Embed Size (px)
Citation preview
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 .
