Transcript
Lesson 15 (Section 3.5)The Chain Rule
Math 1a
October 29, 2007
Announcements
I Come to office hours if you don’t have your midterm yet
Analogy
Think about riding a bike. Togo faster you can either:
I pedal faster
I change gears
The angular position of the back wheel depends on the position ofthe front wheel:
ϕ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
Analogy
Think about riding a bike. Togo faster you can either:
I pedal faster
I change gears
The angular position of the back wheel depends on the position ofthe front wheel:
ϕ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
Analogy
Think about riding a bike. Togo faster you can either:
I pedal faster
I change gears
The angular position of the back wheel depends on the position ofthe front wheel:
ϕ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
Analogy
Think about riding a bike. Togo faster you can either:
I pedal faster
I change gears
The angular position of the back wheel depends on the position ofthe front wheel:
ϕ(θ) =Rθ
r
And so the angular speed of the back wheel depends on thederivative of this function and the speed of the front wheel.
Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007
Page1of6
Theorem of the day: The chain rule
TheoremLet f and g be functions, with g differentiable at a and fdifferentiable at g(a). Then f ◦ g is differentiable at a and
(f ◦ g)′(a) = f ′(g(a))g ′(a)
In Leibnizian notation, let y = f (u) and u = g(x). Then
dy
dx=
dy
du
du
dx
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f (u) =
√u and g(x) = 3x2 + 1. Then
f ′(u) = 12u−1/2, and g ′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g.
Let f (u) =√
u and g(x) = 3x2 + 1. Thenf ′(u) = 1
2u−1/2, and g ′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f (u) =
√u and g(x) = 3x2 + 1.
Thenf ′(u) = 1
2u−1/2, and g ′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f (u) =
√u and g(x) = 3x2 + 1. Then
f ′(u) = 12u−1/2, and g ′(x) = 6x. So
h′(x) = 12u−1/2(6x)
= 12(3x2 + 1)−1/2(6x) =
3x√3x2 + 1
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
SolutionFirst, write h as f ◦ g. Let f (u) =
√u and g(x) = 3x2 + 1. Then
f ′(u) = 12u−1/2, and g ′(x) = 6x. So
h′(x) = 12u−1/2(6x) = 1
2(3x2 + 1)−1/2(6x) =3x√
3x2 + 1
Example
Let f (x) =(
3√
x5 − 2 + 8)2
. Find f ′(x).
Solution
d
dx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 5)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=10
3x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007
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Example
Let f (x) =(
3√
x5 − 2 + 8)2
. Find f ′(x).
Solution
d
dx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 5)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=10
3x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
Example
Let f (x) =(
3√
x5 − 2 + 8)2
. Find f ′(x).
Solution
d
dx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 5)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=10
3x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
Example
Let f (x) =(
3√
x5 − 2 + 8)2
. Find f ′(x).
Solution
d
dx
(3√
x5 − 2 + 8)2
= 2(
3√
x5 − 2 + 8) d
dx
(3√
x5 − 2 + 8)
= 2(
3√
x5 − 2 + 8) d
dx3√
x5 − 2
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3 d
dx(x5 − 5)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=10
3x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
Math 1a - October 29, 2007.GWBMonday, Oct 29, 2007
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A metaphor
Think about peeling an onion:
f (x) =
(3√
x5︸︷︷︸�5
−2
︸ ︷︷ ︸3√�
+8
︸ ︷︷ ︸�+8
)2
︸ ︷︷ ︸�2
f ′(x) = 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
QuestionThe area of a circle, A = πr2, changes as its radius changes. If theradius changes with respect to time, the change in area withrespect to time is
A.dA
dr= 2πr
B.dA
dt= 2πr +
dr
dt
C.dA
dt= 2πr
dr
dtD. not enough information
QuestionThe area of a circle, A = πr2, changes as its radius changes. If theradius changes with respect to time, the change in area withrespect to time is
A.dA
dr= 2πr
B.dA
dt= 2πr +
dr
dt
C.dA
dt= 2πr
dr
dtD. not enough information
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