Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
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.
This document contains notes for a lesson on the chain rule from a Calculus 1 class. It defines the chain rule formula and provides an example of applying the chain rule to find the derivative of a function. It also includes another example problem and its step-by-step solution using the chain rule. The document concludes with a metaphor to help understand applying the chain rule.
1) The document provides definitions and formulas for calculating derivatives, including the derivative of a function at a point, the derivative function, and common derivatives of basic functions like polynomials, exponentials, logarithms, trigonometric functions, and inverse trigonometric functions.
2) Examples are given of calculating derivatives using the definition of the derivative, for functions like f(x)=3x^2, f(x)=x+1, and f(x)=1/x.
3) Rules are listed for calculating the derivatives of sums, products, quotients of functions using properties of 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.
The document provides examples of using implicit differentiation to find derivatives. It demonstrates:
1) Taking the derivative of each side of an equation and putting all terms with dy/dx on one side.
2) Factoring out dy/dx from the resulting equation.
3) Solving for dy/dx.
Several examples are worked through, including finding the derivative of equations, finding the slope of a tangent line at a point, and using the product rule. The document concludes with an exercise involving implicit differentiation.
The document discusses properties of limits of functions in algebra. It presents 9 properties of limits, including: (1) the limit of a constant k is equal to k; (2) the limit of x as x approaches a is equal to a; (3) the limit of kf(x) is equal to k times the limit of f(x); (4) the limit of the sum of two functions is equal to the sum of their individual limits. It also provides examples of calculating limits using these properties, such as finding the limit of 7x - 4 as x approaches 2.
1. A differential equation is an equation that relates an unknown function with some of its derivatives. The document provides a step-by-step example of solving a differential equation to find the xy-equation of a curve with a given gradient condition.
2. The key steps are: (1) write the derivative term as a fraction, (2) integrate both sides, (3) apply the initial condition to determine the constant term, (4) write the final function relationship.
3. Common types of differential equations discussed are separable first order equations, where the derivative terms can be isolated by dividing both sides.
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.
This document contains notes for a lesson on the chain rule from a Calculus 1 class. It defines the chain rule formula and provides an example of applying the chain rule to find the derivative of a function. It also includes another example problem and its step-by-step solution using the chain rule. The document concludes with a metaphor to help understand applying the chain rule.
1) The document provides definitions and formulas for calculating derivatives, including the derivative of a function at a point, the derivative function, and common derivatives of basic functions like polynomials, exponentials, logarithms, trigonometric functions, and inverse trigonometric functions.
2) Examples are given of calculating derivatives using the definition of the derivative, for functions like f(x)=3x^2, f(x)=x+1, and f(x)=1/x.
3) Rules are listed for calculating the derivatives of sums, products, quotients of functions using properties of 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.
The document provides examples of using implicit differentiation to find derivatives. It demonstrates:
1) Taking the derivative of each side of an equation and putting all terms with dy/dx on one side.
2) Factoring out dy/dx from the resulting equation.
3) Solving for dy/dx.
Several examples are worked through, including finding the derivative of equations, finding the slope of a tangent line at a point, and using the product rule. The document concludes with an exercise involving implicit differentiation.
The document discusses properties of limits of functions in algebra. It presents 9 properties of limits, including: (1) the limit of a constant k is equal to k; (2) the limit of x as x approaches a is equal to a; (3) the limit of kf(x) is equal to k times the limit of f(x); (4) the limit of the sum of two functions is equal to the sum of their individual limits. It also provides examples of calculating limits using these properties, such as finding the limit of 7x - 4 as x approaches 2.
1. A differential equation is an equation that relates an unknown function with some of its derivatives. The document provides a step-by-step example of solving a differential equation to find the xy-equation of a curve with a given gradient condition.
2. The key steps are: (1) write the derivative term as a fraction, (2) integrate both sides, (3) apply the initial condition to determine the constant term, (4) write the final function relationship.
3. Common types of differential equations discussed are separable first order equations, where the derivative terms can be isolated by dividing both sides.
This document contains 11 questions regarding differential equations. It asks the student to classify equations as order, linear or non-linear; formulate differential equations by eliminating constants; solve specific differential equations; prove properties of curves; find general solutions by substitution; and determine conditions for exactness and solve exact differential equations.
The document discusses methods for solving first order ordinary differential equations (ODEs). It covers:
1) Finding the integrating factor for exact differential equations.
2) Solving homogeneous first order linear ODEs by making a substitution to reduce it to a separable equation.
3) Solving inhomogeneous first order linear ODEs using an integrating factor.
Examples are provided to demonstrate each method step-by-step.
Pt 3&4 turunan fungsi implisit dan cyclometrilecturer
This document discusses implicit differentiation and provides examples of taking the derivative of implicit functions. It begins by presenting the general form of an implicit function as f(x,y)=0 and provides some examples. It then shows how to take the derivative dy/dx of several implicit functions by applying the chain rule and implicit differentiation. Formulas for the derivatives of inverse trigonometric functions are also provided, along with examples of finding dy/dx for functions involving inverse trigonometric functions.
1. The limit as x approaches 4 of x4-16 is 0. When factored, the expression becomes (x-4)(x+4)(x2+4) which equals 0 as x approaches 4.
2. The limit as x approaches infinity of x7-x2+1 is 1. When factored, the leading terms are x7 for both the top and bottom expressions, which equals 1 as x approaches infinity.
3. The limit as x approaches -1 of x2-1 is 0. When factored, the expression becomes (x+1)(x-1) which equals 0 as the factors are 0 when x is -1.
This document contains the work of a student on a calculus test. It includes:
1) Solving limits, finding derivatives, and applying L'Hopital's rule.
2) Using induction to prove an identity.
3) Providing epsilon-delta proofs of limits.
4) Finding where a tangent line is parallel to a secant line.
5) Proving statements about limits of functions.
The student provides detailed solutions showing their work for each problem on the test.
This document introduces integration as the reverse process of differentiation. It establishes the rule that if dy/dx = axn, then the integral is y = axn+1/(n+1) + c, where c is the constant of integration. Several examples are worked out applying this rule. The document also introduces notation for integration, including the integral sign ∫ and discusses some cases where the power is not a whole number. It emphasizes that the power in the integral must be positive.
This document contains 7 limit problems worked out step-by-step with explanations of the reasoning. The problems involve various algebraic expressions containing polynomials, rational expressions, and variables approaching different limits. The solutions find the limits by simplifying the expressions, factorizing where possible, and evaluating the behavior of terms as the variable approaches the limit.
This document provides an overview of engineering mathematics II with a focus on first order ordinary differential equations (ODEs). It explains what first order ODEs are, how to solve separable and reducible first order ODEs, and provides examples of applying first order ODEs to model real-world scenarios like population growth, decay, and radioactive decay. The objectives are to explain first order ODEs, separable equations, and apply the concepts to real life applications.
Matematika Ekonomi Diferensiasi fungsi sederhanalia170494
This document discusses differentiation rules for simple functions including:
- Constant functions have a derivative of 0
- Polynomial functions have derivatives that are the polynomial with the exponent decreased by 1 and multiplied by the exponent
- The product rule, quotient rule, and chain rule for differentiation
- Examples of applying these rules to differentiate a variety of functions
This document discusses various techniques for factoring polynomials, including:
1. Factoring using the greatest common factor (GCF).
2. Factoring polynomials with 4 or more terms by grouping.
3. Factoring trinomials using factors that add up to the coefficient of the middle term.
4. Using the "box method" to factor trinomials where the coefficient of the x^2 term is not 1.
5. Factoring the difference of two squares using the formula a^2 - b^2 = (a + b)(a - b).
1) An antiderivative of a function f(x) is any function F(x) whose derivative is equal to f(x).
2) The general antiderivative of a function f(x) is written as F(x) + C, where F(x) is a particular antiderivative and C is an arbitrary constant.
3) The indefinite integral notation ∫f(x)dx represents the entire family of antiderivatives for a function f(x), since each value of C defines a different antiderivative.
Solving second order ordinary differential equations (boundary value problems) using the Least Squares Technique. Contains one numerical examples from Shah, Eldho, Desai
Lesson 8: Derivatives of Logarithmic and Exponential Functions (worksheet sol...Matthew Leingang
This document provides solutions to derivatives of exponential, logarithmic, and other functions. It includes:
1) The derivatives of functions such as y=e^2x, y=6^x, y=ln(x^3 + 9), and y=log_3(e^x).
2) Using logarithmic differentiation to find the derivatives of functions like y=x^x^2-1 and y=(x-1)(x-2)(x-3).
3) Taking the derivative of functions involving logarithms, exponents, and square roots such as y=sin^2(x)+2sin(x) and y=x(x-1)^3/
This document is an exam for a differential equations course. It contains 4 problems to solve in 60 minutes. Problem 1 asks students to determine if functions are linearly dependent or independent on an interval. Problem 2 is an initial value problem to solve. Problem 3 is to solve a differential equation. Problem 4 is to find the general solution of another differential equation. Formulas for trigonometric integrals are provided.
This document contains 26 math problems involving derivatives of logarithmic, exponential, and inverse trigonometric functions. The problems include finding derivatives of expressions, setting up and solving differential equations, and determining relationships between derivatives.
1) There are several methods for solving quadratic equations, including factoring, graphing, using the quadratic formula, and completing the square.
2) Factoring involves expressing the quadratic expression as a product of two linear factors. Methods for factoring include finding the greatest common factor, using factor diamonds, grouping, and the borrowing method.
3) The document provides examples of solving quadratics using various factoring techniques and practicing additional problems.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
This document discusses implicit differentiation and finding the slope of tangent lines using implicit differentiation. It begins with an example problem of finding the slope of the tangent line to the curve x^2 + y^2 = 1 at the point (3/5, -4/5). It then explains how to set up and solve the implicit differentiation problem to find the slope. The document emphasizes that even when a relation is not explicitly a function, it can often be treated as locally functional to apply implicit differentiation and find tangent slopes. It provides another example problem and discusses horizontal and vertical tangent lines.
This document contains 11 questions regarding differential equations. It asks the student to classify equations as order, linear or non-linear; formulate differential equations by eliminating constants; solve specific differential equations; prove properties of curves; find general solutions by substitution; and determine conditions for exactness and solve exact differential equations.
The document discusses methods for solving first order ordinary differential equations (ODEs). It covers:
1) Finding the integrating factor for exact differential equations.
2) Solving homogeneous first order linear ODEs by making a substitution to reduce it to a separable equation.
3) Solving inhomogeneous first order linear ODEs using an integrating factor.
Examples are provided to demonstrate each method step-by-step.
Pt 3&4 turunan fungsi implisit dan cyclometrilecturer
This document discusses implicit differentiation and provides examples of taking the derivative of implicit functions. It begins by presenting the general form of an implicit function as f(x,y)=0 and provides some examples. It then shows how to take the derivative dy/dx of several implicit functions by applying the chain rule and implicit differentiation. Formulas for the derivatives of inverse trigonometric functions are also provided, along with examples of finding dy/dx for functions involving inverse trigonometric functions.
1. The limit as x approaches 4 of x4-16 is 0. When factored, the expression becomes (x-4)(x+4)(x2+4) which equals 0 as x approaches 4.
2. The limit as x approaches infinity of x7-x2+1 is 1. When factored, the leading terms are x7 for both the top and bottom expressions, which equals 1 as x approaches infinity.
3. The limit as x approaches -1 of x2-1 is 0. When factored, the expression becomes (x+1)(x-1) which equals 0 as the factors are 0 when x is -1.
This document contains the work of a student on a calculus test. It includes:
1) Solving limits, finding derivatives, and applying L'Hopital's rule.
2) Using induction to prove an identity.
3) Providing epsilon-delta proofs of limits.
4) Finding where a tangent line is parallel to a secant line.
5) Proving statements about limits of functions.
The student provides detailed solutions showing their work for each problem on the test.
This document introduces integration as the reverse process of differentiation. It establishes the rule that if dy/dx = axn, then the integral is y = axn+1/(n+1) + c, where c is the constant of integration. Several examples are worked out applying this rule. The document also introduces notation for integration, including the integral sign ∫ and discusses some cases where the power is not a whole number. It emphasizes that the power in the integral must be positive.
This document contains 7 limit problems worked out step-by-step with explanations of the reasoning. The problems involve various algebraic expressions containing polynomials, rational expressions, and variables approaching different limits. The solutions find the limits by simplifying the expressions, factorizing where possible, and evaluating the behavior of terms as the variable approaches the limit.
This document provides an overview of engineering mathematics II with a focus on first order ordinary differential equations (ODEs). It explains what first order ODEs are, how to solve separable and reducible first order ODEs, and provides examples of applying first order ODEs to model real-world scenarios like population growth, decay, and radioactive decay. The objectives are to explain first order ODEs, separable equations, and apply the concepts to real life applications.
Matematika Ekonomi Diferensiasi fungsi sederhanalia170494
This document discusses differentiation rules for simple functions including:
- Constant functions have a derivative of 0
- Polynomial functions have derivatives that are the polynomial with the exponent decreased by 1 and multiplied by the exponent
- The product rule, quotient rule, and chain rule for differentiation
- Examples of applying these rules to differentiate a variety of functions
This document discusses various techniques for factoring polynomials, including:
1. Factoring using the greatest common factor (GCF).
2. Factoring polynomials with 4 or more terms by grouping.
3. Factoring trinomials using factors that add up to the coefficient of the middle term.
4. Using the "box method" to factor trinomials where the coefficient of the x^2 term is not 1.
5. Factoring the difference of two squares using the formula a^2 - b^2 = (a + b)(a - b).
1) An antiderivative of a function f(x) is any function F(x) whose derivative is equal to f(x).
2) The general antiderivative of a function f(x) is written as F(x) + C, where F(x) is a particular antiderivative and C is an arbitrary constant.
3) The indefinite integral notation ∫f(x)dx represents the entire family of antiderivatives for a function f(x), since each value of C defines a different antiderivative.
Solving second order ordinary differential equations (boundary value problems) using the Least Squares Technique. Contains one numerical examples from Shah, Eldho, Desai
Lesson 8: Derivatives of Logarithmic and Exponential Functions (worksheet sol...Matthew Leingang
This document provides solutions to derivatives of exponential, logarithmic, and other functions. It includes:
1) The derivatives of functions such as y=e^2x, y=6^x, y=ln(x^3 + 9), and y=log_3(e^x).
2) Using logarithmic differentiation to find the derivatives of functions like y=x^x^2-1 and y=(x-1)(x-2)(x-3).
3) Taking the derivative of functions involving logarithms, exponents, and square roots such as y=sin^2(x)+2sin(x) and y=x(x-1)^3/
This document is an exam for a differential equations course. It contains 4 problems to solve in 60 minutes. Problem 1 asks students to determine if functions are linearly dependent or independent on an interval. Problem 2 is an initial value problem to solve. Problem 3 is to solve a differential equation. Problem 4 is to find the general solution of another differential equation. Formulas for trigonometric integrals are provided.
This document contains 26 math problems involving derivatives of logarithmic, exponential, and inverse trigonometric functions. The problems include finding derivatives of expressions, setting up and solving differential equations, and determining relationships between derivatives.
1) There are several methods for solving quadratic equations, including factoring, graphing, using the quadratic formula, and completing the square.
2) Factoring involves expressing the quadratic expression as a product of two linear factors. Methods for factoring include finding the greatest common factor, using factor diamonds, grouping, and the borrowing method.
3) The document provides examples of solving quadratics using various factoring techniques and practicing additional problems.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
This document discusses implicit differentiation and finding the slope of tangent lines using implicit differentiation. It begins with an example problem of finding the slope of the tangent line to the curve x^2 + y^2 = 1 at the point (3/5, -4/5). It then explains how to set up and solve the implicit differentiation problem to find the slope. The document emphasizes that even when a relation is not explicitly a function, it can often be treated as locally functional to apply implicit differentiation and find tangent slopes. It provides another example problem and discusses horizontal and vertical tangent lines.
Lesson 29: Integration by Substition (worksheet solutions)Matthew Leingang
The document provides solutions to 11 integrals using substitution methods. The integrals involve trigonometric, exponential, logarithmic and algebraic functions. Substitutions are given for some integrals and need to be determined for others. The solutions show setting up the substitution, finding the differential, integrating and solving for the antiderivative. Some integrals have multiple valid substitution options. One integral is solved using a trigonometric identity in addition to substitution.
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.
1. Graph and analyze the critical points, extrema, inflection points, intervals of increasing/decreasing, and intervals of concave up/down for 10 functions.
2. Review homework on finding derivatives using the definition of the difference quotient and evaluating limits. Find the derivatives of 6 functions.
3. Use implicit differentiation to find the derivative of one function defined implicitly and to find points with tangent lines of slope 1 for another implicit function.
4. Find the second derivatives of two functions.
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.
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 11: Implicit Differentiation (Section 41 slides)Mel Anthony Pepito
This document provides an overview of implicit differentiation. It begins with a motivating example of finding the slope of the tangent line to the curve x^2 + y^2 = 1 at the point (3/5, -4/5). It is shown that while y is not explicitly defined as a function of x for this curve, it can be treated as such locally using implicit differentiation. The key steps are to take the derivative of the equation with respect to x, which introduces a term for dy/dx, and then solve for dy/dx. This reveals that implicit differentiation allows the derivative of implicitly defined functions to be found.
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 28: Integration by Substitution (worksheet solutions)Matthew Leingang
These problems demonstrate techniques for integration by substitution. Several integrals are solved by making appropriate substitutions to simplify the integrands, including substituting u = 3x - 5, u = x^2 + 9, u = √x, u = 5 + 2sin3x, and u = 4 - 3x. Trigonometric identities are also used to rewrite sin^2x and cos^2x terms before integrating. One integral is solved using long division and another using two different substitutions. The last integral is solved by multiplying the numerator and denominator by secx + tanx before making the substitution u = tanx + secx.
1. The student's name is Azizaty Desiana. She is in class XI IPS 1. Her average score is 41.5. The test had 20 questions.
2. The document provides examples of math problems and questions about functions. It asks for the limit of various expressions as variables approach certain values.
3. The questions are assessing knowledge of functions, inverses, limits, and calculations involving algebraic expressions.
This document section discusses solving exponential and logarithmic equations. It provides examples of solving different types of exponential equations using properties of exponents and logarithms. It also gives examples of solving logarithmic equations by rewriting the equations in exponential form and then solving. The examples demonstrate solving equations with variables in exponents, logarithmic expressions with a single variable, and more complex equations involving logarithms of expressions. Graphing calculators can also be used to approximate solutions.
This document provides examples of calculus problems involving taking derivatives of functions written in terms of y and finding derivatives implicitly. It includes:
1) Three practice problems for taking the derivative of functions written in terms of x and rewriting them in terms of y.
2) Examples and steps for taking the derivative implicitly when given an equation involving both x and y: find the derivative of y by isolating y' and solving for it.
3) The objective is to practice taking derivatives implicitly.
This document provides instructions and examples for taking derivatives of functions written implicitly and explicitly. It begins with examples of taking derivatives of explicitly defined functions written in terms of x and rewriting them in terms of y. Next, it discusses taking derivatives of implicitly defined functions, using the product rule and solving for the derivative of y, y'. Examples are worked through, such as finding that the derivative of xy = 5 is y' = -y/x. The document aims to teach how to take derivatives of various functions written in different forms.
This document discusses solving quadratic equations by graphing. It explains that a quadratic equation has the form y=ax^2 + bx + c, with the quadratic, linear, and constant terms. A quadratic equation will have 0, 1, or 2 real solutions depending on the graph. The solutions are the x-intercepts where the graph crosses the x-axis. Examples show how to identify the solutions by graphing and finding the x-intercepts. The graph of a quadratic is a parabola with roots/zeros at the x-intercepts and a vertex as its maximum or minimum point. One method for graphing uses a table to plot points and sketch the parabola.
This document discusses inverses of functions. It provides examples of finding the inverse of various functions by switching the x and y coordinates, solving for y, and determining if the inverse is a function. Key points made are: to find the inverse change the coordinate pair; a function and its inverse are reflections over y=x; when composing a function with its inverse, you get back the original function. Examples are worked through and conclusions are drawn about the domains and ranges of inverses.
This document contains the answer key for an exam on differential equations. It provides the solutions to 4 problems:
1) Solving a separable ODE to find an implicit solution.
2) Solving a Bernoulli equation by substitution to find the general solution.
3) Finding an integrating factor to solve an exact ODE and determine the general implicit solution.
4) Solving a linear ODE initial value problem to model salt in a tank over time.
Similar to Lesson 11: Implicit Differentiation (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 (slides)Matthew Leingang
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Lesson 11: Implicit Differentiation
1. Section 2.6
Implicit Differentiation
V63.0121.027, Calculus I
October 8, 2009
Announcements
Midterm next Thursday, covering §§1.1–2.4.
.
.
Image credit: Telstar Logistics
. . . . . .
2. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
3. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5).
. . . . . .
4. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5).
. . . . . .
5. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5). .
. . . . . .
6. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
. . . . . .
7. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
. . . . . .
8. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
dy 3 /5 3/5 3
Evaluate: =√ = = .
dx x=3/5 1 − (3 /5 )2 4/5 4
. . . . . .
9. Motivating Example y
.
Problem
Find the slope of the line which
is tangent to the curve
. x
.
2 2
x +y =1
at the point (3/5, −4/5). .
Solution (Explicit)
√
Isolate: y2 = 1 − x2 =⇒ y = − 1 − x2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1−x 2 1 − x2
dy 3 /5 3/5 3
Evaluate: =√ = = .
dx x=3/5 1 − (3 /5 )2 4/5 4
. . . . . .
10. Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x,
but suppose it did.
Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
. . . . . .
11. Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x,
but suppose it did.
Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
We could differentiate this equation to get
2x + 2f(x) · f′ (x) = 0
. . . . . .
12. Motivating Example, another way
We know that x2 + y2 = 1 does not define y as a function of x,
but suppose it did.
Suppose we had y = f(x), so that
x2 + (f(x))2 = 1
We could differentiate this equation to get
2x + 2f(x) · f′ (x) = 0
We could then solve to get
x
f ′ (x ) = −
f(x)
. . . . . .
13. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. x
.
.
. . . . . .
14. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. x
.
.
. . . . . .
15. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. x
.
.
l
.ooks like a function
. . . . . .
16. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the .
graph of a function.
. x
.
. . . . . .
17. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the .
graph of a function.
. x
.
. . . . . .
18. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the .
graph of a function.
l
.ooks like a function
. x
.
. . . . . .
19. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. . x
.
. . . . . .
20. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. . x
.
. . . . . .
21. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
. . x
.
.
does not look like a
function, but that’s
OK—there are only
two points like this
. . . . . .
22. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
So f(x) is defined
“locally”, almost
. x
.
everywhere and is
differentiable
.
l
.ooks like a function
. . . . . .
23. Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
“Near” most points on y
.
the curve x2 + y2 = 1,
the curve resembles the
graph of a function.
So f(x) is defined
“locally”, almost
. x
.
everywhere and is
differentiable
The chain rule then
.
applies for this local
choice.
l
.ooks like a function
. . . . . .
25. Motivating Example, again, with Leibniz notation
Problem
Find the slope of the line which is tangent to the curve
x2 + y2 = 1 at the point (3/5, −4/5).
Solution
dy
Differentiate: 2x + 2y =0
dx
. . . . . .
26. Motivating Example, again, with Leibniz notation
Problem
Find the slope of the line which is tangent to the curve
x2 + y2 = 1 at the point (3/5, −4/5).
Solution
dy
Differentiate: 2x + 2y =0
dx
Remember y is assumed to be a function of x!
. . . . . .
27. Motivating Example, again, with Leibniz notation
Problem
Find the slope of the line which is tangent to the curve
x2 + y2 = 1 at the point (3/5, −4/5).
Solution
dy
Differentiate: 2x + 2y =0
dx
Remember y is assumed to be a function of x!
dy x
Isolate: =− .
dx y
. . . . . .
28. Motivating Example, again, with Leibniz notation
Problem
Find the slope of the line which is tangent to the curve
x2 + y2 = 1 at the point (3/5, −4/5).
Solution
dy
Differentiate: 2x + 2y =0
dx
Remember y is assumed to be a function of x!
dy x
Isolate: =− .
dx y
dy 3 /5 3
Evaluate: = = .
dx ( 3 ,− 4 ) 4/5 4
5 5
. . . . . .
29. Summary
If a relation is given between x and y which isn’t a function:
“Most of the time”, i.e., “at
most places” y can be y
.
assumed to be a function of .
x
we may differentiate the . x
.
relation as is
dy
Solving for does give the
dx
slope of the tangent line to
the curve at a point on the
curve.
. . . . . .
30. Mnemonic
Explicit Implicit
y = f(x) F(x, y) = k
. . . . . .
31. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
33. Example
Find the equation of the line
tangent to the curve
.
y 2 = x 2 (x + 1 ) = x 3 + x 2
at the point (3, −6).
.
Solution
Differentiating the expression implicitly with respect to x gives
dy dy 3x2 + 2x
2y = 3x2 + 2x, so = , and
dx dx 2y
dy 3 · 32 + 2 · 3 33 11
= =− =− .
dx (3,−6) 2(−6) 12 4
. . . . . .
34. Example
Find the equation of the line
tangent to the curve
.
y 2 = x 2 (x + 1 ) = x 3 + x 2
at the point (3, −6).
.
Solution
Differentiating the expression implicitly with respect to x gives
dy dy 3x2 + 2x
2y = 3x2 + 2x, so = , and
dx dx 2y
dy 3 · 32 + 2 · 3 33 11
= =− =− .
dx (3,−6) 2(−6) 12 4
11
Thus the equation of the tangent line is y + 6 = − (x − 3).
4
. . . . . .
35. Line equation forms
slope-intercept form
y = mx + b
where the slope is m and (0, b) is on the line.
point-slope form
y − y0 = m(x − x0 )
where the slope is m and (x0 , y0 ) is on the line.
. . . . . .
38. Solution, continued
Solving the second equation gives
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
(as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
. . . . . .
39. Solution, continued
Solving the second equation gives
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
(as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down
that road.
. . . . . .
40. Solution, continued
Solving the second equation gives
3x2 + 2x
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
2y
(as long as y ̸= 0). So x = 0 or 3x + 2 = 0.
Substituting x = 0 into the first equation gives
y2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down
that road.
Substituting x = −2/3 into the first equation gives
2 2 2
y2 = (− )3 + (− )2 =⇒ y = ± √ ,
3 3 3 3
so there are two horizontal tangents.
. . . . . .
45. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
dx
Tangent lines are vertical when = 0.
dy
Differentiating x implicitly as a function of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (notice this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).
. . . . . .
46. Example
Find the vertical tangent lines to the same curve: y2 = x3 + x2
Solution
dx
Tangent lines are vertical when = 0.
dy
Differentiating x implicitly as a function of y gives
dx dx dx 2y
2y = 3x2 + 2x , so = 2 (notice this is the
dy dy dy 3x + 2x
reciprocal of dy/dx).
We must solve
. .
2y
y2 = x 3 + x2 =0
1
. [(x, y). is on
the curve]
2
.
3x2 + 2x
[tangent line
is vertical]
. . . . . .
47. Solution, continued
Solving the second equation gives
2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x
(as long as 3x2 + 2x ̸= 0).
. . . . . .
48. Solution, continued
Solving the second equation gives
2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x
(as long as 3x2 + 2x ̸= 0).
Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2 (x + 1)
So x = 0 or x = −1.
. . . . . .
49. Solution, continued
Solving the second equation gives
2y
= 0 =⇒ 2y = 0 =⇒ y = 0
3x2 + 2x
(as long as 3x2 + 2x ̸= 0).
Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2 (x + 1)
So x = 0 or x = −1.
x = 0 is not allowed by the first equation, but
dx
= 0,
dy (−1,0)
so here is a vertical tangent.
. . . . . .
64. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
In the first curve,
y
y + xy′ = 0 =⇒ y′ = −
x
. . . . . .
65. Examples
Example
Show that the families of curves
xy = c x2 − y2 = k
are orthogonal, that is, they intersect at right angles.
Solution
In the first curve,
y
y + xy′ = 0 =⇒ y′ = −
x
In the second curve,
x
2x − 2yy′ = 0 = =⇒ y′ =
y
The product is −1, so the tangent lines are perpendicular
wherever they intersect.
. . . . . .
66. Music Selection
“The Curse of Curves” by Cute is What We Aim For
. . . . . .
67. Ideal gases
The ideal gas law relates
temperature, pressure, and
volume of a gas:
PV = nRT
(R is a constant, n is the
amount of gas in moles)
.
.
Image credit: Scott Beale / Laughing Squid
. . . . . .
68. Compressibility
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
. . . . . .
69. Compressibility
Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.
Approximately we have
∆V dV ∆V
≈ = −β V =⇒ ≈ −β∆P
∆P dP V
The smaller the β , the “harder” the fluid.
. . . . . .
71. Example
Find the isothermic compressibility of an ideal gas.
Solution
If PV = k (n is constant for our purposes, T is constant because of
the word isothermic, and R really is constant), then
dP dV dV V
·V+P = 0 =⇒ =−
dP dP dP P
So
1 dV 1
β=− · =
V dP P
Compressibility and pressure are inversely related.
. . . . . .
72. Nonideal gasses
Not that there’s anything wrong with that
Example
The van der Waals equation
makes fewer simplifications: H
..
( ) O .
. xygen . .
n2 H
P + a 2 (V − nb) = nRT, .
V
H
..
where P is the pressure, V the O .
. xygen H
. ydrogen bonds
volume, T the temperature, n H
..
the number of moles of the .
gas, R a constant, a is a O .
. xygen . .
H
measure of attraction
between particles of the gas, H
..
and b a measure of particle
size.
. . . . . .
73. Nonideal gasses
Not that there’s anything wrong with that
Example
The van der Waals equation
makes fewer simplifications:
( )
n2
P + a 2 (V − nb) = nRT,
V
where P is the pressure, V the
volume, T the temperature, n
the number of moles of the
gas, R a constant, a is a
measure of attraction
between particles of the gas,
and b a measure of particle
.
Image
size. .
. . . . . .
79. Nasty derivatives
dβ (2abn3 − an2 V + PV3 )(nV2 ) − (nbV2 − V3 )(2an3 )
=−
db (2abn3 − an2 V + PV3 )2
( 2 )
nV3 an + PV2
= −( )2 < 0
PV3 + an2 (2bn − V)
dβ n2 (bn − V)(2bn − V)V2
= ( )2 > 0
da PV3 + an2 (2bn − V)
(as long as V > 2nb, and it’s probably true that V ≫ 2nb).
. . . . . .
80. Outline
The big idea, by example
Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry
The power rule for rational powers
. . . . . .
84. The power rule for rational powers
Theorem
p p/q−1
If y = xp/q , where p and q are integers, then y′ = x .
q
Proof.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y
. . . . . .
85. The power rule for rational powers
Theorem
p p/q−1
If y = xp/q , where p and q are integers, then y′ = x .
q
Proof.
We have
dy dy p x p −1
yq = xp =⇒ qyq−1 = pxp−1 =⇒ = · q−1
dx dx q y
Now yq−1 = x(p/q)(q−1) = xp−p/q so
x p −1
= xp−1−(p−p/q) = xp/q−1
y q −1
. . . . . .
86. Summary
If a relation is given between x and y which isn’t a function:
“Most of the time”, i.e., “at
most places” y can be y
.
assumed to be a function of .
x
we may differentiate the . x
.
relation as is
dy
Solving for does give the
dx
slope of the tangent line to
the curve at a point on the
curve.
. . . . . .