Definition of the Derivative – Unit 2

Posted on August 25, 2020 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

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

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 The h in the definition has been replaced by x. The function is tan(x) at the point where $\displaystyle a=\tfrac{\pi }{4}$ .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 $\displaystyle {{\sec }^{2}}\left( {\tfrac{\pi }{4}} \right)=2$ .

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

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

The Chain Rule

Posted on September 18, 2018 by Lin McMullin

Most of the function students are faced with in beginning calculus are compositions of the Elementary Functions. The Chain Rule allows you to differentiate composite functions easily. The posts listed below are ways to introduce and use the Chain Rule.

Experimenting with a CAS – Chain Rule Using a CAS to discover the Chain Rule

Power Rule Implies Chain Rule and Foreshadowing the Chain Rule the same ideas.

The Chain Rule

Revised from 9-19-2017

Derivative Formulae

Posted on September 11, 2018 by Lin McMullin

Maria Gaetana Agnesi

So, no one wants to do complicated limits to find derivatives. There are easier ways of course. There are a number of quick ways (rules, formulas) for finding derivatives of the Elementary Functions and their compositions. Here are some ways to introduce these rules; these are the subject of this week’s review of past posts.

Why Radians?

The Derivative I Guessing the derivatives from the definition

The Derivative II Using difference quotients to graph and guess

The Derivative Rules I The Power Rule

The Derivative Rules II Another approach to the Product Rule from my friend Paul Foerster

The Derivative Rules III The Quotient Rule developed using the Power Rule, an approach first suggested by Maria Gaetana Agnesi (1718 – 1799) who was helping her brother learn the calculus.

Next week: The Chain Rule.

Revised from 9-12-2017

Implicit Differentiation

Posted on September 28, 2017 by Lin McMullin

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.

The Chain Rule

Posted on September 19, 2017 by Lin McMullin

Most of the function students are faced with in beginning calculus are compositions of the Elementary Functions. The Chain Rule allows you to differentiate composite functions easily. The posted listed below are ways to introduce and then use the Chain Rule.

Experimenting with a CAS – Chain Rule Using a CAS to discover the Chain Rule

Power Rule Implies Chain Rule and Foreshadowing the Chain Rule the same ideas.

The Chain Rule

Differentiation Techniques

Posted on September 12, 2017 by Lin McMullin

Maria Gaetana Agnesi

Why Radians?

The Derivative I Guessing the derivatives from the definition

The Derivative II Using difference Quotient to graph and guess

The Derivative Rules I The Power Rule

The Derivative Rules II Another approach to the Product Rule from my friend Paul Foerster

The Derivative Rules III The Quotient Rule developed using the Power Rule, an approach first suggested by Maria Gaetana Agnesi (1718 – 1799) who was helping her brother learn the calculus.

Next week: The Chain Rule.

Working up to the derivative.

Posted on August 29, 2017 by Lin McMullin

While limit is what makes all of the calculus work, people usually think of calculus as starting with the derivative. The first problem in calculus is finding the slope of a line tangent to a graph at a point and then writing the equation of that tangent line.

Local Linearity is the graphical manifestation of differentiability. If you zoom-in of the graph of a function (at a point where we will soon say the function is differentiable), the graph eventually looks like a line: the graph appears to be straight, and its slope is the number we will call its derivative.

To do this we need to zoom-in numerically. Zooming-in numerically is accomplished by finding the slope of a secant line, a line that intersects the graph twice, and then finding the limit of that slope as the two points come closer together.

This week’s posts start with local linearity and tangent lines. They lead to the difference quotient and the equation of the tangent line.

Local Linearity I

Local Linearity II Working up to difference quotient. The next post explains this in more detail.

Tangent Lines approaching difference quotients on calculator by graphing tan line.

Next week: Difference Quotients.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Tag Archives: derivative

Definition of the Derivative – Unit 2