )=𝑑𝑠𝑑𝑡

The velocity function, , is obtain by differentiating the position function with respect to time.

𝑠 (𝑡 )=4 𝑡2+𝑡𝑣 (𝑡 )=𝑠 ′ (𝑡)=8 𝑡+1

𝑠 (𝑡 )=5 𝑡3−6 𝑡 2+6𝑣 (𝑡 )=𝑠 ′ (𝑡)=15 𝑡2−12 𝑡

Position, Velocity, and Acceleration

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 )=𝑑𝑠𝑑𝑡

The velocity function, , is obtain by differentiating the position function with respect to time.

𝑠 (𝑡 )=4 𝑡2+𝑡𝑣 (𝑡 )=𝑠 ′ (𝑡)=8 𝑡+1

𝑠 (𝑡 )=5 𝑡3−6 𝑡 2+6𝑣 (𝑡 )=𝑠 ′ (𝑡)=15 𝑡2−12 𝑡

Position, Velocity, and Acceleration

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.

𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑑𝑣𝑑𝑡

=𝑠′ ′ (𝑡 )= 𝑑2𝑠𝑑 𝑡2

𝑠 (𝑡 )=4 𝑡2+𝑡

𝑣 (𝑡 )=𝑠 ′ (𝑡)=8 𝑡+1

𝑠 (𝑡 )=5 𝑡3−6 𝑡 2+6

𝑣 (𝑡 )=𝑠 ′ (𝑡)=15 𝑡2−12 𝑡

𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑠′ ′ (𝑡 )=8 𝑎 (𝑡 )=𝑣 ′ (𝑡 )=𝑠 ′ ′ (𝑡)=30 𝑡−12

1.8 –Higher-Order Derivatives

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?

𝑣 (𝑡 )=𝑑𝑠𝑑𝑡

=4 𝑡+8a)

b)

1.8 –Higher-Order Derivatives

𝑎 (𝑡 )= 𝑑𝑣𝑑𝑡

=4

𝑓𝑒𝑒𝑡

𝑣 (5 )=4 (5 )+8 𝑓𝑒𝑒𝑡 / 𝑠𝑒𝑐

𝑎 (5 )=4feet/sec/sec or

1.8 –Higher-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 – 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 )=𝑑𝑠𝑑𝑡

=4 𝑡3−6 𝑡 2−8 𝑡+12a)

b)

c)

d)

𝑣 (1 )=2 h𝑖𝑛𝑐 𝑒𝑠 /𝑠𝑒𝑐

𝑣 (2 )=4 h𝑖𝑛𝑐 𝑒𝑠 /𝑠𝑒𝑐

𝑣 (4 )=140 h𝑖𝑛𝑐 𝑒𝑠 /𝑠𝑒𝑐

𝑎 (𝑡 )= 𝑑𝑣𝑑𝑡

=12 𝑡2−12 𝑡−8

𝑎 (4 )=136 𝑓𝑒𝑒𝑡 /𝑠𝑒𝑐2

𝑣 (𝑡 )=0𝑎𝑡 𝑟𝑒𝑠𝑡

0=4 𝑡 3−6 𝑡2−8 𝑡+12

0=2 𝑡 2 (2 𝑡−3 )−4 (2𝑡−3   )

0=(2 𝑡−3   ) (2 𝑡 2−4 )

𝑡=32,1.414 𝑠𝑒𝑐 .