The document discusses the chain rule for derivatives. It begins with examples and analogies to build intuition. It then states the chain rule theorem - that the derivative of the composition of two functions f and g is the product of the derivatives of f and g, evaluated at the point where the functions are composed. Examples are provided to illustrate the chain rule. The key points are that the derivative of a composition is the product of the individual derivatives, and these must be evaluated at the same point where the functions are composed.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Analysis and Design of One Dimensional Periodic Foundations for Seismic Base ...IJERA Editor
Periodic foundationis a new type of seismic base isolation system. It is inspired by the periodic material crystal
lattice in the solid state physics. This kind of material has a unique property, which is termed as frequency band
gap that is capable of blocking incoming waves having frequencies falling within the band gap. Consequently,
seismic waves having frequencies falling within the frequency band gap are blocked by the periodic foundation.
The ability to block the seismic waveshas put this kind of foundation as a prosperous next generation of seismic
base isolators. This paper provides analytical study on the one dimensional (1D) type periodic foundations to
investigate their seismic performance. The general idea of basic theory of one dimensional (1D) periodic
foundations is first presented.Then, the parametric studies considering infinite and finite boundary conditions are
discussed. The effect of superstructure on the frequency band gap is investigated as well. Based on the analytical
study, a set of equations is proposed for the design guidelines of 1D periodic foundations for seismic base
isolation of structures.
SCRUTINY TO THE NON-AXIALLY DEFORMATIONS OF AN ELASTIC FOUNDATION ON A CYLIND...P singh
This paper is devoted to homogenization of partial differential operators to use in special structure that is a plate allied to an elastic foundation when it is situated through the basic loads (especially with the harmonic forces) with a Non-axially deformation of the cantilever. Furthermore, it contains the Equations of motion that they can be derived from degenerate Non-linear elliptic ones. Through the mentioned processes, there exists many excess works related to computing the bounded conditions for this special application form of study (when the deformation phenomenon has occurred). At the end of the article whole results of the study on a circular plate are debated and new ways assigned to them are discussed. Afterwards all the processes are formulised with the collection of contracting sequences and expanding sequences integrable functions that are intrinsically joints with the characteristic functions to expanding the behaviour of an elastic foundation. Thenceforth all the resultant functions are sets and compared with the other ones (without the loads). Sample pictures and analysis of the study were employed with the ANSYS software to obtain the better observations and conclusions.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.
The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.
Analysis and Design of One Dimensional Periodic Foundations for Seismic Base ...IJERA Editor
Periodic foundationis a new type of seismic base isolation system. It is inspired by the periodic material crystal
lattice in the solid state physics. This kind of material has a unique property, which is termed as frequency band
gap that is capable of blocking incoming waves having frequencies falling within the band gap. Consequently,
seismic waves having frequencies falling within the frequency band gap are blocked by the periodic foundation.
The ability to block the seismic waveshas put this kind of foundation as a prosperous next generation of seismic
base isolators. This paper provides analytical study on the one dimensional (1D) type periodic foundations to
investigate their seismic performance. The general idea of basic theory of one dimensional (1D) periodic
foundations is first presented.Then, the parametric studies considering infinite and finite boundary conditions are
discussed. The effect of superstructure on the frequency band gap is investigated as well. Based on the analytical
study, a set of equations is proposed for the design guidelines of 1D periodic foundations for seismic base
isolation of structures.
SCRUTINY TO THE NON-AXIALLY DEFORMATIONS OF AN ELASTIC FOUNDATION ON A CYLIND...P singh
This paper is devoted to homogenization of partial differential operators to use in special structure that is a plate allied to an elastic foundation when it is situated through the basic loads (especially with the harmonic forces) with a Non-axially deformation of the cantilever. Furthermore, it contains the Equations of motion that they can be derived from degenerate Non-linear elliptic ones. Through the mentioned processes, there exists many excess works related to computing the bounded conditions for this special application form of study (when the deformation phenomenon has occurred). At the end of the article whole results of the study on a circular plate are debated and new ways assigned to them are discussed. Afterwards all the processes are formulised with the collection of contracting sequences and expanding sequences integrable functions that are intrinsically joints with the characteristic functions to expanding the behaviour of an elastic foundation. Thenceforth all the resultant functions are sets and compared with the other ones (without the loads). Sample pictures and analysis of the study were employed with the ANSYS software to obtain the better observations and conclusions.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.
The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.
The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 10: The Chain Rule
1. Section 2.5
The Chain Rule
V63.0121.002.2010Su, Calculus I
New York University
May 25, 2010
Announcements
Quiz 2 Thursday in class on Sections 1.5–2.5
Assignment 2 is on Blackboard
. . . . . .
2. Announcements
Quiz 2 Thursday in class
on Sections 1.5–2.5
Assignment 2 is on
Blackboard
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 2 / 36
3. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 3 / 36
4. 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.”
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
5. 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.”
x
. g
. (x)
g
. .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
6. 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.”
x
. g
. (x)
g
. . f
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
7. 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.”
x
. g
. (x) f
.(g(x))
g
. . f
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
8. 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.”
g
. (x) f
.(g(x))
. ◦ g
x
.
g
. f . f
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
9. 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.”
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 4 / 36
10. Outline
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 5 / 36
11. Analogy
Think about riding a bike. To go
faster you can either:
.
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
12. Analogy
Think about riding a bike. To go
faster you can either:
pedal faster
.
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
13. Analogy
Think about riding a bike. To go
faster you can either:
pedal faster
change gears
.
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
14. Analogy
Think about riding a bike. To go
faster you can either:
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.
.
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
15. Analogy
Think about riding a bike. To go
faster you can either:
pedal faster
change gears
r
. adius of front sprocket .
The angular position (φ) of the back wheel depends on the position of
the front sprocket (θ):
R.θ
.
φ(θ) =
r.
.
r
. adius of back sprocket
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
16. Analogy
Think about riding a bike. To go
faster you can either:
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.
.
And so the angular speed of the back wheel depends on the derivative
of this function and the speed of the front sprocket.
. . . . . .
.
Image credit: SpringSun
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 6 / 36
17. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 7 / 36
18. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 7 / 36
19. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 7 / 36
20. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
The slope of the composition is the product of the slopes of the
two functions.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 7 / 36
21. The Linear Case
Question
Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the
composition?
Answer
f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
The composition is also linear
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 7 / 36
22. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 8 / 36
23. Outline
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 9 / 36
24. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 10 / 36
25. Observations
Succinctly, the derivative of a
composition is the product of
the derivatives
.
. . . . . .
.
Image credit: ooOJasonOoo
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 11 / 36
26. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 12 / 36
27. Observations
Succinctly, the derivative of a
composition is the product of
the derivatives
The only complication is where
these derivatives are
evaluated: at the same point
the functions are
.
. . . . . .
.
Image credit: ooOJasonOoo
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 13 / 36
28. 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.”
g
. (x) f
.(g(x))
. ◦ g
x
.
g
. f . f
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 14 / 36
29. Observations
Succinctly, the derivative of a
composition is the product of
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
.
. . . . . .
.
Image credit: ooOJasonOoo
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 15 / 36
30. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 16 / 36
31. 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)
dy
In Leibnizian notation, let y = f(u) and u = g(x).du
Then
. .
dx
du
dy dy du
=
dx du dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 16 / 36
32. Outline
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 17 / 36
33. Example
Example
√
let h(x) = 3x2 + 1. Find h′ (x).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 18 / 36
34. Example
Example
√
let h(x) = 3x2 + 1. Find h′ (x).
Solution
First, write h as f ◦ g.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 18 / 36
35. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 18 / 36
36. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 18 / 36
37. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 18 / 36
38. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 19 / 36
39. Does order matter?
Example
d d
Find (sin 4x) and compare it to (4 sin x).
dx dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 20 / 36
40. Does order matter?
Example
d d
Find (sin 4x) and compare it to (4 sin x).
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 20 / 36
41. Does order matter?
Example
d d
Find (sin 4x) and compare it to (4 sin x).
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
For the second, let u = sin x and y = 4u. Then
dy dy du
= · = 4 · cos x
dx du dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 20 / 36
42. Order matters!
Example
d d
Find (sin 4x) and compare it to (4 sin x).
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
For the second, let u = sin x and y = 4u. Then
dy dy du
= · = 4 · cos x
dx du dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 20 / 36
43. Example
(√ )2
. Find f′ (x).
3
Let f(x) = x5 − 2 + 8
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 21 / 36
44. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 21 / 36
49. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 22 / 36
50. Combining techniques
Example
d ( 3 )
Find (x + 1)10 sin(4x2 − 7)
dx
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 23 / 36
51. Combining techniques
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:
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 23 / 36
52. Combining techniques
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:
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 23 / 36
53. Combining techniques
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:
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)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 23 / 36
54. 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))))))
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 24 / 36
55. Solution to #1
Example
Find the derivative of y = (1 − x2 )10 .
Solution
y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 25 / 36
56. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 26 / 36
57. Solution to #3
Example
√
Find the derivative of y = sin x.
Solution
(√ )
d cos x
y′ = sin(x1/2 ) = cos(x1/2 ) 2 x−1/2 =
1 √
dx 2 x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 27 / 36
58. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 28 / 36
59. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 29 / 36
60. Solution to #6
Example
Find the derivative of y = tan(cos x).
Solution
y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 30 / 36
61. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 31 / 36
62. 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))
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 32 / 36
63. Outline
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 33 / 36
64. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 34 / 36
65. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 34 / 36
66. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 35 / 36
67. 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
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 35 / 36
68. 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!
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.5 The Chain Rule May 25, 2010 36 / 36