# 2019 CED – Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Unit 3 covers the Chain Rule, differentiation techniques that follow from it, and higher order derivatives. (CED – 2019 p. 67 – 77). These topics account for about 9 – 13% of questions on the AB exam and 4 – 7% of the BC questions.

### Topics 3.1 – 3.6

Topic 3.1 The Chain Rule. Students learn how to apply the Chain Rule in basic situations.

Topic 3.2 Implicit Differentiation. The Chain Rule is used to find the derivative of implicit relations.

Topic 3.3 Differentiation Inverse Functions.  The Chain Rule is used to differentiate inverse functions.

Topic 3.4 Differentiating Inverse Trigonometric Functions. Continuing the previous section, the ideas of the derivative of the inverse are applied to the inverse trigonometric functions.

Topic 3.5 Selecting Procedures for Calculating Derivatives. Students need to be able to choose which differentiation procedure should be used for any function they are given. This is where you can review (spiral) techniques from Unit 2 and practice those from this unit.

Topic 3.6 Calculating Higher Order Derivatives. Second and higher order derivatives are considered. Also, the notations for higher order derivatives are included here.

Topics 3.2, 3.4, and 3.5 will require more than one class period. You may want to do topic 3.6 before 3.5 and use 3.5 to practice all the differentiated techniques learned so far. The suggested number of 40 – 50-minute class periods is about 10 – 11 for AB and 8 – 9 for BC. This includes time for testing etc.
Posts on these topics include:

The Power Rule Implies Chain Rule

The Chain Rule

Seeing the Chain Rule

Derivative Practice – Numbers

Derivative Practice – Graphs

Experimenting with CAS – Chain Rule

Implicit Differentiation of Parametric Equations

This series of posts reviews and expands what students know from pre-calculus about inverses. This leads to finding the derivative of exponential functions, ax, and the definition of e, from which comes the definition of the natural logarithm.

Inverses Graphically and Numerically

The Range of the Inverse

The Calculus of Inverses

The Derivatives of Exponential Functions and the Definition of e and This pair of posts shows how to find the derivative of an exponential function, how and why e is chosen to help this differentiation.

Logarithms Inverses are used to define the natural logarithm function as the inverse of ex. This follow naturally from the work on inverses. However, integration is involved and this is best saved until later. I will mention it then.
Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions

2019 CED Unit 10 Infinite Sequences and Series

Updated to include the series on inverses – July 7, 2020

# Type 7 Questions: Miscellaneous

Any topic in the Course and Exam Description may be the subject of a free-response or multiple-choice question. There are topics that are not asked often enough to be classified as a type of their own. The two topics listed here have been the subject of full free-response questions or major parts of them. Other topics occasionally asked are mentioned in the question list at the end of the post.

Implicitly defined relations and implicit differentiation

These questions may ask students to find the first or second derivative of an implicitly defined relation. Often the derivative is given and students are required to show that it is correct. (This is because without the correct derivative the rest of the question cannot be done.) The follow-up is to answer questions about the function such as finding an extreme value, second derivative test, or find where the tangent is horizontal or vertical.

What students should know how to do

• Know how to find the first derivative of an implicit relation using the product rule, quotient rule, chain rule, etc.
• Know how to find the second derivative, including substituting for the first derivative.
• Know how to evaluate the first and second derivative by substituting both coordinates of a given point. (Note: If all that is needed is the numerical value of the derivative then the substitution is often easier if done before solving for dy/dx or d2y/dx2, and as usual the arithmetic need not be done.)
• Analyze the derivative to determine where the relation has horizontal and/or vertical tangents.
• Write and work with lines tangent to the relation.
• Find extreme values. It may also be necessary to show that the point where the derivative is zero is actually on the graph and to justify the answer.

Simpler questions about implicit differentiation my appear on the multiple-choice sections of the exam.

Related Rates

Derivatives are rates and when more than one variable is changing over time the relationships among the rates can be found by differentiating with respect to time. The time variable may not appear in the equations. These questions appear occasionally on the free-response sections; if not there, then a simpler version may appear in the multiple-choice sections. In the free-response sections they may be an entire problem, but more often appear as one or two parts of a longer question.

What students should know how to do

• Set up and solve related rate problems.
• Be familiar with the standard type of related rate situations, but also be able to adapt to different contexts.
• Know how to differentiate with respect to time. That is, find dy/dt even if there is no time variable in the given equations using any of the differentiation techniques.
• Interpret the answer in the context of the problem.
• Unit analysis.

Shorter questions on this concept also appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on related rate see October 8, and 10, 2012 and for implicit relations see November 14, 2012.

Free response questions (many of the BC questions are suitable for AB)

• Finding derivatives using the chain rule, the quotient rule, etc. from tables of values: 2016 AB 6 and 2015 AB 6
• Implicit differentiation 2004 AB and 2016 BC 4
• L’Hospital’s Rule 2016 BC 4
• Continuity and piecewise defined functions: 2012 AB 4, 2011 AB 6 and 2014 BC 5
• Related rate: 2014 AB4/BC4, 2016 AB5/BC5
• Arc length (BC Topic) 2014 BC 5
• Partial fractions (BC Topic) 2015 BC 5
• Improper integrals (BC topic): 2017 BC 5

Multiple-choice questions from non-secure exams:

• 2012 AB 27 (implicit differentiation), 77 (IVT), 88 (related rate)
• 2012 BC 4 (Curve length), 7 (Implicit differentiation), 11 (continuity/differentiability), 12 (Implicit differentiation), 77 (dominance), 82 (average value), 85 (related rate) , 92 (compositions)

Schedule of review postings:

# An Exploration in Differential Equations

This is an exploration based on the AP Calculus question 2018 AB 6. I originally posed it for teachers last summer. This will make, I hope, a good review of many of the concepts and techniques students have learned during the year. The exploration, which will take an hour or more, includes these topics:

• Finding the general solution of the differential equation by separating the variables
• Checking the solution by substitution
• Using a graphing utility to explore the solutions for all values of the constant of integration, C
• Finding the solutions’ horizontal and vertical asymptotes
• Finding several particular solutions
• Finding the domains of the particular solutions
• Finding the extreme value of all solutions in terms of C
• Finding the second derivative (implicit differentiation)
• Considering concavity
• Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles.

The exploration is here in a PDF file. Here are the solutions.

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam:  In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and$35/non-members. You can register for the event by following this link: http://eventreg.collegeboard.org/d/xbqbjz

# Implicit Differentiation

Often a relation (an expression in x and y), that has a graph but is not a function, needs to be analyzed. But the relation is not or cannot be solved for y. What to do? The answer is to use the technique of implicit differentiation. Assume there is a way to solve for y and differentiate using the Chain Rule. Whenever you get to the y,“differentiate” it by writing dy/dx. Then solve for dy/dx

Here are several previous posts on this topic and how to go about using it.

Implicit Differentiation

Implicit Differentiation and Inverses

Implicit differentiation of parametric equations   These are BC topics

A Vector’s Derivative  These are BC topics

_____________________________________________________

# Summer Fun

Every Spring I have a lot of fun proofreading Audrey Weeks’ new Calculus in Motion illustrations for the most recent AP Calculus Exam questions. These illustrations run on Geometers’ Sketchpad. In addition to the exam questions Calculus in Motion (and its companion Algebra in Motion) include separate animations illustrating most of the concepts in calculus and algebra. This is a great resource for your classes.

This year, I really got into 2018 AB 6, the differential equation question. I wrote an exploration (or as the kids would say “worksheet”) on a function very similar to the differential equation in that question. The exploration, which is rather long, includes these topics:

• Finding the general solution of the differential equation by separating the variables
• Checking the solution by substitution
• Using a graphing utility to explore the solutions for all values of the constant of integration, C
• Finding the solutions’ horizontal and vertical asymptotes
• Finding several particular solutions
• Finding the domains of the particular solutions
• Finding the extreme value of all solutions in terms of C
• Finding the second derivative (implicit differentiation)
• Considering concavity
• Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles.

Students will need to have studied calculus through differential equations before they start the exploration. I will repost it next January for them.

The exploration is here for you to try. Try it before you look at the solutions. It will give you something to do over the summer – well not all summer, only an hour or so.

There will be only occasional, very occasional, posts over the Summer. More regular posting will begin again in August. Enjoy the Explorations, and, more important, enjoy the Summer!

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# Implicit Differentiation and Inverses

Implicit differentiation of relations is done using the Chain Rule.

Implicit Differentiation (from last Friday’s post. I discovered I never did a post on this topic before!)

Implicit differentiation of parametric equations

A Vector’s Derivative

The inverse series

This series of posts reviews and expands what students know from pre-calculus about inverses. This leads to finding the derivative of exponential functions, ax, and the definition of e, from which comes the definition of the natural logarithm.

Inverses Graphically and Numerically

The Range of the Inverse

The Calculus of Inverses

The Derivatives of Exponential Functions and the Definition of e  and This pair of posts shows how to find the derivative of an exponential function, how and why e is chosen to help this differentiation.

Logarithms Inverses are used to define the natural logarithm function as the inverse of ex. This follow naturally from the work on inverses. However, integration is involved and this is best saved until later. I will mention it then.

# Implicit Differentiation

I discovered in doing next week’s post that I apparently never wrote about implicit differentiation. So here goes – an extra post this week!

Implicit Differentiation

The technique of implicit differentiation allows you to find slopes of relations given by equations that are not written as functions or may even be impossible to write as functions.

Example 1: A good way to start investigating this idea is to give your class the equation of a circle, say ${{x}^{2}}+{{y}^{2}}=25$ and ask them to find the slope of the tangent line (the derivative) where x = 3. No hints, just let them try.

Most students will hit upon solving for y and then differentiating:

$y=\pm \sqrt{{25-{{x}^{2}}}}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{-2x}}{{\pm 2\sqrt{{25-{{x}^{2}}}}}}=\frac{{-x}}{{\pm \sqrt{{25-{{x}^{2}}}}}}$

There are two points where x = 3: (3, 4) and (3, –4) at the first point the slope is – ¾ and at the second ¾.

Then show them another way – implicit differentiation.

To use this technique, assume that y is a function of x, but do not bother to find that function. Then using the chain rule on any terms containing a y. For${{x}^{2}}+{{y}^{2}}=25$ , we have

$\displaystyle 2x+2y\frac{{dy}}{{dx}}=0$

Then solve for the derivative

$\displaystyle \frac{{dy}}{{dx}}=-\frac{x}{y}$

We see that this is the same as we found the first time, since $y=\pm \sqrt{{25-{{x}^{2}}}}$! There is a slight advantage here: we can now find the slopes from the coordinates without solving or dealing with the plus/minus sign. *

Example 2: Now let’s consider a more difficult example. Find the derivative of ${{x}^{2}}+4{{y}^{2}}=7+3xy$. To solve for y here is possible but somewhat difficult (Hint: use the quadratic formula). We can continue writing ${y}'$ for dy/dx.

$2x+8y{y}'=0+3x{y}'+3y$

Note that the last term on the right is differentiated using the product rule.  Since this happens fairly often, students need to be reminded of it.

Now solving for ${y}'$gives

$\displaystyle {y}'=\frac{{3y-2x}}{{8y-3x}}$

Then we can find the derivatives at specific points by substituting the coordinates of the point. At the point (3,2) on the curve, the slope is $\displaystyle {y}'=\frac{{3\left( 2 \right)-2\left( 3 \right)}}{{8\left( 2 \right)-3\left( 3 \right)}}=\frac{0}{7}=0$

Note: the derivative of an implicit relation usually involves both the x and y coordinates.

Second Derivatives

This idea can be repeated to find second and higher derivatives.

Example 1 continued: In the first example with $\displaystyle \frac{{dy}}{{dx}}=\frac{{-x}}{y}$  we differentiate using the quotient rule:

$\displaystyle {{y}'}'=\frac{{y(-1)-(-x){y}'}}{{{{y}^{2}}}}$

The second derivative is a function, not just of x and y, but also of ${y}'$. We can replace it with the first derivative and simplify.

$\displaystyle {{y}'}'=\frac{{-y+x\left( {\frac{{-x}}{y}} \right)}}{{{{y}^{2}}}}=\frac{{-{{y}^{2}}-{{x}^{2}}}}{{{{y}^{3}}}}=-\frac{{25}}{{{{y}^{3}}}}$

(This might be a good time to do a quick review of simplifying complex fractions; they occur often in implicit differentiation problems.)

To find the value of the second derivative at a given point we can substitute into either of the two expressions above. At (3, –4) where the derivative has been previously found to be ¾ we have $\displaystyle {{y}'}'=\frac{{y(-1)-(-x){y}'}}{{{{y}^{2}}}}=\frac{{\left( {-4} \right)\left( {-1} \right)-\left( {-3} \right)\left( {\frac{3}{{-4}}} \right)}}{{{{{\left( {-4} \right)}}^{2}}}}=-\frac{{-16-9}}{{-48}}=\frac{{25}}{{48}}$

Or we can use the second form

$\displaystyle {{y}'}'=-\frac{{25}}{{{{y}^{3}}}}=-\frac{{25}}{{{{{\left( {-4} \right)}}^{3}}}}=\frac{{25}}{{48}}$

Example 2 continued: The second example was taken from an AB Calculus exam (2004 AB 4). The first part gave the first derivative and asked students to show that it was correct. This was done (instead of just asking the students to find the first derivative) so that students would be sure to have the correct derivative to use later in the question.

The second part asked students to show that the tangent line is horizontal at the point where x = 3. This included finding the coordinates of the point, (3, 2) and showing that it is on the curve.

The third part of the question asked students to determine whether the point from part (b) was a relative maximum, a relative minimum or neither, and to justify their answer. Since there is no way to determine how the sign of the first derivative changes at the point the First Derivative Test cannot be used. Likewise, the Candidates’ Test (a/k/a the closed interval test) cannot be used without solving for y, and determining the domain of each part. That leaves the Second Derivative Test as the easiest choice.

$\displaystyle {y}'=\frac{{3y-2x}}{{8y-3x}}$  at (3,2) $\displaystyle {y}'=\frac{{3\left( 2 \right)-2\left( 3 \right)}}{{8\left( 2 \right)-3\left( 3 \right)}}=0$

$\displaystyle {{y}'}'=\frac{{\left( {8y-3x} \right)\left( {3{y}'-2} \right)-\left( {3y-2x} \right)\left( {8{y}'-3} \right)}}{{{{{\left( {8y-3x} \right)}}^{2}}}}$

Substituting the values into this without doing the algebra to remove the first derivative gives

$\displaystyle \begin{array}{l}{{y}'}'=\frac{{\left( {8\left( 2 \right)-3\left( 3 \right)} \right)\left( {0-2} \right)-\left( {3\left( 2 \right)-2\left( 3 \right)} \right)\left( {8\left( 0 \right)-3} \right)}}{{{{{\left( {8\left( 2 \right)-3\left( 3 \right)} \right)}}^{2}}}}=\frac{{\left( {16-9} \right)\left( {-2} \right)-0}}{{{{{\left( {16-9} \right)}}^{2}}}}=-\frac{2}{7}\\\end{array}$

So, the point (3, 2) is a relative maximum.

The graph of the relation, an ellipse is shown below.

* Incidentally, there is another clever way of doing example 1: The radius to any point on a circle centered at the origin has a slope of y/x.  Since tangents to circles are perpendicular to the radii drawn to the point of tangency, the slope of the tangent must be –x/y.