This document is a lecture summary for Calculus I at New York University. It covers the objectives and outline for sections 2.1 and 2.2 on the derivative and rates of change. The summary uses examples and graphics to demonstrate how to find the slope of the tangent line to a curve at a given point, both graphically and numerically using the definition of the derivative.
The document outlines topics to be covered in Sections 2.1-2.2 of a Calculus I course, including defining the derivative, finding derivatives of functions, relating the graph of a function to its derivative, and finding tangent lines. It provides examples of how derivatives can model real-world rates of change and outlines the objectives to understand the definition of the derivative and how to apply it to functions.
The document is from a Calculus I course at New York University and covers the topic of the derivative and rates of change. It discusses finding the slope of the tangent line to a curve at a given point, using the example of finding the slope of the tangent line to the curve y=x^2 at the point (2,4). It then shows this problem solved graphically and numerically by calculating the limit of the difference quotient as Δx approaches 0.
The document outlines lecture notes for Calculus I on the topic of the derivative and rates of change. It includes objectives for Sections 2.1 and 2.2 such as understanding the definition of the derivative and using it to find derivatives of functions. It also provides examples of how derivatives can model real-world phenomena like velocity, population growth, and marginal costs. The notes give an outline of topics to be covered including tangent lines, various notations for derivatives, and how to find second derivatives.
This document contains notes from a Calculus I class lecture on the derivative. The lecture covered the definition of the derivative and examples of how it can be used to model rates of change in various contexts like velocity, population growth, and marginal costs. It also discussed properties of the derivative like how the derivative of a function relates to whether the function is increasing or decreasing over an interval.
The derivative is a major tool for investigating the behavior of a function. Since functions are ubiquitous, so are their derivatives. Velocity, growth rates, marginal costs, and material strain are all examples of derivatives. We motivate and define the derivative and compute a few examples, then discuss how features of a function are manifested in its derivative.
Lesson 7-8: Derivatives and Rates of Change, The Derivative as a functionMatthew Leingang
The derivative is one of the fundamental quantities in calculus, partly because it is ubiquitous in nature. We give examples of it coming about, a few calculations, and ways information about the function an imply information about the derivative
The document discusses key concepts in calculus including functions, limits, derivatives, and derivatives of trigonometric functions. It provides examples of calculating derivatives from first principles using the definition of the derivative and common derivative rules like the product rule and quotient rule. Formulas are also derived for the derivatives of the sine, cosine, and tangent functions.
This document provides an introduction to derivatives. It has four main goals: 1) introduce risk and the role of derivatives in managing risk, 2) discuss general finance terms, 3) introduce three major classes of derivatives (forwards, futures, options), and 4) introduce how these securities are analyzed and the major traders. It defines derivatives as financial instruments whose value depends on underlying variables. The document discusses the basics of risk, positions (long/short), commissions, bid-ask spreads, and market efficiency. It provides an example of Disney using earthquake bonds to transfer risk from shareholders to bondholders.
The document outlines topics to be covered in Sections 2.1-2.2 of a Calculus I course, including defining the derivative, finding derivatives of functions, relating the graph of a function to its derivative, and finding tangent lines. It provides examples of how derivatives can model real-world rates of change and outlines the objectives to understand the definition of the derivative and how to apply it to functions.
The document is from a Calculus I course at New York University and covers the topic of the derivative and rates of change. It discusses finding the slope of the tangent line to a curve at a given point, using the example of finding the slope of the tangent line to the curve y=x^2 at the point (2,4). It then shows this problem solved graphically and numerically by calculating the limit of the difference quotient as Δx approaches 0.
The document outlines lecture notes for Calculus I on the topic of the derivative and rates of change. It includes objectives for Sections 2.1 and 2.2 such as understanding the definition of the derivative and using it to find derivatives of functions. It also provides examples of how derivatives can model real-world phenomena like velocity, population growth, and marginal costs. The notes give an outline of topics to be covered including tangent lines, various notations for derivatives, and how to find second derivatives.
This document contains notes from a Calculus I class lecture on the derivative. The lecture covered the definition of the derivative and examples of how it can be used to model rates of change in various contexts like velocity, population growth, and marginal costs. It also discussed properties of the derivative like how the derivative of a function relates to whether the function is increasing or decreasing over an interval.
The derivative is a major tool for investigating the behavior of a function. Since functions are ubiquitous, so are their derivatives. Velocity, growth rates, marginal costs, and material strain are all examples of derivatives. We motivate and define the derivative and compute a few examples, then discuss how features of a function are manifested in its derivative.
Lesson 7-8: Derivatives and Rates of Change, The Derivative as a functionMatthew Leingang
The derivative is one of the fundamental quantities in calculus, partly because it is ubiquitous in nature. We give examples of it coming about, a few calculations, and ways information about the function an imply information about the derivative
The document discusses key concepts in calculus including functions, limits, derivatives, and derivatives of trigonometric functions. It provides examples of calculating derivatives from first principles using the definition of the derivative and common derivative rules like the product rule and quotient rule. Formulas are also derived for the derivatives of the sine, cosine, and tangent functions.
This document provides an introduction to derivatives. It has four main goals: 1) introduce risk and the role of derivatives in managing risk, 2) discuss general finance terms, 3) introduce three major classes of derivatives (forwards, futures, options), and 4) introduce how these securities are analyzed and the major traders. It defines derivatives as financial instruments whose value depends on underlying variables. The document discusses the basics of risk, positions (long/short), commissions, bid-ask spreads, and market efficiency. It provides an example of Disney using earthquake bonds to transfer risk from shareholders to bondholders.
The document discusses the concept of derivative as a rate of change. It defines the average rate of change and instantaneous rate of change over an interval and as the limit as the interval approaches zero respectively. It provides examples of calculating average and instantaneous rates of change from graphs and data tables.
This document provides questions about finding the equations of tangents and normals to various polynomial functions at given points. It contains 8 parts with multiple questions each about finding the equations of tangents to polynomial curves and normals to polynomial curves at specified points using differentiation.
1) The document discusses the concept of congruence in geometry, specifically how congruence relates to triangles. It states that two triangles are congruent if specifying two sides and the angle between them (SAS), two angles and the side between them (ASA), or two angles and a corresponding adjacent side (AAS).
2) It also discusses properties of congruence, including that corresponding parts of congruent triangles (sides and angles) are congruent to each other.
3) The document provides examples justifying statements about angles and sides of triangles using properties of equality and congruence.
This document contains lecture notes on derivatives from a Calculus I class at New York University. It discusses the derivative as a function, finding the derivative of other functions, and the relationship between a function and its derivative. The notes include examples of finding the derivative of the reciprocal function and state that if a function is decreasing on an interval, its derivative will be nonpositive on that interval, while if it is increasing the derivative will be nonnegative. It also contains proofs and graphs related to derivatives.
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.
To summarize:
1) To find the derivative of an implicit function y=y(x) defined by an equation F(x,y)=0, take the derivative of both sides with respect to x.
2) This will give a new equation involving x, y, and dy/dx that can be solved for dy/dx.
3) The examples show applying this process to find derivatives and tangent lines for various implicit equations.
The document discusses vector spaces and related concepts:
1) It defines a vector space as a set V with vector addition and scalar multiplication operations that satisfy certain properties. Examples of vector spaces include R2, the plane in R3, and the space of real polynomials.
2) A subspace is a subset of a vector space that is closed under vector addition and scalar multiplication and thus forms a vector space with the inherited operations. Examples given include the x-axis in Rn and solution spaces of linear differential equations.
3) The span of a set of vectors is the smallest subspace that contains those vectors, consisting of all possible linear combinations of the vectors in the set.
Lesson 20: Derivatives and the Shapes of Curves (slides)Matthew Leingang
This document contains lecture notes on derivatives and the shapes of curves from a Calculus I class taught by Professor Matthew Leingang at New York University. The notes cover using derivatives to determine the intervals where a function is increasing or decreasing, classifying critical points as maxima or minima, using the second derivative to determine concavity, and applying the first and second derivative tests. Examples are provided to illustrate finding intervals of monotonicity for various functions.
The document explains the chain rule, which provides a method for finding the derivative of a composite function. The chain rule states that the derivative of a composite function f(g(x)) is equal to the derivative of the outside function f'(g(x)) multiplied by the derivative of the inside function g'(x). This allows the calculation of derivatives of more complex functions that cannot be solved using basic derivative rules. Several examples are provided to demonstrate how to use the chain rule to calculate derivatives of various composite functions.
ANURAG TYAGI CLASSES (ATC) is an organisation destined to orient students into correct path to achieve
success in IIT-JEE, AIEEE, PMT, CBSE & ICSE board classes. The organisation is run by a competitive staff comprising of Ex-IITians. Our goal at ATC is to create an environment that inspires students to recognise and explore their own potentials and build up confidence in themselves.ATC was founded by Mr. ANURAG TYAGI on 19 march, 2001.
MEET US AT:
www.anuragtyagiclasses.com
the derivative measure the instantaneous rate of change of a function, or the slope of the line tangent to its graph. It has countless applications.
[Note: We did not do this entire show in class. We will finish it on Thursday]
The document discusses the chain rule, which is used to find the derivative of composite functions. It provides examples of applying the chain rule to functions of the form f(g(x)) by taking the derivative of the outside function with respect to the inside function, and multiplying by the derivative of the inside function with respect to x. The chain rule can be used repeatedly when a function is composed of multiple nested functions. Derivative formulas themselves incorporate the chain rule. The chain rule is essential for finding derivatives and is the most common mistake made by students on tests.
The document discusses higher order derivatives. It defines the nth derivative of a function f(x) as f(n)(x). The first example finds the first five derivatives of f(x)=2x^4 - x^3 - 2. The second example finds the first three derivatives of f(x)=-x^2/3. The third example finds the first four derivatives of f(x)=ln(x) and discusses how derivatives of rational functions become more complicated with higher orders. It also provides examples of finding derivatives of other functions like sin(x).
This document is a section from a Calculus I course at New York University that discusses basic differentiation rules. It provides objectives for understanding rules like the derivative of a constant function, the constant multiple rule, sum rule, and derivatives of sine and cosine functions. It then gives examples of finding the derivative of a squaring function using the definition, and introduces the concept of the second derivative.
The document provides rules to determine if a number is divisible by certain integers from 2 to 12. It gives examples of numbers that satisfy each rule and the reasoning for why they are divisible. The key points are:
1) A number is divisible by 2 if the last digit is even
2) A number is divisible by 3 if the sum of its digits is divisible by 3
3) A number is divisible by 4 if the last two digits form a number divisible by 4
4) A number is divisible by 5 if it ends in 0 or 5
5) A number is divisible by 6 if it is divisible by both 2 and 3
6) A number is divisible by 8 if the last three digits form
This document discusses the chain rule for functions of multiple variables. It begins by reviewing the chain rule for single-variable functions, then extends it to functions of more variables. The chain rule is presented for cases where the dependent variable z is a function of intermediate variables x and y, which are themselves functions of independent variables s and t. General formulas are given using partial derivatives. Examples are worked out, such as finding the derivative of a function defined implicitly by an equation. Diagrams are used to illustrate the relationships between variables.
The document discusses functions and their derivatives. It defines functions, different types of functions, and notation used for functions. It then covers the concept of limits, theorems on limits, and limits at infinity. The document defines the slope of a tangent line to a curve and increments. It provides definitions and rules for derivatives, including differentiation from first principles and various differentiation rules. It includes examples of finding derivatives using these rules and taking multiple derivatives.
Trigonometric functions are used extensively in calculus. When using trig functions in calculus, radian measure must be used for angles. Even trig functions like cosine are symmetric about the y-axis, while odd functions like sine change sign when x changes sign. Trig functions can be shifted, stretched, or shrunk by applying transformations to their graphs.
This document provides a summary of a stock trading project report on Nirmal Bang Securities Pvt Ltd, an online stock trading company in India.
1) It outlines the objectives of studying Nirmal Bang's marketing strategies and online trading platform, as well as improving various aspects of their services.
2) The research methodology section describes using primary data collection through interviews with brokers, agents and investors, as well as secondary data. 100 samples were obtained through judgement sampling.
3) The findings note areas for improvement like increasing exposure for distributors' clients, reducing high brokerage charges, and providing more promotional offers to attract new customers.
The document outlines topics to be covered in Calculus I class sessions on the derivative and rates of change, including: defining the derivative at a point and using it to find the slope of the tangent line to a curve at that point; examples of derivatives modeling rates of change; and how to find the derivative function and second derivative of a given function. It provides learning objectives, an outline of topics, and an example problem worked out graphically and numerically to illustrate finding the slope of the tangent line.
The document contains lecture notes for a Calculus I class covering Sections 2.1-2.2 on the derivative. It includes announcements about upcoming quizzes and assignments. The notes cover objectives like defining the derivative, relating it to rates of change, examples like velocity and population growth, and the formal definition of the derivative. Formulas are given for instantaneous rates of change like velocity, growth, and marginal cost in terms of limits and the derivative.
The document discusses the concept of derivative as a rate of change. It defines the average rate of change and instantaneous rate of change over an interval and as the limit as the interval approaches zero respectively. It provides examples of calculating average and instantaneous rates of change from graphs and data tables.
This document provides questions about finding the equations of tangents and normals to various polynomial functions at given points. It contains 8 parts with multiple questions each about finding the equations of tangents to polynomial curves and normals to polynomial curves at specified points using differentiation.
1) The document discusses the concept of congruence in geometry, specifically how congruence relates to triangles. It states that two triangles are congruent if specifying two sides and the angle between them (SAS), two angles and the side between them (ASA), or two angles and a corresponding adjacent side (AAS).
2) It also discusses properties of congruence, including that corresponding parts of congruent triangles (sides and angles) are congruent to each other.
3) The document provides examples justifying statements about angles and sides of triangles using properties of equality and congruence.
This document contains lecture notes on derivatives from a Calculus I class at New York University. It discusses the derivative as a function, finding the derivative of other functions, and the relationship between a function and its derivative. The notes include examples of finding the derivative of the reciprocal function and state that if a function is decreasing on an interval, its derivative will be nonpositive on that interval, while if it is increasing the derivative will be nonnegative. It also contains proofs and graphs related to derivatives.
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.
To summarize:
1) To find the derivative of an implicit function y=y(x) defined by an equation F(x,y)=0, take the derivative of both sides with respect to x.
2) This will give a new equation involving x, y, and dy/dx that can be solved for dy/dx.
3) The examples show applying this process to find derivatives and tangent lines for various implicit equations.
The document discusses vector spaces and related concepts:
1) It defines a vector space as a set V with vector addition and scalar multiplication operations that satisfy certain properties. Examples of vector spaces include R2, the plane in R3, and the space of real polynomials.
2) A subspace is a subset of a vector space that is closed under vector addition and scalar multiplication and thus forms a vector space with the inherited operations. Examples given include the x-axis in Rn and solution spaces of linear differential equations.
3) The span of a set of vectors is the smallest subspace that contains those vectors, consisting of all possible linear combinations of the vectors in the set.
Lesson 20: Derivatives and the Shapes of Curves (slides)Matthew Leingang
This document contains lecture notes on derivatives and the shapes of curves from a Calculus I class taught by Professor Matthew Leingang at New York University. The notes cover using derivatives to determine the intervals where a function is increasing or decreasing, classifying critical points as maxima or minima, using the second derivative to determine concavity, and applying the first and second derivative tests. Examples are provided to illustrate finding intervals of monotonicity for various functions.
The document explains the chain rule, which provides a method for finding the derivative of a composite function. The chain rule states that the derivative of a composite function f(g(x)) is equal to the derivative of the outside function f'(g(x)) multiplied by the derivative of the inside function g'(x). This allows the calculation of derivatives of more complex functions that cannot be solved using basic derivative rules. Several examples are provided to demonstrate how to use the chain rule to calculate derivatives of various composite functions.
ANURAG TYAGI CLASSES (ATC) is an organisation destined to orient students into correct path to achieve
success in IIT-JEE, AIEEE, PMT, CBSE & ICSE board classes. The organisation is run by a competitive staff comprising of Ex-IITians. Our goal at ATC is to create an environment that inspires students to recognise and explore their own potentials and build up confidence in themselves.ATC was founded by Mr. ANURAG TYAGI on 19 march, 2001.
MEET US AT:
www.anuragtyagiclasses.com
the derivative measure the instantaneous rate of change of a function, or the slope of the line tangent to its graph. It has countless applications.
[Note: We did not do this entire show in class. We will finish it on Thursday]
The document discusses the chain rule, which is used to find the derivative of composite functions. It provides examples of applying the chain rule to functions of the form f(g(x)) by taking the derivative of the outside function with respect to the inside function, and multiplying by the derivative of the inside function with respect to x. The chain rule can be used repeatedly when a function is composed of multiple nested functions. Derivative formulas themselves incorporate the chain rule. The chain rule is essential for finding derivatives and is the most common mistake made by students on tests.
The document discusses higher order derivatives. It defines the nth derivative of a function f(x) as f(n)(x). The first example finds the first five derivatives of f(x)=2x^4 - x^3 - 2. The second example finds the first three derivatives of f(x)=-x^2/3. The third example finds the first four derivatives of f(x)=ln(x) and discusses how derivatives of rational functions become more complicated with higher orders. It also provides examples of finding derivatives of other functions like sin(x).
This document is a section from a Calculus I course at New York University that discusses basic differentiation rules. It provides objectives for understanding rules like the derivative of a constant function, the constant multiple rule, sum rule, and derivatives of sine and cosine functions. It then gives examples of finding the derivative of a squaring function using the definition, and introduces the concept of the second derivative.
The document provides rules to determine if a number is divisible by certain integers from 2 to 12. It gives examples of numbers that satisfy each rule and the reasoning for why they are divisible. The key points are:
1) A number is divisible by 2 if the last digit is even
2) A number is divisible by 3 if the sum of its digits is divisible by 3
3) A number is divisible by 4 if the last two digits form a number divisible by 4
4) A number is divisible by 5 if it ends in 0 or 5
5) A number is divisible by 6 if it is divisible by both 2 and 3
6) A number is divisible by 8 if the last three digits form
This document discusses the chain rule for functions of multiple variables. It begins by reviewing the chain rule for single-variable functions, then extends it to functions of more variables. The chain rule is presented for cases where the dependent variable z is a function of intermediate variables x and y, which are themselves functions of independent variables s and t. General formulas are given using partial derivatives. Examples are worked out, such as finding the derivative of a function defined implicitly by an equation. Diagrams are used to illustrate the relationships between variables.
The document discusses functions and their derivatives. It defines functions, different types of functions, and notation used for functions. It then covers the concept of limits, theorems on limits, and limits at infinity. The document defines the slope of a tangent line to a curve and increments. It provides definitions and rules for derivatives, including differentiation from first principles and various differentiation rules. It includes examples of finding derivatives using these rules and taking multiple derivatives.
Trigonometric functions are used extensively in calculus. When using trig functions in calculus, radian measure must be used for angles. Even trig functions like cosine are symmetric about the y-axis, while odd functions like sine change sign when x changes sign. Trig functions can be shifted, stretched, or shrunk by applying transformations to their graphs.
This document provides a summary of a stock trading project report on Nirmal Bang Securities Pvt Ltd, an online stock trading company in India.
1) It outlines the objectives of studying Nirmal Bang's marketing strategies and online trading platform, as well as improving various aspects of their services.
2) The research methodology section describes using primary data collection through interviews with brokers, agents and investors, as well as secondary data. 100 samples were obtained through judgement sampling.
3) The findings note areas for improvement like increasing exposure for distributors' clients, reducing high brokerage charges, and providing more promotional offers to attract new customers.
The document outlines topics to be covered in Calculus I class sessions on the derivative and rates of change, including: defining the derivative at a point and using it to find the slope of the tangent line to a curve at that point; examples of derivatives modeling rates of change; and how to find the derivative function and second derivative of a given function. It provides learning objectives, an outline of topics, and an example problem worked out graphically and numerically to illustrate finding the slope of the tangent line.
The document contains lecture notes for a Calculus I class covering Sections 2.1-2.2 on the derivative. It includes announcements about upcoming quizzes and assignments. The notes cover objectives like defining the derivative, relating it to rates of change, examples like velocity and population growth, and the formal definition of the derivative. Formulas are given for instantaneous rates of change like velocity, growth, and marginal cost in terms of limits and the derivative.
This document is a section from a Calculus I course at New York University that discusses basic differentiation rules. It covers objectives like differentiating constant, sum, and difference functions. It also reviews derivatives of sine and cosine. Examples are provided, like finding the derivative of a squaring function x^2 using the definition of a derivative. The document outlines the topics and provides explanations and step-by-step solutions.
This document contains lecture notes on differentiation rules from a Calculus I class at New York University. It begins with objectives to understand basic differentiation rules like the derivative of a constant function, the Constant Multiple Rule, the Sum Rule, and derivatives of sine and cosine. It then provides examples of using the definition of the derivative to find the derivatives of squaring and cubing functions. It illustrates the functions and their derivatives on graphs and discusses properties like a function being increasing when its derivative is positive.
This document is a section from a Calculus I course at New York University that discusses basic differentiation rules. It covers objectives like differentiating constant, sum, and difference functions. It also covers derivatives of sine and cosine. Examples are provided, like finding the derivative of the squaring function x^2, which is 2x. Notation for derivatives is explained, including Leibniz notation. The concept of the second derivative is also introduced.
This document contains lecture notes on basic differentiation rules from a Calculus I course at New York University. It begins with announcements about an extra credit opportunity. The objectives and outline describe rules that will be covered, including the derivatives of constant, sum, difference, sine and cosine functions. Examples are provided to derive the derivatives of square, cube, square root, cube root and other power functions using the definition of the derivative. The Power Rule is stated and explained using concepts like Pascal's triangle.
Lesson 16: Inverse Trigonometric Functions (Section 041 slides)Mel Anthony Pepito
The document is a lecture note on inverse trigonometric functions from a Calculus I class at New York University. It defines inverse trigonometric functions like arcsin, arccos, arctan and gives their domains, ranges and other properties. It also provides examples of calculating the values of inverse trig functions like arcsin(1/2) = π/6 and arctan(-1) = -π/4.
Lesson20 -derivatives_and_the_shape_of_curves_021_slidesMel Anthony Pepito
f is decreasing on (−∞, −4/5] and increasing on [−4/5, ∞).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
This document discusses the definite integral and its properties. It begins by stating the objectives of computing definite integrals using Riemann sums and limits, estimating integrals using approximations like the midpoint rule, and reasoning about integrals using their properties. The outline then reviews the integral as a limit of Riemann sums and how to estimate integrals. It also discusses properties of the integral and comparison properties. Finally, it restates the theorem that if a function is continuous, the limit of Riemann sums is the same regardless of the choice of sample points.
The document discusses the definite integral, including computing it using Riemann sums, estimating it using approximations like the midpoint rule, and reasoning about its properties. It outlines the topics to be covered, such as recalling previous concepts and comparing properties of integrals. Formulas are provided for calculating Riemann sums using different representative points within the intervals.
This document outlines the key rules for differentiation that will be covered in Calculus I class. It introduces the objectives of understanding derivatives of constant functions, the constant multiple rule, sum and difference rules, and derivatives of sine and cosine. It then provides examples of finding the derivatives of squaring and cubing functions using the definition of a derivative. Finally, it discusses properties of the derivatives of these functions.
The document is a lecture note from an NYU Calculus I class covering the definite integral. It provides announcements about upcoming quizzes and exams. The content discusses computing definite integrals using Riemann sums and the limit of Riemann sums, as well as properties of the definite integral. Examples are given of calculating Riemann sums using different representative points in each interval.
Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)Mel Anthony Pepito
Describe the monotonicity of f(x) = arctan(x).
The derivative of arctan(x) is f'(x) = 1/(1+x^2), which is always positive. Therefore, f(x) = arctan(x) is an increasing function on (-∞,∞).
Lesson 19: The Mean Value Theorem (Section 021 slides)Mel Anthony Pepito
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.
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.
The document provides information about an upcoming calculus exam, quiz, and lecture on curve sketching. It outlines the procedure for sketching curves, including using the increasing/decreasing test and concavity test. It then provides an example of sketching a cubic function, showing the steps of finding critical points, inflection points, and putting together a sign chart to determine the function's monotonicity and concavity over intervals.
Similar to Lesson 7: The Derivative (Section 21 slide) (20)
This document provides guidance on developing effective lesson plans for calculus instructors. It recommends starting by defining specific learning objectives and assessments. Examples should be chosen carefully to illustrate concepts and engage students at a variety of levels. The lesson plan should include an introductory problem, definitions, theorems, examples, and group work. Timing for each section should be estimated. After teaching, the lesson can be improved by analyzing what was effective and what needs adjustment for the next time. Advanced preparation is key to looking prepared and ensuring students learn.
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.
This document discusses electronic grading of paper assessments using PDF forms. Key points include:
- Various tools for creating fillable PDF forms using LaTeX packages or desktop software.
- Methods for stamping completed forms onto scanned documents including using pdftk or overlaying in TikZ.
- Options for grading on tablets or desktops including GoodReader, PDFExpert, Adobe Acrobat.
- Extracting data from completed forms can be done in Adobe Acrobat or via command line with pdftk.
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 26: The Fundamental Theorem of Calculus (slides)Matthew Leingang
g(x) represents the area under the curve of f(t) between 0 and x.
.
x
What can you say about g? 2 4 6 8 10f
The First Fundamental Theorem of Calculus
Theorem (First Fundamental Theorem of Calculus)
Let f be a con nuous func on on [a, b]. Define the func on F on [a, b] by
∫ x
F(x) = f(t) dt
a
Then F is con nuous on [a, b] and differentiable on (a, b) and for all x in (a, b),
F′(x
Lesson 26: The Fundamental Theorem of Calculus (slides)Matthew Leingang
The document discusses the Fundamental Theorem of Calculus, which has two parts. The first part states that if a function f is continuous on an interval, then the derivative of the integral of f is equal to f. This is proven using Riemann sums. The second part relates the integral of a function f to the integral of its derivative F'. Examples are provided to illustrate how the area under a curve relates to these concepts.
Lesson 27: Integration by Substitution (handout)Matthew Leingang
This document contains lecture notes on integration by substitution from a Calculus I class. It introduces the technique of substitution for both indefinite and definite integrals. For indefinite integrals, the substitution rule is presented, along with examples of using substitutions to evaluate integrals involving polynomials, trigonometric, exponential, and other functions. For definite integrals, the substitution rule is extended and examples are worked through both with and without first finding the indefinite integral. The document emphasizes that substitution often simplifies integrals and makes them easier to evaluate.
Lesson 26: The Fundamental Theorem of Calculus (handout)Matthew Leingang
1) The document discusses lecture notes on Section 5.4: The Fundamental Theorem of Calculus from a Calculus I course. 2) It covers stating and explaining the Fundamental Theorems of Calculus and using the first fundamental theorem to find derivatives of functions defined by integrals. 3) The lecture outlines the first fundamental theorem, which relates differentiation and integration, and gives examples of applying it.
This document contains notes from a calculus class lecture on evaluating definite integrals. It discusses using the evaluation theorem to evaluate definite integrals, writing derivatives as indefinite integrals, and interpreting definite integrals as the net change of a function over an interval. The document also contains examples of evaluating definite integrals, properties of integrals, and an outline of the key topics covered.
This document contains lecture notes from a Calculus I class covering Section 5.3 on evaluating definite integrals. The notes discuss using the Evaluation Theorem to calculate definite integrals, writing derivatives as indefinite integrals, and interpreting definite integrals as the net change of a function over an interval. Examples are provided to demonstrate evaluating definite integrals using the midpoint rule approximation. Properties of integrals such as additivity and the relationship between definite and indefinite integrals are also outlined.
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.
This document contains lecture notes from a Calculus I class discussing optimization problems. It begins with announcements about upcoming exams and courses the professor is teaching. It then presents an example problem about finding the rectangle of a fixed perimeter with the maximum area. The solution uses calculus techniques like taking the derivative to find the critical points and determine that the optimal rectangle is a square. The notes discuss strategies for solving optimization problems and summarize the key steps to take.
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.
The document discusses curve sketching of functions by analyzing their derivatives. It provides:
1) A checklist for graphing a function which involves finding where the function is positive/negative/zero, its monotonicity from the first derivative, and concavity from the second derivative.
2) An example of graphing the cubic function f(x) = 2x^3 - 3x^2 - 12x through analyzing its derivatives.
3) Explanations of the increasing/decreasing test and concavity test to determine monotonicity and concavity from a function's derivatives.
The document contains lecture notes on curve sketching from a Calculus I class. It discusses using the first and second derivative tests to determine properties of a function like monotonicity, concavity, maxima, minima, and points of inflection in order to sketch the graph of the function. It then provides an example of using these tests to sketch the graph of the cubic function f(x) = 2x^3 - 3x^2 - 12x.
Lesson 20: Derivatives and the Shapes of Curves (handout)Matthew Leingang
This document contains lecture notes on calculus from a Calculus I course. It covers determining the monotonicity of functions using the first derivative test. Key points include using the sign of the derivative to determine if a function is increasing or decreasing over an interval, and using the first derivative test to classify critical points as local maxima, minima, or neither. Examples are provided to demonstrate finding intervals of monotonicity for various functions and applying the first derivative test.
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.
1. Section 2.1–2.2
The Derivative and Rates of Change
The Derivative as a Function
V63.0121.021, Calculus I
New York University
September 28, 2010
Announcements
Quiz this week in recitation on §§1.1–1.4
Get-to-know-you/photo due Friday October 1
. . . . . .
2. Announcements
Quiz this week in recitation
on §§1.1–1.4
Get-to-know-you/photo
due Friday October 1
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 2 / 49
3. Format of written work
Please:
Use scratch paper and
copy your final work onto
fresh paper.
Use loose-leaf paper (not
torn from a notebook).
Write your name, lecture
section, assignment
number, recitation, and
date at the top.
Staple your homework
together.
See the website for more information.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 3 / 49
4. Objectives for Section 2.1
Understand and state the
definition of the derivative
of a function at a point.
Given a function and a
point in its domain, decide
if the function is
differentiable at the point
and find the value of the
derivative at that point.
Understand and give
several examples of
derivatives modeling rates
of change in science.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 4 / 49
5. Objectives for Section 2.2
Given a function f, use the
definition of the derivative
to find the derivative
function f’.
Given a function, find its
second derivative.
Given the graph of a
function, sketch the graph
of its derivative.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 5 / 49
6. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 6 / 49
7. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
8. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x2 at the point (2, 4).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 7 / 49
9. Graphically and numerically
y
.
x2 − 22
x m=
x−2
. .
4 .
. . x
.
2
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
10. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3
. .
9 .
. .
4 .
. . . x
.
2
. 3
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
11. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
. .
9 .
. .
4 .
. . . x
.
2
. 3
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
12. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5
. .25 .
6 .
. .
4 .
. . . x
.
22
. . .5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
13. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
. .25 .
6 .
. .
4 .
. . . x
.
22
. . .5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
14. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1
. .41 .
4 .
. .
4 .
. .. x
.
2
.. .1
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
15. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
. .41 .
4 .
. .
4 .
. .. x
.
2
.. .1
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
16. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01
. .0401 .
4 4
. .
. . x
.
2.
. .01
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
17. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .0401 .
4 4
. .
. . x
.
2.
. .01
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
18. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
1
. .
1 .
. . . x
.
1
. 2
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
19. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
1 3
. .
1 .
. . . x
.
1
. 2
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
20. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
1.5
. .25 .
2 .
1 3
. . . x
.
1 2
. .5 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
21. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
1.5 3.5
. .25 .
2 .
1 3
. . . x
.
1 2
. .5 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
22. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
. .61 .
3 . 1.9
1.5 3.5
1 3
. .. x
.
1.
. .9
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
23. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
. .
4 .
. .61 .
3 . 1.9 3.9
1.5 3.5
1 3
. .. x
.
1.
. .9
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
24. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
1.99
. .9601 .
3 4
. .
1.9 3.9
1.5 3.5
1 3
. . x
.
1.
. .99
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
25. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
1.99 3.99
. .9601 .
3 4
. .
1.9 3.9
1.5 3.5
1 3
. . x
.
1.
. .99
2
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
26. Graphically and numerically
y
.
x2 − 22
x m=
x−2
3 5
2.5 4.5
2.1 4.1
2.01 4.01
1.99 3.99
. .
4 .
1.9 3.9
1.5 3.5
1 3
. . x
.
2
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 8 / 49
29. The tangent problem
Problem
Given a curve and a point on the curve, find the slope of the line
tangent to the curve at that point.
Example
Find the slope of the line tangent to the curve y = x2 at the point (2, 4).
Upshot
If the curve is given by y = f(x), and the point on the curve is (a, f(a)),
then the slope of the tangent line is given by
f(x) − f(a)
mtangent = lim
x→a x−a
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 9 / 49
30. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 10 / 49
31. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
32. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
33. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
34. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
35. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
36. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
37. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
38. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
39. Numerical evidence
h(t) = 50 − 5t2
Fill in the table:
h(t) − h(1)
t vave =
t−1
2 − 15
1.5 − 12.5
1.1 − 10.5
1.01 − 10.05
1.001 − 10.005
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 11 / 49
40. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45
v = lim
t→1 t−1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
41. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2
v = lim = lim
t→1 t−1 t→1 t − 1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
42. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
43. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
= (−5) lim (1 + t)
t→1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
44. Velocity
.
Problem
Given the position function of a moving object, find the velocity of the object at
a certain instant in time.
Example
Drop a ball off the roof of the Silver Center so that its height can be described
by
h(t) = 50 − 5t2
where t is seconds after dropping it and h is meters above the ground. How
fast is it falling one second after we drop it?
Solution
The answer is
(50 − 5t2 ) − 45 5 − 5t2 5(1 − t)(1 + t)
v = lim = lim = lim
t→1 t−1 t→1 t − 1 t→1 t−1
= (−5) lim (1 + t) = −5 · 2 = −10
t→1
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 12 / 49
45. Velocity in general
. . = h(t)
y
Upshot
. (t0 ) .
h .
If the height function is given by
h(t), the instantaneous velocity . h
∆
at time t0 is given by
. (t0 + ∆t) .
h .
h(t) − h(t0 )
v = lim
t→t0 t − t0
h(t0 + ∆t) − h(t0 )
= lim
∆t→0 ∆t
. . . t .
∆ ..
t
t
.0 t
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 13 / 49
46. Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
47. Population growth
Problem
Given the population function of a group of organisms, find the rate of
growth of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the function
3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish
population growing fastest in 1990, 2000, or 2010? (Estimate
numerically)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 14 / 49
48. Derivation
Solution
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so ideally we would want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
49. Derivation
Solution
Let ∆t be an increment in time and ∆P the corresponding change in
population:
∆P = P(t + ∆t) − P(t)
This depends on ∆t, so ideally we would want
( )
∆P 1 3et+∆t 3et
lim = lim −
∆t→0 ∆t ∆t→0 ∆t 1 + et+∆t 1 + et
But rather than compute a complicated limit analytically, let us
approximate numerically. We will try a small ∆t, for instance 0.1.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 15 / 49
50. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
r1990
r2000
r2010
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
51. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
P(−10 + 0.1) − P(−10)
r1990 ≈
0.1
P(0.1) − P(0)
r2000 ≈
0.1
P(10 + 0.1) − P(10)
r2010 ≈
0.1
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
52. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
53. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
54. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
= 0.749376
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
55. Numerical evidence
Solution (Continued)
To approximate the population change in year n, use the difference
P(t + ∆t) − P(t)
quotient , where ∆t = 0.1 and t = n − 2000.
∆t
( )
P(−10 + 0.1) − P(−10) 1 3e−9.9 3e−10
r1990 ≈ = −
0.1 0.1 1 + e−9.9 1 + e−10
= 0.000143229
( )
P(0.1) − P(0) 1 3e0.1 3e0
r2000 ≈ = −
0.1 0.1 1 + e0.1 1 + e0
= 0.749376
( )
P(10 + 0.1) − P(10) 1 3e10.1 3e10
r2010 ≈ = −
0.1 0.1 1 + e10.1 1 + e10
= 0.0001296
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 16 / 49
56. Population growth
.
Problem
Given the population function of a group of organisms, find the rate of growth
of the population at a particular instant.
Example
Suppose the population of fish in the East River is given by the function
3et
P(t) =
1 + et
where t is in years since 2000 and P is in millions of fish. Is the fish population
growing fastest in 1990, 2000, or 2010? (Estimate numerically)
Answer
We estimate the rates of growth to be 0.000143229, 0.749376, and 0.0001296.
So the population is growing fastest in 2000.
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 17 / 49
57. Population growth in general
Upshot
The instantaneous population growth is given by
P(t + ∆t) − P(t)
lim
∆t→0 ∆t
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 18 / 49
58. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
59. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a year is
C(q) = q3 − 12q2 + 60q
We are currently producing 5 tons a year. Should we change that?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 19 / 49
60. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4
5
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
61. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
62. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5 125
6
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
63. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q)
4 112
5 125
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
64. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112
5 125
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
65. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
66. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
67. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q
4 112 28
5 125 25
6 144 24
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
68. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28
5 125 25
6 144 24
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
69. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25
6 144 24
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
70. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25 19
6 144 24
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
71. Comparisons
Solution
C(q) = q3 − 12q2 + 60q
Fill in the table:
q C(q) AC(q) = C(q)/q ∆C = C(q + 1) − C(q)
4 112 28 13
5 125 25 19
6 144 24 31
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 20 / 49
72. Marginal costs
Problem
Given the production cost of a good, find the marginal cost of
production after having produced a certain quantity.
Example
Suppose the cost of producing q tons of rice on our paddy in a year is
C(q) = q3 − 12q2 + 60q
We are currently producing 5 tons a year. Should we change that?
Answer
If q = 5, then C = 125, ∆C = 19, while AC = 25. So we should
produce more to lower average costs.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 21 / 49
73. Marginal Cost in General
Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but is still only an average rate of change.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
74. Marginal Cost in General
Upshot
The incremental cost
∆C = C(q + 1) − C(q)
is useful, but is still only an average rate of change.
The marginal cost after producing q given by
C(q + ∆q) − C(q)
MC = lim
∆q→0 ∆q
is more useful since it’s an instantaneous rate of change.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 22 / 49
75. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 23 / 49
76. The definition
All of these rates of change are found the same way!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
77. The definition
All of these rates of change are found the same way!
Definition
Let f be a function 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 function is said to be differentiable at a and f′ (a) is the
derivative of f at a.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 24 / 49
78. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
79. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a)
f′ (a) = lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
80. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
81. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2
= lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
82. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
83. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h)
h→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
84. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (a).
Solution
f(a + h) − f(a) (a + h)2 − a2
f′ (a) = lim = lim
h→0 h h→0 h
(a2 + 2ah + h2 ) − a2 2ah + h2
= lim = lim
h→0 h h→0 h
= lim (2a + h) = 2a
h→0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 25 / 49
85. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
86. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
87. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
88. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1
= lim
x→2 2x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
89. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 26 / 49
90. “Can you do it the other way?"
Same limit, different form
Solution
f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
91. “Can you do it the other way?"
Same limit, different form
Solution
f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h)
= lim
h→0 2h(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
92. “Can you do it the other way?"
Same limit, different form
Solution
f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
93. “Can you do it the other way?"
Same limit, different form
Solution
f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
−1
= lim
h→0 2(2 + h)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
94. “Can you do it the other way?"
Same limit, different form
Solution
f(2 + h) − f(2)
f′ (2) = lim
h→0 h
1
−1
= lim 2+h 2
h→0 h
2 − (2 + h) −h
= lim = lim
h→0 2h(2 + h) h→0 2h(2 + h)
−1 1
= lim =−
h→0 2(2 + h) 4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 27 / 49
95. “How did you get that?"
The Sure-Fire Sally Rule (SFSR) for adding Fractions
Fact
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x−2 x−2
2−x
=
2x(x − 2)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
96. “How did you get that?"
The Sure-Fire Sally Rule (SFSR) for adding Fractions
Fact
a c ad ± bc
± =
b d bd
So
1 1 2−x
−
x 2 = 2x
x−2 x−2
2−x
=
2x(x − 2)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 28 / 49
97. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each point
x where f is differentiable, and come up with another function, the
derivative function.
What can we say about this function f′ ?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
98. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each point
x where f is differentiable, and come up with another function, the
derivative function.
What can we say about this function f′ ?
If f is decreasing on an interval, f′ is negative (technically,
nonpositive) on that interval
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 29 / 49
99. Derivative of the reciprocal function
Example
1
Suppose f(x) = . Use the
x
definition of the derivative to
find f′ (2). x
.
Solution
1/x − 1/2
f′ (2) = lim .
x→2 x−2 . x
.
2−x
= lim
x→2 2x(x − 2)
−1 1
= lim =−
x→2 2x 4
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 30 / 49
100. What does f tell you about f′ ?
If f is a function, we can compute the derivative f′ (x) at each point
x where f is differentiable, and come up with another function, the
derivative function.
What can we say about this function f′ ?
If f is decreasing on an interval, f′ is negative (technically,
nonpositive) on that interval
If f is increasing on an interval, f′ is positive (technically,
nonnegative) on that interval
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 31 / 49
102. What does f tell you about f′ ?
Fact
If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
103. What does f tell you about f′ ?
Fact
If f is decreasing on the open interval (a, b), then f′ ≤ 0 on (a, b).
Picture Proof. .
y
.
If f is decreasing, then all
secant lines point downward,
hence have negative slope.
The derivative is a limit of
slopes of secant lines, which
are all negative, so the limit .
must be ≤ 0. . .
.
.
x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 33 / 49
104. What does f tell you about f′ ?
.
Fact
If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).
The Real Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
105. What does f tell you about f′ ?
.
Fact
If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).
The Real Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
still!
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
106. What does f tell you about f′ ?
.
Fact
If f is decreasing on on the open interval (a, b), then f′ ≤ 0 on (a, b).
The Real Proof.
If f is decreasing on (a, b), and ∆x > 0, then
f(x + ∆x) − f(x)
f(x + ∆x) < f(x) =⇒ <0
∆x
But if ∆x < 0, then x + ∆x < x, and
f(x + ∆x) − f(x)
f(x + ∆x) > f(x) =⇒ <0
∆x
f(x + ∆x) − f(x)
still! Either way, < 0, so
∆x
f(x + ∆x) − f(x)
f′ (x) = lim ≤0
∆x→0 ∆x
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 34 / 49
107. Going the Other Way?
Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
108. Going the Other Way?
Question
If a function has a negative derivative on an interval, must it be
decreasing on that interval?
Answer
Maybe.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 35 / 49
109. Outline
Rates of Change
Tangent Lines
Velocity
Population growth
Marginal costs
The derivative, defined
Derivatives of (some) power functions
What does f tell you about f′ ?
How can a function fail to be differentiable?
Other notations
The second derivative
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 36 / 49
110. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
111. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
112. Differentiability is super-continuity
Theorem
If f is differentiable at a, then f is continuous at a.
Proof.
We have
f(x) − f(a)
lim (f(x) − f(a)) = lim · (x − a)
x→a x→a x−a
f(x) − f(a)
= lim · lim (x − a)
x→a x−a x→a
′
= f (a) · 0 = 0
Note the proper use of the limit law: if the factors each have a limit at
a, the limit of the product is the product of the limits.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 37 / 49
113. Differentiability FAIL
Kinks
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
114. Differentiability FAIL
Kinks
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
115. Differentiability FAIL
Kinks
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
.
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 38 / 49
116. Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
117. Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
118. Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
119. Differentiability FAIL
Cusps
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 39 / 49
120. Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x)
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
121. Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
122. Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
123. Differentiability FAIL
Vertical Tangents
Example
Let f have the graph on the left-hand side below. Sketch the graph of
the derivative f′ .
f
.(x) .′ (x)
f
. x
. . x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 40 / 49
124. Differentiability FAIL
Weird, Wild, Stuff
Example
f
.(x)
. x
.
This function is differentiable
at 0.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.1–2.2 The Derivative September 28, 2010 41 / 49