This document provides an outline for a calculus lecture on basic differentiation rules. It includes objectives to understand key rules like the constant multiple rule, sum rule, and derivatives of sine and cosine. Examples are worked through to find the derivatives of functions like squaring, cubing, square root, and cube root using the definition of the derivative. Graphs and properties of derived functions are also discussed.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
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.
Lesson 15: Exponential Growth and Decay (slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
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.
Lesson 15: Exponential Growth and Decay (slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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.
Lesson 8: Derivatives of Polynomials and Exponential functionsMatthew Leingang
Some of the most famous rules of the calculus of derivatives: the power rule, the sum rule, the constant multiple rule, and the number e defined so that e^x is its own derivative!
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
Continuous function have an important property that small changes in input do not produce large changes in output. The Intermediate Value Theorem shows that a continuous function takes all values between any two values. From this we know that your height and weight were once the same, and right now there are two points on opposite sides of the world with the same temperature!
Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
Many functions in nature are described as the rate of change of another function. The concept is called the derivative. Algebraically, the process of finding the derivative involves a limit of difference quotients.
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 12: Linear Approximations and Differentials (slides)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
Lesson 8: Derivatives of Polynomials and Exponential functionsMatthew Leingang
Some of the most famous rules of the calculus of derivatives: the power rule, the sum rule, the constant multiple rule, and the number e defined so that e^x is its own derivative!
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
Continuous function have an important property that small changes in input do not produce large changes in output. The Intermediate Value Theorem shows that a continuous function takes all values between any two values. From this we know that your height and weight were once the same, and right now there are two points on opposite sides of the world with the same temperature!
Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
Many functions in nature are described as the rate of change of another function. The concept is called the derivative. Algebraically, the process of finding the derivative involves a limit of difference quotients.
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 12: Linear Approximations and Differentials (slides)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.
JEE Physics/ Lakshmikanta Satapathy/ Wave Motion QA part 3/ JEE question on fundamental frequencies of Open and Closed pipes solved with the related concepts
The basic rules of differentiation, including the power rule, the sum rule, the constant multiple rule, and the rule for differentiating sine and cosine.
After defining the limit and calculating a few, we introduced the limit laws. Today we do the same for the derivative. We calculate a few and introduce laws which allow us to computer more. The Power Rule shows us how to compute derivatives of polynomials, and we can also find directly the derivative of sine and cosine.
After defining the limit and calculating a few, we introduced the limit laws. Today we do the same for the derivative. We calculate a few and introduce laws which allow us to computer more. The Power Rule shows us how to compute derivatives of polynomials, and we can also find directly the derivative of sine and cosine.
Lesson 16: Derivatives of Logarithmic and Exponential FunctionsMatthew Leingang
We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated differentiation problems easier.
Lesson 16: Derivatives of Logarithmic and Exponential FunctionsMatthew Leingang
We show the the derivative of the exponential function is itself! And the derivative of the natural logarithm function is the reciprocal function. We also show how logarithms can make complicated differentiation problems easier.
Lesson 14: Derivatives of Logarithmic and Exponential Functions (slides)Mel Anthony Pepito
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Continuity says that the limit of a function at a point equals the value of the function at that point, or, that small changes in the input give only small changes in output. This has important implications, such as the Intermediate Value Theorem.
The second Fundamental Theorem of Calculus makes calculating definite integrals a problem of antidifferentiation!
(the slideshow has extra examples based on what happened in class)
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.
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.
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.
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.
Lesson 15: Exponential Growth and Decay (handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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This talk is aimed at encouraging a more independent approach to using PHP frameworks, moving towards a more flexible and future-proof approach to PHP development.
Transcript: Selling digital books in 2024: Insights from industry leaders - T...BookNet Canada
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Link to video recording: https://bnctechforum.ca/sessions/selling-digital-books-in-2024-insights-from-industry-leaders/
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Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
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https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
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Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
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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.
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Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
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Lesson 8: Basic Differentation Rules (slides)
1. Sec on 2.3
Basic Differenta on Rules
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
. NYUMathematics
2. Announcements
Quiz 1 this week on
1.1–1.4
Quiz 2 March 3/4 on 1.5,
1.6, 2.1, 2.2, 2.3
Midterm Monday March
7 in class
3. Objectives
Understand and use
these differen a on
rules:
the deriva ve of a
constant func on (zero);
the Constant Mul ple
Rule;
the Sum Rule;
the Difference Rule;
the deriva ves of sine
and cosine.
4. Recall: the derivative
Defini on
Let f be a func on and a a point in the domain of f. If the limit
f(a + h) − f(a) f(x) − f(a)
f′ (a) = lim = lim
h→0 h x→a x−a
exists, the func on is said to be differen able at a and f′ (a) is the
deriva ve of f at a.
5. The deriva ve …
…measures the slope of the line through (a, f(a)) tangent to
the curve y = f(x);
…represents the instantaneous rate of change of f at a
…produces the best possible linear approxima on to f near a.
6. Notation
Newtonian nota on Leibnizian nota on
dy d df
f′ (x) y′ (x) y′ f(x)
dx dx dx
7. Link between the notations
f(x + ∆x) − f(x) ∆y dy
f′ (x) = lim = lim =
∆x→0 ∆x ∆x→0 ∆x dx
dy
Leibniz thought of as a quo ent of “infinitesimals”
dx
dy
We think of as represen ng a limit of (finite) difference
dx
quo ents, not as an actual frac on itself.
The nota on suggests things which are true even though they
don’t follow from the nota on per se
8. Outline
Deriva ves so far
Deriva ves of power func ons by hand
The Power Rule
Deriva ves of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Mul ple Rule
Deriva ves of sine and cosine
9. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
10. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
Solu on
f(x + h) − f(x)
f′ (x) = lim
h→0 h
11. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
12. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
2
x2
x2
= lim
h→0 h
13. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
x2
2
x2
2x + h2
h
¡
= lim = lim
h→0 h h→0 h
14. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)2 − x2
f (x) = lim = lim
h→0 h h→0 h
+ 2xh + h −
x2
2
x2
2x + h2
h
¡
= lim = lim
h→0 h h→0 h
= lim (2x + h) = 2x.
h→0
15. The second derivative
If f is a func on, so is f′ , and we can seek its deriva ve.
f′′ = (f′ )′
It measures the rate of change of the rate of change!
16. The second derivative
If f is a func on, so is f′ , and we can seek its deriva ve.
f′′ = (f′ )′
It measures the rate of change of the rate of change! Leibnizian
nota on:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2
19. The squaring function and its derivatives
y
f′
f increasing =⇒ f′ ≥ 0
f decreasing =⇒ f′ ≤ 0
. f x horizontal tangent at 0
=⇒ f′ (0) = 0
20. The squaring function and its derivatives
y
f′
f′′ f increasing =⇒ f′ ≥ 0
f decreasing =⇒ f′ ≤ 0
. f x horizontal tangent at 0
=⇒ f′ (0) = 0
21. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x).
22. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
23. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
x3
+ 3x2 h + 3xh2 + h3 − 3
x
= lim
h→0 h
24. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
1 2
!
¡ !
¡
x3
+
2 2
3x h + 3xh + h 3
− 3
x 3x2 +
h
¡
3xh2 + ¡
h3
= lim = lim
h→0 h h→0 h
25. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the defini on of deriva ve to find f′ (x).
Solu on
′ f(x + h) − f(x) (x + h)3 − x3
f (x) = lim = lim
h→0 h h→0 h
1 2
!
¡ !
¡
x3
2 2
3x h + 3xh + h
+
3
− 3
x 3x2 +
h
¡
3xh2 + ¡
h3
= lim = lim
h→0
( 2 h ) h→0 h
= lim 3x + 3xh + h2 = 3x2 .
h→0
28. The cubing function and its derivatives
No ce that f is increasing,
y and f′ 0 except
f′ f′ (0) = 0
f
. x
29. The cubing function and its derivatives
No ce that f is increasing,
y and f′ 0 except
f′′ f′ f′ (0) = 0
f
. x
30. The cubing function and its derivatives
No ce that f is increasing,
y and f′ 0 except
f′′ f′ f′ (0) = 0
No ce also that the
f tangent line to the graph
. x of f at (0, 0) crosses the
graph (contrary to a
popular “defini on” of
the tangent line)
31. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
32. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
33. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
34. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡
= lim (√ √ )
h→0 h x+h+ x
35. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡ h
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h x+h+ x
h→0
36. Derivative of the square root
Example
√
Suppose f(x) = x = x1/2 . Fnd f′ (x) with the defini on.
Solu on
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0
√ h h→0 h
√ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(x + h) − x
¡ ¡ h
1
= lim (√ √ ) = lim (√ √ )= √
h→0 h x+h+ x h x+h+ x
h→0 2 x
49. The cube root and its derivative
y
Here lim f′ (x) = ∞ and f
f x→0
is not differen able at 0
f′
. x
50. The cube root and its derivative
y
Here lim f′ (x) = ∞ and f
f x→0
is not differen able at 0
f′
. x No ce also
lim f′ (x) = 0
x→±∞
51. One more
Example
Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x).
52. One more
Example
Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x).
Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
53. One more
Example
Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x).
Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0 h
54. One more
Example
Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x).
Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0
( h)
1 −2/3
= 3x 2x1/3
55. One more
Example
Suppose f(x) = x2/3 . Use the defini on of deriva ve to find f′ (x).
Solu on
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
1/3 ( )
(x + h) − x
1/3
= lim · (x + h) + x
1/3 1/3
h→0
( h)
1 −2/3
= 3x 2x1/3 = 2 x−1/3
3
66. Recap: The Tower of Power
y y′
x2 2x1 The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
67. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
68. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
69. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
70. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
71. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
72. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x2 one in each deriva ve
1 −1/2 The coefficient in the
x1/2 2x deriva ve is the power of
1 −2/3
x1/3 3x
the original func on
2 −1/3
x2/3 3x
73. The Power Rule
There is moun ng evidence for
Theorem (The Power Rule)
Let r be a real number and f(x) = xr . Then
f′ (x) = rxr−1
as long as the expression on the right-hand side is defined.
Perhaps the most famous rule in calculus
We will assume it as of today
We will prove it many ways for many different r.
75. Outline
Deriva ves so far
Deriva ves of power func ons by hand
The Power Rule
Deriva ves of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Mul ple Rule
Deriva ves of sine and cosine
76. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
77. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
78. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
79. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
80. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
81. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
82. Remember your algebra
Fact
Let n be a posi ve whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
∑
n
(x + h) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
87. Proving the Power Rule
Theorem (The Power Rule)
d n
Let n be a posi ve whole number. Then x = nxn−1 .
dx
88. Proving the Power Rule
Theorem (The Power Rule)
d n
Let n be a posi ve whole number. Then x = nxn−1 .
dx
Proof.
As we showed above,
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
(x + h)n − xn nxn−1 h + (stuff with at least two hs in it)
So =
h h
= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h → 0.
89. The Power Rule for constants?
Theorem
d
Let c be a constant. Then c=0
dx
90. The Power Rule for constants?
d 0
Theorem like x = 0x−1
d dx
Let c be a constant. Then c = 0.
.
dx
91. The Power Rule for constants?
d 0
Theorem like x = 0x−1
d dx
Let c be a constant. Then c = 0.
.
dx
Proof.
Let f(x) = c. Then
f(x + h) − f(x) c − c
= =0
h h
So f′ (x) = lim 0 = 0.
h→0
93. Recall the Limit Laws
Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
x→a x→a
1. lim [f(x) + g(x)] = L + M
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL
x→a
4. . . .
94. Adding functions
Theorem (The Sum Rule)
Let f and g be func ons and define
(f + g)(x) = f(x) + g(x)
Then if f and g are differen able at x, then so is f + g and
(f + g)′ (x) = f′ (x) + g′ (x).
Succinctly, (f + g)′ = f′ + g′ .
95. Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
96. Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
97. Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
98. Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
′ ′
= f (x) + g (x)
99. Scaling functions
Theorem (The Constant Mul ple Rule)
Let f be a func on and c a constant. Define
(cf)(x) = cf(x)
Then if f is differen able at x, so is cf and
(cf)′ (x) = c · f′ (x)
Succinctly, (cf)′ = cf′ .
100. Proof of Constant Multiple Rule
Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
101. Proof of Constant Multiple Rule
Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
(cf)′ (x) = lim = lim
h→0 h h→0 h
102. Proof of Constant Multiple Rule
Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
(cf)′ (x) = lim = lim
h→0 h h→0 h
f(x + h) − f(x)
= c lim
h→0 h
103. Proof of Constant Multiple Rule
Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x) cf(x + h) − cf(x)
(cf)′ (x) = lim = lim
h→0 h h→0 h
f(x + h) − f(x)
= c lim = c · f′ (x)
h→0 h
105. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on
d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
106. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on
d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
107. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on
d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3
108. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solu on
d ( 3 ) d ( 3) d d ( ) d
2x + x4 − 17x12 + 37 = 2x + x4 + −17x12 + (37)
dx dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3
= 6x2 + 4x3 − 204x11
109. Outline
Deriva ves so far
Deriva ves of power func ons by hand
The Power Rule
Deriva ves of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Mul ple Rule
Deriva ves of sine and cosine
110. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
111. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
112. Angle addition formulas
See Appendix A
sin(A + B) = . A cos B + cos A sin B
sin
cos(A + B) = cos A cos B − sin A sin B
113. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h
114. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
116. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1
117. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the defini on:
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1
118. Derivatives of Sine and Cosine
Fact
d
sin x = cos x
dx
Proof.
From the defini on:
d sin(x + h) − sin x ( sin x cos h + cos x sin h) − sin x
sin x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1 = cos x
120. Illustration of Sine and Cosine
y
. x
π −π 0 π π cos x
2 2
sin x
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
121. Illustration of Sine and Cosine
y
. x
π −π 0 π π cos x
2 2
sin x
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
what happens at the horizontal tangents of cos?
123. Derivative of Cosine
Fact
d
cos x = − sin x
dx
Proof.
We already did the first. The second is similar (muta s mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h
124. Derivative of Cosine
Fact
d
cos x = − sin x
dx
Proof.
We already did the first. The second is similar (muta s mutandis):
d cos(x + h) − cos x (cos x cos h − sin x sin h) − cos x
cos x = lim = lim
dx h→0 h h→0 h
125. Derivative of Cosine
Fact
d
cos x = − sin x
dx
Proof.
We already did the first. The second is similar (muta s mutandis):
d cos(x + h) − cos x (cos x cos h − sin x sin h) − cos x
cos x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
126. Derivative of Cosine
Fact
d
cos x = − sin x
dx
Proof.
We already did the first. The second is similar (muta s mutandis):
d cos(x + h) − cos x (cos x cos h − sin x sin h) − cos x
cos x = lim = lim
dx h→0 h h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
= cos x · 0 − sin x · 1 = − sin x
128. Summary
What have we learned today?
The Power Rule
The deriva ve of a sum is the sum of the deriva ves
The deriva ve of a constant mul ple of a func on is that
constant mul ple of the deriva ve
129. Summary
What have we learned today?
The Power Rule
The deriva ve of a sum is the sum of the deriva ves
The deriva ve of a constant mul ple of a func on is that
constant mul ple of the deriva ve
The deriva ve of sine is cosine
The deriva ve of cosine is the opposite of sine.