Section 2.5The Chain Rule
V63.0121.021, Calculus I
New York University
October 7, 2010
Announcements
I Quiz 2 in recitation next week (October 11-15)
I Midterm in class Tuesday, october 19 on §§1.1–2.5
Announcements
I Quiz 2 in recitation nextweek (October 11-15)
I Midterm in class Tuesday,october 19 on §§1.1–2.5
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 2 / 36
Objectives
I Given a compoundexpression, write it as acomposition of functions.
I Understand and apply theChain Rule for the derivativeof a composition offunctions.
I Understand and useNewtonian and Leibniziannotations for the Chain Rule.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 3 / 36
Notes
Notes
Notes
1
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
CompositionsSee Section 1.2 for review
Definition
If f and g are functions, the composition (f ◦ g)(x) = f (g(x)) means “dog first, then f .”
g fx g(x) f (g(x))
f ◦ g
Our goal for the day is to understand how the derivative of the compositionof two functions depends on the derivatives of the individual functions.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 5 / 36
Analogy
Think about riding a bike. To gofaster you can either:
I pedal faster
I change gears
Image credit: SpringSun
The angular position (ϕ) of the back wheel depends on the position of thefront sprocket (θ):
ϕ(θ) =R
radius of front sprocket
θ
r
radius of back sprocket
And so the angular speed of the back wheel depends on the derivative ofthis function and the speed of the front sprocket.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
Notes
Notes
Notes
2
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
The Linear Case
Question
Let f (x) = mx + b and g(x) = m′x + b′. What can you say about thecomposition?
Answer
I f (g(x)) = m(m′x + b′) + b = (mm′)x + (mb′ + b)
I The composition is also linear
I The slope of the composition is the product of the slopes of the twofunctions.
The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
The Nonlinear Case
Let u = g(x) and y = f (u). Suppose x is changed by a small amount ∆x .Then
∆y ≈ f ′(y)∆u
and∆u ≈ g ′(u)∆x .
So
∆y ≈ f ′(y)g ′(u)∆x =⇒ ∆y
∆x≈ f ′(y)g ′(u)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 8 / 36
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 9 / 36
Notes
Notes
Notes
3
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Theorem of the day: The chain rule
Theorem
Let f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f ′(g(x))g ′(x)
In Leibnizian notation, let y = f (u) and u = g(x). Then
dy
dx=
dy
du
du
dx
dy
��du��du
dx
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 10 / 36
Observations
I Succinctly, the derivative of acomposition is the product ofthe derivatives
I The only complication is wherethese derivatives are evaluated:at the same point the functionsare
I In Leibniz notation, the ChainRule looks like cancellation of(fake) fractions
Image credit: ooOJasonOooV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 11 / 36
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 17 / 36
Notes
Notes
Notes
4
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Example
Example
let h(x) =√
3x2 + 1. Find h′(x).
Solution
First, 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
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
Corollary
Corollary (The Power Rule Combined with the Chain Rule)
If n is any real number and u = g(x) is differentiable, then
d
dx(un) = nun−1 du
dx.
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 19 / 36
Order matters!
Example
Findd
dx(sin 4x) and compare it to
d
dx(4 sin x).
Solution
I For the first, let u = 4x and y = sin(u). Then
dy
dx=
dy
du· du
dx= cos(u) · 4 = 4 cos 4x .
I For the second, let u = sin x and y = 4u. Then
dy
dx=
dy
du· du
dx= 4 · cos x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
Notes
Notes
Notes
5
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
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 − 2)
= 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
=10
3x4(
3√
x5 − 2 + 8)
(x5 − 2)−2/3
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
A metaphor
Think about peeling an onion:
f (x) =
(3√
x5︸︷︷︸�5
−2
︸ ︷︷ ︸3√�
+8
︸ ︷︷ ︸�+8
)2
︸ ︷︷ ︸�2
Image credit: photobunny
f ′(x) = 2(
3√
x5 − 2 + 8)
13(x5 − 2)−2/3(5x4)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 22 / 36
Combining techniques
Example
Findd
dx
((x3 + 1)10 sin(4x2 − 7)
)Solution
The “last” part of the function is the product, so we apply the productrule. Each factor’s derivative requires the chain rule:
d
dx
((x3 + 1)10 · sin(4x2 − 7)
)=
(d
dx(x3 + 1)10
)· sin(4x2 − 7) + (x3 + 1)10 ·
(d
dxsin(4x2 − 7)
)= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
Notes
Notes
Notes
6
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Your Turn
Find derivatives of these functions:
1. y = (1− x2)10
2. y =√
sin x
3. y = sin√
x
4. y = (2x − 5)4(8x2 − 5)−3
5. F (z) =
√z − 1
z + 16. y = tan(cos x)
7. y = csc2(sin θ)
8. y = sin(sin(sin(sin(sin(sin(x))))))
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 24 / 36
Solution to #1
Example
Find the derivative of y = (1− x2)10.
Solution
y ′ = 10(1− x2)9(−2x) = −20x(1− x2)9
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 25 / 36
Solution to #2
Example
Find the derivative of y =√
sin x .
Solution
Writing√
sin x as (sin x)1/2, we have
y ′ = 12 (sin x)−1/2 (cos x) =
cos x
2√
sin x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 26 / 36
Notes
Notes
Notes
7
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Solution to #3
Example
Find the derivative of y = sin√
x .
Solution
y ′ =d
dxsin(x1/2) = cos(x1/2)12x−1/2 =
cos(√
x)
2√
x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 27 / 36
Solution to #4
Example
Find the derivative of y = (2x − 5)4(8x2 − 5)−3
Solution
We need to use the product rule and the chain rule:
y ′ = 4(2x − 5)3(2)(8x2 − 5)−3 + (2x − 5)4(−3)(8x2 − 5)−4(16x)
The rest is a bit of algebra, useful if you wanted to solve the equationy ′ = 0:
y ′ = 8(2x − 5)3(8x2 − 5)−4[(8x2 − 5)− 6x(2x − 5)
]= 8(2x − 5)3(8x2 − 5)−4
(−4x2 + 30x − 5
)= −8(2x − 5)3(8x2 − 5)−4
(4x2 − 30x + 5
)
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 28 / 36
Solution to #5
Example
Find the derivative of F (z) =
√z − 1
z + 1.
Solution
y ′ =1
2
(z − 1
z + 1
)−1/2((z + 1)(1)− (z − 1)(1)
(z + 1)2
)=
1
2
(z + 1
z − 1
)1/2( 2
(z + 1)2
)=
1
(z + 1)3/2(z − 1)1/2
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 29 / 36
Notes
Notes
Notes
8
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Solution to #6
Example
Find the derivative of y = tan(cos x).
Solution
y ′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 30 / 36
Solution to #7
Example
Find the derivative of y = csc2(sin θ).
Solution
Remember the notation:
y = csc2(sin θ) = [csc(sin θ)]2
So
y ′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ)
= −2 csc2(sin θ) cot(sin θ) cos θ
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 31 / 36
Solution to #8
Example
Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))).
Solution
Relax! It’s just a bunch of chain rules. All of these lines are multipliedtogether.
y ′ = cos(sin(sin(sin(sin(sin(x))))))
· cos(sin(sin(sin(sin(x)))))
· cos(sin(sin(sin(x))))
· cos(sin(sin(x)))
· cos(sin(x))
· cos(x))
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 32 / 36
Notes
Notes
Notes
9
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Outline
HeuristicsAnalogyThe Linear Case
The chain rule
Examples
Related rates of change
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 33 / 36
Related rates of change at the Deli
Question
Suppose a deli clerk can slice a stick of pepperoni (assume the taperedends have been removed) by hand at the rate of 2 inches per minute, whilea machine can slice pepperoni at the rate of 10 inches per minute. ThendV
dtfor the machine is 5 times greater than
dV
dtfor the deli clerk. This is
explained by the
A. chain rule
B. product rule
C. quotient Rule
D. addition rule
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 34 / 36
Related rates of change in the ocean
Question
The area of a circle, A = πr2,changes as its radius changes. Ifthe radius changes with respectto 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
Image credit: Jim FrazierV63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36
Notes
Notes
Notes
10
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010
Summary
I The derivative of acomposition is the productof derivatives
I In symbols:(f ◦ g)′(x) = f ′(g(x))g ′(x)
I Calculus is like an onion,and not because it makesyou cry!
V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 36 / 36
Notes
Notes
Notes
11
Section 2.5 : The Chain RuleV63.0121.021, Calculus I October 7, 2010