The product rule can be iterated to find the derivative of products with more than two factors. The derivative of a three-factor product uvw is u'vw + uv'w + uvw'. More generally, the derivative of a product of n factors breaks the product into a sum of n terms by applying the product rule recursively.
There's an obvious guess for the rule about the derivative of a product—which happens to be wrong. We explain the correct product rule and why it has to be true, along with the quotient rule. We show how to compute derivatives of additional trigonometric functions and new proofs of the Power Rule.
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
There's an obvious guess for the rule about the derivative of a product—which happens to be wrong. We explain the correct product rule and why it has to be true, along with the quotient rule. We show how to compute derivatives of additional trigonometric functions and new proofs of the Power Rule.
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 27: Integration by Substitution (Section 4 version)Matthew Leingang
The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples and graphics.
Here's a toy problem: What is the SMALLEST number of unit balls you can fit in a box such that no more will fit?
In this talk, I will show how just thinking about a naive greedy approach to this problem leads to a simple derivation of several of the most important theoretical results in the field of mesh generation.
We'll prove classic upper and lower bounds on both the number of balls and the complexity of their interrelationships.
Then, we'll relate this problem to a similar one called the Fat Voronoi Problem, in which we try to find point sets such that every Voronoi cell is fat
(the ratio of the radii of the largest contained to smallest containing ball is bounded).
This problem has tremendous promise in the future of mesh generation as it can circumvent the classic lowerbounds presented in the first half of the talk.
Unfortunately the simple approach no longer works.
In the end we will show that the number of neighbors of any cell in a Fat Voronoi Diagram in the plane is bounded by a constant
(if you think that's obvious, spend a minute to try to prove it).
We'll also talk a little about the higher dimensional version of the problem and its wide range of applications.
Lesson 27: Integration by Substitution (Section 10 version)Matthew Leingang
The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples.
Supervised machine learning addresses the problem of approximating a function, given the examples of inputs and outputs. The classical tasks of regression and classification deal with functions whose outputs are real numbers. Structured output prediction goes beyond one-dimensional outputs, and allows predicting complex objects, such as sequences, trees, and graphs. In this talk I will show how to apply structured output prediction to building informative summaries of the topic graphs—a problem I encountered in my Ph.D. research. The focus of the talk will be on understanding the intuitions behind the machine learning algorithms. We will start from the basics and walk our way through the inner workings of DAgger—state-of-the-art method of structured output prediction.
This talk was be given at a seminar in Google Krakow.
A Study on Intuitionistic Multi-Anti Fuzzy Subgroups mathsjournal
For any intuitionistic multi-fuzzy set A = { < x , µA(x) , νA(x) > : x∈X} of an universe set X, we study the set [A](α, β) called the (α, β)–lower cut of A. It is the crisp multi-set { x∈X : µi(x) ≤ αi , νi(x) ≥ βi , ∀i } of X. In this paper, an attempt has been made to study some algebraic structure of intuitionistic multi-anti fuzzy subgroups and their properties with the help of their (α, β)–lower cut sets
First principle, power rule, derivative of constant term, product rule, quotient rule, chain rule, derivatives of trigonometric functions and their inverses, derivatives of exponential functions and natural logarithmic functions, implicit differentiation, parametric differentiation, L'Hopital's rule
Lesson 27: Integration by Substitution (Section 4 version)Matthew Leingang
The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples and graphics.
Here's a toy problem: What is the SMALLEST number of unit balls you can fit in a box such that no more will fit?
In this talk, I will show how just thinking about a naive greedy approach to this problem leads to a simple derivation of several of the most important theoretical results in the field of mesh generation.
We'll prove classic upper and lower bounds on both the number of balls and the complexity of their interrelationships.
Then, we'll relate this problem to a similar one called the Fat Voronoi Problem, in which we try to find point sets such that every Voronoi cell is fat
(the ratio of the radii of the largest contained to smallest containing ball is bounded).
This problem has tremendous promise in the future of mesh generation as it can circumvent the classic lowerbounds presented in the first half of the talk.
Unfortunately the simple approach no longer works.
In the end we will show that the number of neighbors of any cell in a Fat Voronoi Diagram in the plane is bounded by a constant
(if you think that's obvious, spend a minute to try to prove it).
We'll also talk a little about the higher dimensional version of the problem and its wide range of applications.
Lesson 27: Integration by Substitution (Section 10 version)Matthew Leingang
The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples.
Supervised machine learning addresses the problem of approximating a function, given the examples of inputs and outputs. The classical tasks of regression and classification deal with functions whose outputs are real numbers. Structured output prediction goes beyond one-dimensional outputs, and allows predicting complex objects, such as sequences, trees, and graphs. In this talk I will show how to apply structured output prediction to building informative summaries of the topic graphs—a problem I encountered in my Ph.D. research. The focus of the talk will be on understanding the intuitions behind the machine learning algorithms. We will start from the basics and walk our way through the inner workings of DAgger—state-of-the-art method of structured output prediction.
This talk was be given at a seminar in Google Krakow.
A Study on Intuitionistic Multi-Anti Fuzzy Subgroups mathsjournal
For any intuitionistic multi-fuzzy set A = { < x , µA(x) , νA(x) > : x∈X} of an universe set X, we study the set [A](α, β) called the (α, β)–lower cut of A. It is the crisp multi-set { x∈X : µi(x) ≤ αi , νi(x) ≥ βi , ∀i } of X. In this paper, an attempt has been made to study some algebraic structure of intuitionistic multi-anti fuzzy subgroups and their properties with the help of their (α, β)–lower cut sets
First principle, power rule, derivative of constant term, product rule, quotient rule, chain rule, derivatives of trigonometric functions and their inverses, derivatives of exponential functions and natural logarithmic functions, implicit differentiation, parametric differentiation, L'Hopital's rule
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
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Generating a custom Ruby SDK for your web service or Rails API using Smithyg2nightmarescribd
Have you ever wanted a Ruby client API to communicate with your web service? Smithy is a protocol-agnostic language for defining services and SDKs. Smithy Ruby is an implementation of Smithy that generates a Ruby SDK using a Smithy model. In this talk, we will explore Smithy and Smithy Ruby to learn how to generate custom feature-rich SDKs that can communicate with any web service, such as a Rails JSON API.
Dev Dives: Train smarter, not harder – active learning and UiPath LLMs for do...UiPathCommunity
💥 Speed, accuracy, and scaling – discover the superpowers of GenAI in action with UiPath Document Understanding and Communications Mining™:
See how to accelerate model training and optimize model performance with active learning
Learn about the latest enhancements to out-of-the-box document processing – with little to no training required
Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
This is a hands-on session specifically designed for automation developers and AI enthusiasts seeking to enhance their knowledge in leveraging the latest intelligent document processing capabilities offered by UiPath.
Speakers:
👨🏫 Andras Palfi, Senior Product Manager, UiPath
👩🏫 Lenka Dulovicova, Product Program Manager, UiPath
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The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
JMeter webinar - integration with InfluxDB and GrafanaRTTS
Watch this recorded webinar about real-time monitoring of application performance. See how to integrate Apache JMeter, the open-source leader in performance testing, with InfluxDB, the open-source time-series database, and Grafana, the open-source analytics and visualization application.
In this webinar, we will review the benefits of leveraging InfluxDB and Grafana when executing load tests and demonstrate how these tools are used to visualize performance metrics.
Length: 30 minutes
Session Overview
-------------------------------------------
During this webinar, we will cover the following topics while demonstrating the integrations of JMeter, InfluxDB and Grafana:
- What out-of-the-box solutions are available for real-time monitoring JMeter tests?
- What are the benefits of integrating InfluxDB and Grafana into the load testing stack?
- Which features are provided by Grafana?
- Demonstration of InfluxDB and Grafana using a practice web application
To view the webinar recording, go to:
https://www.rttsweb.com/jmeter-integration-webinar
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We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on the notifications, alerts, and approval requests using Slack for Bonterra Impact Management. The solutions covered in this webinar can also be deployed for Microsoft Teams.
Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
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Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
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Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
Knowledge engineering: from people to machines and back
Lesson 9: The Product and Quotient Rules (slides)
1. Sec on 2.4
The Product and Quo ent Rules
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
February 23, 2011
.
2. Announcements
Quiz 2 next week on
§§1.5, 1.6, 2.1, 2.2
Midterm March 7 on all
sec ons in class (covers
all sec ons up to 2.5)
3. Help!
Free resources:
Math Tutoring Center
(CIWW 524)
College Learning Center
(schedule on Blackboard)
TAs’ office hours
my office hours
each other!
4. Objectives
Understand and be able
to use the Product Rule
for the deriva ve of the
product of two func ons.
Understand and be able
to use the Quo ent Rule
for the deriva ve of the
quo ent of two
func ons.
5. Outline
Deriva ve of a Product
Deriva on
Examples
The Quo ent Rule
Deriva on
Examples
More deriva ves of trigonometric func ons
Deriva ve of Tangent and Cotangent
Deriva ve of Secant and Cosecant
More on the Power Rule
Power Rule for Nega ve Integers
6. Recollection and extension
We have shown that if u and v are func ons, that
(u + v)′ = u′ + v′
(u − v)′ = u′ − v′
What about uv?
7. Is the derivative of a product the
product of the derivatives?
(uv)′ = u′ v′ ?
.
8. Is the derivative of a product the
product of the derivatives?
(uv)′ = u′ v′ !
.
Try this with u = x and v = x2 .
9. Is the derivative of a product the
product of the derivatives?
(uv)′ = u′ v′ !
.
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
10. Is the derivative of a product the
product of the derivatives?
(uv)′ = u′ v′ !
.
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
But u′ v′ = 1 · 2x = 2x.
11. Is the derivative of a product the
product of the derivatives?
(uv)′ = u′ v′ !
.
Try this with u = x and v = x2 .
Then uv = x3 =⇒ (uv)′ = 3x2 .
But u′ v′ = 1 · 2x = 2x.
So we have to be more careful.
12. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
.
13. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
.
14. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.
.
15. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.
Say you get a 25 cent raise in
your hourly wages and work 5
hours more per week. How
much extra money do you
make?
.
16. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.
Say you get a 25 cent raise in
your hourly wages and work 5
hours more per week. How
much extra money do you
make?
.
∆I = 5 × $0.25 = $1.25?
17. Mmm...burgers
Say you work in a fast-food joint. You want to make more money.
What are your choices?
Work longer hours.
Get a raise.
Say you get a 25 cent raise in
your hourly wages and work 5
hours more per week. How
much extra money do you
make?
.
∆I = 5 × $0.25 = $1.25?
18. Money money money money
The answer depends on how much you work already and your
current wage. Suppose you work h hours and are paid w. You get a
me increase of ∆h and a wage increase of ∆w. Income is wages
mes hours, so
∆I = (w + ∆w)(h + ∆h) − wh
FOIL
= w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh
= w · ∆h + ∆w · h + ∆w · ∆h
20. A geometric argument
Draw a box:
∆h w ∆h ∆w ∆h
h wh ∆w h
.
w ∆w
∆I = w ∆h + h ∆w + ∆w ∆h
21. Cash flow
Supose wages and hours are changing con nuously over me. Over
a me interval ∆t, what is the average rate of change of income?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
22. Cash flow
Supose wages and hours are changing con nuously over me. Over
a me interval ∆t, what is the average rate of change of income?
∆I w ∆h + h ∆w + ∆w ∆h
=
∆t ∆t
∆h ∆w ∆h
=w +h + ∆w
∆t ∆t ∆t
What is the instantaneous rate of change of income?
dI ∆I dh dw
= lim =w +h +0
dt ∆t→0 ∆t dt dt
23. Eurekamen!
We have discovered
Theorem (The Product Rule)
Let u and v be differen able at x. Then
(uv)′ (x) = u(x)v′ (x) + u′ (x)v(x)
in Leibniz nota on
d du dv
(uv) = ·v+u
dx dx dx
25. Sanity Check
Example
Apply the product rule to u = x and v = x2 .
Solu on
(uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2
This is what we get the “normal” way.
26. Which is better?
Example
Find this deriva ve two ways: first by direct mul plica on and then
by the product rule:
d [ ]
(3 − x2 )(x3 − x + 1)
dx
28. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by direct mul plica on:
d [ ] FOIL d [ 5 ]
(3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3
dx dx
29. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by direct mul plica on:
d [ ] FOIL d [ 5 ]
(3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3
dx dx
= −5x4 + 12x2 − 2x − 3
30. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
31. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
32. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
33. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
34. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
35. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
36. Which is better?
Example
d [ ]
(3 − x2 )(x3 − x + 1)
dx
Solu on
by the product rule:
( ) ( )
dy d d 3
= (3 − x ) (x − x + 1) + (3 − x )
2 3 2
(x − x + 1)
dx dx dx
= (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
= −5x4 + 12x2 − 2x − 3
38. One more
Example
d
Find x sin x.
dx
Solu on
( ) ( )
d d d
x sin x = x sin x + x sin x
dx dx dx
39. One more
Example
d
Find x sin x.
dx
Solu on
( ) ( )
d d d
x sin x = x sin x + x sin x
dx dx dx
= 1 · sin x + x · cos x
40. One more
Example
d
Find x sin x.
dx
Solu on
( ) ( )
d d d
x sin x = x sin x + x sin x
dx dx dx
= 1 · sin x + x · cos x
= sin x + x cos x
41. Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
42. Musical interlude
jazz bandleader and
singer
hit song “Minnie the
Moocher” featuring “hi
de ho” chorus
played Cur s in The Blues
Brothers
Cab Calloway
1907–1994
43. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.
44. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ .
45. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′ .
46. Iterating the Product Rule
Example
Apply the product
Use the product rule to find the deriva ve of a three-fold product uvw.
rule to uv and w
Solu on
(uvw)′ = ((uv)w)′ .
47. Iterating the Product Rule
Example
Apply the product
Use the product rule to find the deriva ve of a three-fold product uvw.
rule to uv and w
Solu on
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
48. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.product
Apply the
rule to u and v
Solu on
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
49. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.product
Apply the
rule to u and v
Solu on
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
50. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
= u′ vw + uv′ w + uvw′
51. Iterating the Product Rule
Example
Use the product rule to find the deriva ve of a three-fold product uvw.
Solu on
(uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
= (u′ v + uv′ )w + (uv)w′
= u′ vw + uv′ w + uvw′
So we write down the product three mes, taking the deriva ve of each factor
once.
52. Outline
Deriva ve of a Product
Deriva on
Examples
The Quo ent Rule
Deriva on
Examples
More deriva ves of trigonometric func ons
Deriva ve of Tangent and Cotangent
Deriva ve of Secant and Cosecant
More on the Power Rule
Power Rule for Nega ve Integers
54. The Quotient Rule
What about the deriva ve of a quo ent?
u
Let u and v be differen able func ons and let Q = . Then
v
u = Qv
55. The Quotient Rule
What about the deriva ve of a quo ent?
u
Let u and v be differen able func ons and let Q = . Then
v
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′ v + Qv′
56. The Quotient Rule
What about the deriva ve of a quo ent?
u
Let u and v be differen able func ons and let Q = . Then
v
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′ v + Qv′
′ u′ − Qv′ u′ u v′
=⇒ Q = = − ·
v v v v
57. The Quotient Rule
What about the deriva ve of a quo ent?
u
Let u and v be differen able func ons and let Q = . Then
v
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′ v + Qv′
′ u′ − Qv′ u′ u v′
=⇒ Q = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
58. The Quotient Rule
What about the deriva ve of a quo ent?
u
Let u and v be differen able func ons and let Q = . Then
v
u = Qv
If Q is differen able, we have
u′ = (Qv)′ = Q′ v + Qv′
′ u′ − Qv′ u′ u v′
=⇒ Q = = − ·
v v v v
( u )′ u′ v − uv′
=⇒ Q′ = =
v v2
This is called the Quo ent Rule.
59. The Quotient Rule
We have discovered
Theorem (The Quo ent Rule)
u
Let u and v be differen able at x, and v(x) ̸= 0. Then is
v
differen able at x, and
( u )′ u′ (x)v(x) − u(x)v′ (x)
(x) =
v v(x)2
60. Verifying Example
Example
( )
d x2 d
Verify the quo ent rule by compu ng and comparing it to (x).
dx x dx
61. Verifying Example
Example
( )
d x2 d
Verify the quo ent rule by compu ng and comparing it to (x).
dx x dx
Solu on
( 2) ( )
d x x dx x2 − x2 dx (x) x · 2x − x2 · 1
d d
= =
dx x x2 x2
x2 d
= 2 =1= (x)
x dx
62. Mnemonic
Let u = “hi” and v = “lo”. Then
( u )′ vu′ − uv′
= = “lo dee hi minus hi dee lo over lo lo”
v v2
63. Examples
Example
d 2x + 5
1.
dx 3x − 2
d sin x
2.
dx x2
d 1
3. 2+t+2
dt t
82. Solution to second example
Solu on
d sin x x2 dx sin x − sin x
d
=
dx x2
83. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2
84. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
85. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
=
86. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2
=
87. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2 cos x
=
88. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2 cos x − 2x
=
89. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2 cos x − 2x sin x
=
90. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2 cos x − 2x sin x
=
x4
91. Solution to second example
Solu on
d sin x x2 dx sin x − sin x dx x2
d d
=
dx x2 (x2 )2
x2 cos x − 2x sin x
=
x4
x cos x − 2 sin x
=
x3
92. Another way to do it
Find the deriva ve with the product rule instead.
Solu on
d sin x d ( )
2
= sin x · x−2
dx x dx
( ) ( )
d −2 d −2
= sin x · x + sin x · x
dx dx
= cos x · x−2 + sin x · (−2x−3 )
= x−3 (x cos x − 2 sin x)
No ce the technique of factoring out the largest nega ve power,
leaving posi ve powers.
93. Examples
Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1
3. 2+t+2
dt t
95. Solution to third example
Solu on
d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
96. Solution to third example
Solu on
d 1 (t2 + t + 2)(0) − (1)(2t + 1)
=
dt t2 + t + 2 (t2 + t + 2)2
2t + 1
=− 2
(t + t + 2)2
97. A nice little takeaway
Fact
1
Let v be differen able at x, and v(x) ̸= 0. Then is differen able at
v
0, and ( )′
1 v′
=− 2
v v
Proof.
( )
d 1 v· d
dx (1)−1· d
dx v v · 0 − 1 · v′ v′
= = =− 2
dx v v2 v2 v
98. Examples
Example Answers
d 2x + 5 19
1. 1. −
dx 3x − 2 (3x − 2)2
d sin x x cos x − 2 sin x
2. 2.
dx x2 x3
d 1 2t + 1
3. 3. − 2
dt t2+t+2 (t + t + 2)2
99. Outline
Deriva ve of a Product
Deriva on
Examples
The Quo ent Rule
Deriva on
Examples
More deriva ves of trigonometric func ons
Deriva ve of Tangent and Cotangent
Deriva ve of Secant and Cosecant
More on the Power Rule
Power Rule for Nega ve Integers
101. Derivative of Tangent
Example
d
Find tan x
dx
Solu on
( )
d d sin x
tan x =
dx dx cos x
102. Derivative of Tangent
Example
d
Find tan x
dx
Solu on
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
103. Derivative of Tangent
Example
d
Find tan x
dx
Solu on
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x
=
cos2 x
104. Derivative of Tangent
Example
d
Find tan x
dx
Solu on
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= =
cos2 x cos2 x
105. Derivative of Tangent
Example
d
Find tan x
dx
Solu on
( )
d d sin x cos x · cos x − sin x · (− sin x)
tan x = =
dx dx cos x cos2 x
cos2 x + sin2 x 1
= 2x
= 2x
= sec2 x
cos cos
108. Derivative of Cotangent
Example
d
Find cot x
dx
Solu on
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
109. Derivative of Cotangent
Example
d
Find cot x
dx
Solu on
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x
=
sin2 x
110. Derivative of Cotangent
Example
d
Find cot x
dx
Solu on
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x 1
= =− 2
sin2 x sin x
111. Derivative of Cotangent
Example
d
Find cot x
dx
Solu on
d d ( cos x ) sin x · (− sin x) − cos x · cos x
cot x = =
dx dx sin x sin2 x
− sin2 x − cos2 x 1
= = − 2 = − csc2 x
sin2 x sin x
113. Derivative of Secant
Example
d
Find sec x
dx
Solu on
( )
d d 1
sec x =
dx dx cos x
114. Derivative of Secant
Example
d
Find sec x
dx
Solu on
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
115. Derivative of Secant
Example
d
Find sec x
dx
Solu on
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x
=
cos2 x
116. Derivative of Secant
Example
d
Find sec x
dx
Solu on
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= = ·
cos2 x cos x cos x
117. Derivative of Secant
Example
d
Find sec x
dx
Solu on
( )
d d 1 cos x · 0 − 1 · (− sin x)
sec x = =
dx dx cos x cos2 x
sin x 1 sin x
= = · = sec x tan x
cos2 x cos x cos x
120. Derivative of Cosecant
Example
d
Find csc x
dx
Solu on
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
121. Derivative of Cosecant
Example
d
Find csc x
dx
Solu on
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x
=− 2
sin x
122. Derivative of Cosecant
Example
d
Find csc x
dx
Solu on
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− ·
sin x sin x sin x
123. Derivative of Cosecant
Example
d
Find csc x
dx
Solu on
( )
d d 1 sin x · 0 − 1 · (cos x)
csc x = =
dx dx sin x sin2 x
cos x 1 cos x
=− 2 =− · = − csc x cot x
sin x sin x sin x
124. Recap: Derivatives of
trigonometric functions
y y′
sin x cos x Func ons come in pairs
(sin/cos, tan/cot, sec/csc)
cos x − sin x
Deriva ves of pairs follow
tan x sec2 x similar pa erns, with
cot x − csc2 x func ons and
co-func ons switched
sec x sec x tan x and an extra sign.
csc x − csc x cot x
125. Outline
Deriva ve of a Product
Deriva on
Examples
The Quo ent Rule
Deriva on
Examples
More deriva ves of trigonometric func ons
Deriva ve of Tangent and Cotangent
Deriva ve of Secant and Cosecant
More on the Power Rule
Power Rule for Nega ve Integers
126. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
127. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
128. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
d −n d 1
x =
dx dx xn
129. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
d n
d −n d 1 dx x
x = =− n 2
dx dx xn (x )
130. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
d n
d −n d 1 dx x nxn−1
x = = − n 2 = − 2n
dx dx xn (x ) x
131. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
d n
d −n d 1 dx x nxn−1
x = n
= − n 2 = − 2n = −nxn−1−2n
dx dx x (x ) x
132. Power Rule for Negative Integers
We will use the quo ent rule to prove
Theorem
d −n
x = (−n)x−n−1
dx
for posi ve integers n.
Proof.
d n
d −n d 1 dx x nxn−1
x = n
= − n 2 = − 2n = −nxn−1−2n = −nx−n−1
dx dx x (x ) x
133. Summary
The Product Rule: (uv)′ = u′ v + uv′
( u )′ vu′ − uv′
The Quo ent Rule: =
v v2
Deriva ves of tangent/cotangent, secant/cosecant
d d
tan x = sec2 x sec x = sec x tan x
dx dx
d d
cot x = − csc2 x csc x = − csc x cot x
dx dx
The Power Rule is true for all whole number powers, including
nega ve powers:
d n
x = nxn−1
dx