Implicit Differentiation of Parametric Equations

Posted on October 31, 2021 by Lin McMullin

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $\displaystyle y=y(t)$ and, $\displaystyle x=x(t)$ then the traditional formulas give

$\displaystyle \frac{{dy}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$ , and

$\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$

It is that last part, where you divide by $\displaystyle {\frac{{dx}}{{dt}}}$ , that bothers me. Where did the $\displaystyle {\frac{{dx}}{{dt}}}$ come from?

Then it occurred to me that dividing by $\displaystyle {\frac{{dx}}{{dt}}}$ is the same as multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$

It’s just implicit differentiation!

Since $\displaystyle \frac{{dy}}{{dx}}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$ gives the second derivative. (There is a technical requirement here that given $\displaystyle x=x(t)$ , then its inverse $\displaystyle t={{x}^{{-1}}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{{dx}}y(t)=\frac{{dy}}{{dt}}\cdot \frac{{dt}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$

The reason you do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $\displaystyle x={{t}^{3}}-3$ and $\displaystyle y=\ln \left( t \right)$

Then $\displaystyle \frac{{dy}}{{dt}}=\frac{1}{t}$ and $\displaystyle \frac{{dx}}{{dt}}=3{{t}^{2}}$ and $\displaystyle \frac{{dt}}{{dx}}=\frac{1}{{3{{t}^{2}}}}$

Then $\displaystyle \frac{{dy}}{{dx}}=\frac{1}{t}\cdot \frac{{dt}}{{dx}}=\frac{1}{t}\cdot \frac{1}{{3{{t}^{2}}}}=\frac{1}{3}{{t}^{{-3}}}$

And $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)} \right)\cdot \frac{{dt}}{{dx}}=\left( {-{{t}^{{-4}}}} \right)\cdot \left( {\frac{1}{{3{{t}^{2}}}}} \right)=-\frac{1}{{3{{t}^{6}}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014. Revised: 2/8/2016. Originally posted May 5, 2014.

Reading the Derivative’s Graph

Posted on October 19, 2021 by Lin McMullin

A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, students find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

Feature	the function
${y}'$ > 0	is increasing
${y}'$ < 0	is decreasing
${y}'$ changes – to +	has a local minimum
${y}'$ changes + to –	has a local maximum
${y}'$ increasing	is concave up
${y}'$ decreasing	is concave down
${y}'$ extreme value	has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked.

Here is the same graph with the second derivative features marked.

The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x = -2 because its derivative changes from positive to negative at x = -2.”

Conclusion	Justification
y is increasing	${y}'$ > 0
y is decreasing	${y}'$ < 0
y has a local minimum	${y}'$ changes – to +
y has a local maximum	${y}'$ changes + to –
y is concave up	${y}'$ increasing
y is concave down	${y}'$ decreasing
y has a point of inflection	${y}'$ extreme values

For notes on asymptotes see Asymptotes and the Derivative and Other Asymptotes.

Originally posted on October 26, 2012, and my single most viewed post over the years.

Unit 2 – Definition of the Derivative

Posted on September 14, 2021 by Lin McMullin

This is a re-post and update of the second in a series of posts from last year. It contains links to posts on this blog about the definition of the derivative for your reference in planning. Other updated post on the 2019 CED will come throughout the year, hopefully, a few weeks before you get to the topic.

Unit 2 contains topics rates of change, difference quotients, and the definition of the derivative (CED – 2019 p. 51 – 66). These topics account for about 10 – 12% of questions on the AB exam and 4 – 7% of the BC questions.

Topics 2.1 – 2.4: Introducing and Defining the Derivative

Topic 2.1: Average and Instantaneous Rate of Change. The forward difference quotient is used to introduce the idea of rate of change over an interval and its limit as the length of the interval approaches zero is the instantaneous rate of change.

Topic 2.2: Defining the derivative and using derivative notation. The derivative is defined as the limit of the difference quotient from topic 1 and several new notations are introduced. The derivative is the slope of the tangent line at a point on the graph. Explain graphically, numerically, and analytically how the three representations relate to each other and the slope.

Topic 2.3 Estimating the derivative at a point. Using tables and technology to approximate derivatives is used in this topic. The two resources in the sidebar will be helpful here.

Topic 2.4: Differentiability and Continuity. An important theorem is that differentiability implies continuity – everywhere a function is differentiable it is continuous. Its converse is false – a function may be continuous at a point, but not differentiable there. A counterexample is the absolute value function, |x|, at x = 0.

One way that the definition of derivative is tested on recent exams which bothers some students is to ask a limit like

$displaystyle underset{{xto 0}}{mathop{{lim }}},frac{{tan left( {tfrac{pi }{4}+x} right)-tan left( {tfrac{pi }{4}} right)}}{x}$ .

From the form of the limit students should realize this as the limit definition of the derivative. The h in the definition has been replaced by x. The function is tan(x) at the point where $displaystyle a=tfrac{pi }{4}$ . The limit is $displaystyle {{sec }^{2}}left( {tfrac{pi }{4}} right)=2$ .

Topics 2.5 – 2.10: Differentiation Rules

The remaining topics in this chapter are the rules for calculating derivatives without using the definition. These rules should be memorized as students will be using them constantly. There will be additional rules in Unit 3 (Chain Rule, Implicit differentiation, higher order derivative) and for BC, Unit 9 (parametric and vector equations).

Topic 2.5: The Power Rule

Topic 2.6: Constant, sum, difference, and constant multiple rules

Topic 2.7: Derivatives of the cos(x), sin(x), e^x, and ln(x). This is where you use the squeeze theorem.

Topic 2.8. The Product Rule

Topic 2.9: The Quotient Rule

Topic 2.10: Derivative of the other trigonometric functions

The rules can be tested directly by just asking for the derivative or its value at a point for a given function. Or they can be tested by requiring the students to use the rule of an general expression and then find the values from a table, or a graph. See 2019 AB 6(b)

The suggested number of 40 – 50 minute class periods is 13 – 14 for AB and 9 – 10 for BC. This includes time for testing etc. Topics 2.1, 2,2, and 2.3 kind of flow together, but are important enough that you should spend time on them so that students develop a good understanding of what a derivative is. Topics 2.5 thru 2.10 can be developed in 2 -3 days, but then time needs to be spent deciding which rule(s) to use and in practice using them. The sidebar resource in the CED on “Selecting Procedures for Derivative” may be helpful here.

Other post on these topics

DEFINITION OF THE DERIVATIVE

Local Linearity 1 The graphical manifestation of differentiability with pathological examples.

Local Linearity 2 Using local linearity to approximate the tangent line. A calculator exploration.

Discovering the Derivative A graphing calculator exploration

The Derivative 1 Definition of the derivative

The Derivative 2 Calculators and difference quotients

Difference Quotients 1

Difference Quotients II

Tangents and Slopes

Differentiability Implies Continuity

Adapting 2021 AB 4 / BC 4

FINDING DERIVATIVES

Why Radians? Don’t do calculus without them

The Derivative Rules 1 Constants, sums and differences, powers.

The Derivative Rules 2 The Product rule

The Derivative Rules 3 The Quotient rule

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.the 2019 versions.

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

Adapting 2021 AB 4 / BC 4

Posted on July 13, 2021 by Lin McMullin

Four of nine. Continuing the series started in the last three posts, this post looks at the AP Calculus 2021 exam question AB 4 / BC 4. The series considers each question with the aim of showing ways to use the question with your class as is, or by adapting and expanding it. Like most of the AP Exam questions there is a lot more you can ask from the stem and a lot of other calculus you can discuss.

2021 AB 4 / BC 4

This is a Graph Analysis Problem (type 3) and contains topics from Units 2, 4, and 6 of the current Course and Exam Description. The things that are asked in these questions should be easy for the students, however each year the scores are low. This may be because some textbooks simply do not give students problems like this. Therefore, supplementing with graph analysis questions from past exams is necessary.

There are many additional questions that can be asked based on this stem and the stems of similar problems. Usually, the graph of the derivative is given, and students are asked questions about the graph of the function. See Reading the Derivative’s Graph.

Some years this question is given a context, such as the graph is the velocity of a moving particle. Occasionally there is no graph and an expression for the derivative or function is given.

Here is the 2021 AB 4 / BC 4 stem:

The first thing students should do when they see $G\left( x \right)=\int_{0}^{x}{{f\left( t \right)}}dt$ is to write prominently on their answer page ${G}'\left( x \right)=f\left( x \right)$ and $\displaystyle {G}''\left( x \right)={f}'\left( t \right)$ . While they may understand and use this, they must say it.

Part (a): Students were asked for the open intervals where the graph is concave up and to give a reason for their answer. (Asking for an open interval is to remove any concern about the endpoints being included or excluded, a place where textbooks differ. See Going Up.)

Discussion and ideas for adapting this question:

Using this or similar graphs go through each of these with your class until the answers and reasons become automatic. There are quite a few other things that may be asked here based on the derivative.
- Where is the function increasing?
- Decreasing?
- Concave down, concave up?
- Where are the local extreme values?
- What are the local extreme values?
- Where are the absolute extreme values?
- What are the absolute extreme values?
There are also integration questions that may be asked, such as finding the value of the functions at various points, such as G(1) = 2 found by using the areas of the regions. Also, questions about the local extreme values and the absolute extreme value including their values. These questions are answered by finding the areas of the regions enclosed by the derivative’s graph and the x-axis. Parts (b) and (c) do some of this.
Choose different graphs, including one that has the derivative’s extreme value on the x-axis. Ask what happens there.

Part (b): A new function is defined as the product of G(x) and f(x) and its derivative is to be found at a certain value of x. To use the product rule students must calculate the value of G(x) by using the area between f(x) and the x-axis and the value of ${f}'\left( x \right)$ by reading the slope of f(x) from the graph.

Discussion and ideas for adapting this question:

This is really practice using the product rule. Adapt the problem by making up functions using the quotient rule, the chain rule etc. Any combination of $\displaystyle G,{G}',{G}'',f,{f}',\text{ or }{f}''$ may be used. Before assigning your own problem, check that all the values can be found from the given graph.
Different values of x may be used.

Part (c): Students are asked to find a limit. The approach is to use L’Hospital’s Rule.

Discussion and ideas for adapting this question:

To use L’Hospital’s Rule, students must first show clearly on their paper that the limit of the numerator and denominator are both zero or +/- infinity. Saying the limit is equal to 0/0 is considered bad mathematics and will not earn this point. Each limit should be shown separately on the paper, before applying L’Hospital’s Rule.
Variations include a limit where L’Hospital’s Rule does not apply. The limit is found by substituting the values from the graph.
Another variation is to use a different expression where L’Hospital’s Rule applies, but still needs values read from the graph.

Part (d): The question asked to find the average rate of change (slope between the endpoints) on an interval and then determine if the Mean Value Theorem guarantees a place where $\displaystyle {G}'$ equals this value. Students also must justify their answer.

Discussion and ideas for adapting this question:

To justify their answer students must check that the hypotheses of the MVT are met and say so in their answer.
Adapt by using a different interval where the MVT applies.
Adapt by using an interval where the MVT does not apply and (1) the conclusion is still true, or (b) where the conclusion is false.

Next week 2021 AB 5.

I would be happy to hear your ideas for other ways to use this questions. Please use the reply box below to share your ideas.

Definition of the Derivative – Unit 2

Posted on August 25, 2020 by Lin McMullin

Topics 2.1 – 2.4: Introducing and Defining the Derivative

Topic 2.3 Estimating the derivative at a point. Using tables and technology to approximate derivatives is used in this topic. The two resources in the sidebar will be helpful here.

One way that the definition of derivative is tested on recent exams which bothers some students is to ask a limit like

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{\tan \left( {\tfrac{\pi }{4}+x} \right)-\tan \left( {\tfrac{\pi }{4}} \right)}}{x}$ .

From the form of the limit students should realize this as the limit definition of the derivative. The h in the definition has been replaced by x. The function is tan(x) at the point where $\displaystyle a=\tfrac{\pi }{4}$ . The limit is $\displaystyle {{\sec }^{2}}\left( {\tfrac{\pi }{4}} \right)=2$ .