Lesson 10: The Chain Rule (Section 41 slides)

Section 2.5
The Chain Rule

V63.0121.041, Calculus I

New York University

October 6, 2010

Announcements

Quiz 2 in recitation next week (October 11-15)
No class Monday, October 11
Midterm in class Monday, October 18 on §§1.1–2.5

. . . . . .

Announcements

Quiz 2 in recitation next
week (October 11-15)
No class Monday, October
11
Midterm in class Monday,
October 18 on §§1.1–2.5

. . . . . .

V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 2 / 36

Objectives

Given a compound
expression, write it as a
composition of functions.
Understand and apply the
Chain Rule for the
derivative of a composition
of functions.
Understand and use
Newtonian and Leibnizian
notations for the Chain
Rule.

. . . . . .


Compositions
See Section 1.2 for review

Definition
If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do
g first, then f.”

.

. . . . . .


Compositions

Definition
g first, then f.”

x
. g
. (x)
g
. .

. . . . . .


Compositions

Definition
g first, then f.”

x
. g
. (x)
g
. . f
.

. . . . . .


Compositions

Definition
g first, then f.”

x
. g
. (x) f
.(g(x))
g
. . f
.

. . . . . .


Compositions

Definition
g first, then f.”

g
. (x) f
.(g(x))
. ◦ g
x
.
g
. f . f
.

. . . . . .


Compositions

Definition
g first, then f.”

g
. (x) f
.(g(x))
. ◦ g
x
.
g
. f . f
.

Our goal for the day is to understand how the derivative of the
composition of two functions depends on the derivatives of the
individual functions.
. . . . . .


Outline

Heuristics
Analogy
The Linear Case

The chain rule

Examples

Related rates of change

. . . . . .


Analogy

Think about riding a bike. To go
faster you can either:

.

. . . . . .
.
Image credit: SpringSun

Analogy

pedal faster

.

. . . . . .
.

Analogy

pedal faster
change gears

.

. . . . . .
.

Analogy

pedal faster
change gears

.
The angular position (φ) of the back wheel depends on the position of
the front sprocket (θ):
R.θ
.
φ(θ) =
r.
.

. . . . . .
.

Analogy

pedal faster
change gears

r
. adius of front sprocket .
R.θ
.
φ(θ) =
r.
.

r
. adius of back sprocket
. . . . . .
.

Analogy

pedal faster
change gears

.
R.θ
.
φ(θ) =
r.
.
And so the angular speed of the back wheel depends on the derivative
of this function and the speed of the front sprocket.
. . . . . .
.

The Linear Case

Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?

. . . . . .


The Linear Case

Question
composition?

Answer

f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)

. . . . . .


The Linear Case

Question
composition?

Answer

f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear

. . . . . .


The Linear Case

Question
composition?

Answer

f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The slope of the composition is the product of the slopes of the
two functions.

. . . . . .


The Linear Case

Question
composition?

Answer

f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The slope of the composition is the product of the slopes of the
two functions.

The derivative is supposed to be a local linearization of a function. So
there should be an analog of this property in derivatives.

. . . . . .


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
∆y ≈ f′ (y)g′ (u)∆x =⇒ ≈ f′ (y)g′ (u)
∆x

. . . . . .


Outline

Heuristics
Analogy
The Linear Case

The chain rule

Examples


. . . . . .


Theorem of the day: The chain rule

Theorem
Let f and g be functions, with g differentiable at x and f differentiable at
g(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 dy du
=
dx du dx

. . . . . .


Observations

Succinctly, the derivative of a
composition is the product of
the derivatives

.

. . . . . .
.
Image credit: ooOJasonOoo


Theorem

(f ◦ g)′ (x) = f′ (g(x))g′ (x)


dy dy du
=
dx du dx

. . . . . .


Observations

the derivatives
The only complication is where
these derivatives are
evaluated: at the same point
the functions are

.

. . . . . .
.

Compositions

Definition
g first, then f.”

g
. (x) f
.(g(x))
. ◦ g
x
.
g
. f . f
.

. . . . . .


Observations

the derivatives
The only complication is where
these derivatives are
evaluated: at the same point
the functions are
In Leibniz notation, the Chain
Rule looks like cancellation of
(fake) fractions
.

. . . . . .
.


Theorem

(f ◦ g)′ (x) = f′ (g(x))g′ (x)


dy dy du
=
dx du dx

. . . . . .



Theorem

(f ◦ g)′ (x) = f′ (g(x))g′ (x)

dy
In Leibnizian notation, let y = f(u) and u = g(x).du
Then
. .
dx
du
dy dy du
=
dx du dx

. . . . . .


Outline

Heuristics
Analogy
The Linear Case

The chain rule

Examples


. . . . . .


Example

Example
√
let h(x) = 3x2 + 1. Find h′ (x).

. . . . . .


Example

Example
√
let h(x) = 3x2 + 1. Find h′ (x).

Solution
First, write h as f ◦ g.

. . . . . .


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.

. . . . . .


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. Then
f′ (u) = 1 u−1/2 , and g′ (x) = 6x. So
2

h′ (x) = 1 u−1/2 (6x)
2

. . . . . .


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. Then
f′ (u) = 1 u−1/2 , and g′ (x) = 6x. So
2

3x
h′ (x) = 1 u−1/2 (6x) = 1 (3x2 + 1)−1/2 (6x) = √
2 2
3x2 + 1

. . . . . .


Corollary

Corollary (The Power Rule Combined with the Chain Rule)
If n is any real number and u = g(x) is differentiable, then

d n du
(u ) = nun−1 .
dx dx

. . . . . .


Does order matter?

Example
d d
Find (sin 4x) and compare it to (4 sin x).
dx dx

. . . . . .


Does order matter?

Example
d d
dx dx

Solution

For the first, let u = 4x and y = sin(u). Then

dy dy du
= · = cos(u) · 4 = 4 cos 4x.
dx du dx

. . . . . .


Does order matter?

Example
d d
dx dx

Solution


dy dy du
= · = cos(u) · 4 = 4 cos 4x.
dx du dx
For the second, let u = sin x and y = 4u. Then

dy dy du
= · = 4 · cos x
dx du dx

. . . . . .


Order matters!

Example
d d
dx dx

Solution


dy dy du
= · = cos(u) · 4 = 4 cos 4x.
dx du dx
For the second, let u = sin x and y = 4u. Then

dy dy du
= · = 4 · cos x
dx du dx

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

Solution

d (√ 5
3
)2 (√
3
) d (√
3
)
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

Solution

d (√ 5
3
)2 (√
3
) d (√
3
)
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
(√ ) d√
3 3
=2 x5 − 2 + 8 x5 − 2
dx

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

Solution

d (√ 5
3
)2 (√
3
) d (√
3
)
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
(√ ) d√
3 3
=2 x5 − 2 + 8 x5 − 2
dx
(√ ) d
x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2)
3
=2 3 dx

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

Solution

d (√ 5
3
)2 (√
3
) d (√
3
)
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
(√ ) d√
3 3
=2 x5 − 2 + 8 x5 − 2
dx
(√ ) d
x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2)
3
=2 3
(√ ) dx
=2
3
x 5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3

. . . . . .


Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8

Solution

d (√ 5
3
)2 (√
3
) d (√
3
)
x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8
dx dx
(√ ) d√
3 3
=2 x5 − 2 + 8 x5 − 2
dx
(√ ) d
x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2)
3
=2 3
(√ ) dx
=2
3
x 5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
(√
10 4 3 5 )
= x x − 2 + 8 (x5 − 2)−2/3
3

. . . . . .


A metaphor

Think about peeling an onion:
(√ )2
3
f(x) = x 5
−2 +8
5

√
3

+8
.
(√ )
2

f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 )
3
3

. . . . . .
.
Image credit: photobunny

Combining techniques

Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
dx

. . . . . .



Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
dx

Solution
The “last” part of the function is the product, so we apply the product
rule. Each factor’s derivative requires the chain rule:

. . . . . .



Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
dx

Solution

d ( 3 )
(x + 1)10 · sin(4x2 − 7)
dx ( ) ( )
d 3 d
= (x + 1) 10
· sin(4x − 7) + (x + 1) ·
2 3 10
sin(4x − 7)
2
dx dx

. . . . . .



Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
dx

Solution

d ( 3 )
(x + 1)10 · sin(4x2 − 7)
dx ( ) ( )
d 3 d
= (x + 1) 10
· sin(4x − 7) + (x + 1) ·
2 3 10
sin(4x − 7)
2
dx dx
= 10(x3 + 1)9 (3x2 ) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

. . . . . .


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
√
z−1
5. F(z) =
z+1
6. y = tan(cos x)
7. y = csc2 (sin θ)
8. y = sin(sin(sin(sin(sin(sin(x))))))

. . . . . .


Solution to #1

Example
Find the derivative of y = (1 − x2 )10 .

Solution
y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9

. . . . . .


Solution to #2

Example
√
Find the derivative of y = sin x.

Solution
√
Writing sin x as (sin x)1/2 , we have
cos x
y′ = 1
2 (sin x)−1/2 (cos x) = √
2 sin x

. . . . . .


Solution to #3

Example
√
Find the derivative of y = sin x.

Solution
(√ )
′ d 1/2 1 −1/2 cos x
y = 1/2
sin(x ) = cos(x ) 2 x = √
dx 2 x

. . . . . .


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 equation
y′ = 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
. . . . . .


Solution to #5

Example
√
z−1
Find the derivative of F(z) = .
z+1

Solution

( )−1/2 ( )
1 z−1(z + 1)(1) − (z − 1)(1)
y′ =
2 z+1 (z + 1)2
( )1/2 ( )
1 z+1 2 1
= =
2 z−1 (z + 1)2 (z + 1)3/2 (z − 1)1/2

. . . . . .


Solution to #6

Example
Find the derivative of y = tan(cos x).

Solution
y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x

. . . . . .


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

. . . . . .


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 multiplied
together.

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))

. . . . . .


Outline

Heuristics
Analogy
The Linear Case

The chain rule

Examples


. . . . . .


Related rates of change at the Deli

Question
Suppose a deli clerk can slice a stick of pepperoni (assume the
tapered ends have been removed) by hand at the rate of 2 inches per
minute, while a machine can slice pepperoni at the rate of 10 inches
dV dV
per minute. Then for the machine is 5 times greater than for
dt dt
the deli clerk. This is explained by the
A. chain rule
B. product rule
C. quotient Rule
D. addition rule

. . . . . .


Related rates of change in the ocean

Question
The area of a circle, A = πr2 ,
changes as its radius changes.
If the radius changes with
respect to time, the change in
area with respect to time is
dA
A. = 2πr
dr
dA dr
B. = 2πr +
dt dt .
dA dr
C. = 2πr
dt dt
D. not enough information

. . . . . .
.
Image credit: Jim Frazier

Summary

The derivative of a
composition is the product
of derivatives
In symbols:
(f ◦ g)′ (x) = f′ (g(x))g′ (x)
Calculus is like an onion,
and not because it makes
you cry!

. . . . . .


Lesson 10: The Chain Rule (Section 41 slides)

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