Category Archives: Derivative Theory

Parametric Equations, Polar Coordinates, and Vector-Valued Functions – Unit 9

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Unit 9 includes all the topics listed in the title. These are BC only topics (CED – 2019 p. 163 – 176). These topics account for about 11 – 12% of questions on the BC exam.

Comments on Prerequisites: In BC Calculus the work with parametric, vector, and polar equations is somewhat limited. I always hoped that students had studied these topics in detail in their precalculus classes and had more precalculus knowledge and experience with them than is required for the BC exam. This will help them in calculus, so see that they are included in your precalculus classes.

Topics 9.1 – 9.3 Parametric Equations

Topic 9.1: Defining and Differentiation Parametric Equations. Finding dy/dx in terms of dy/dt and dx/dt

Topic 9.2: Second Derivatives of Parametric Equations. Finding the second derivative. See Implicit Differentiation of Parametric Equations this discusses the second derivative.

Topic 9.3: Finding Arc Lengths of Curves Given by Parametric Equations. 

Topics 9.4 – 9.6 Vector-Valued Functions and Motion in the plane

Topic 9.4 : Defining and Differentiating Vector-Valued Functions. Finding the second derivative. See this A Vector’s Derivatives which includes a note on second derivatives. 

Topic 9.5: Integrating Vector-Valued Functions

Topic 9.6: Solving Motion Problems Using Parametric and Vector-Valued Functions. Position, Velocity, acceleration, speed, total distance traveled, and displacement extended to motion in the plane. 

Topics 9.7 – 9.9 Polar Equation and Area in Polar Form.

Topic 9.7: Defining Polar Coordinate and Differentiation in Polar Form. The derivatives and their meaning.

Topic 9.8: Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve

Topic 9.9: Finding the Area of the Region Bounded by Two Polar Curves. Students should know how to find the intersections of polar curves to use for the limits of integration. 

Timing

The suggested time for Unit 9 is about 10 – 11 BC classes of 40 – 50-minutes, this includes time for testing etc.

Previous posts on these topics :

Parametric Equations

Vector Valued Functions

Polar Form

Discovering the MVT

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Today’s Blog is an exploration that will lead up to the Mean Value Theorem (MVT) and, I hope, help your students better understand the MVT and why it is true.

While you may do this by hand, using a graphing calculator will make things way easier. This is good calculator practice and can be done on a graphing, non-CAS, calculator without writing anything down. Try it that way.

Here are the steps to follow. My solution with screen pictures is below.

  1. Choose your favorite differentiable function. Call it f(x) and enter it in your calculator as Y1.
  2. Choose two values, a and b, in the domain of your function. Save (store) these on your calculator as a and b.
  3. Find the slope of the line (a, f(a)) and (b, f(b)). It would be best, but not necessary, that the line intersects the function only at (a, f(a)) and (b, f(b)) not between them, and not be horizontal. Store this in your calculator as m.
  4. Write the equation of the line through (a, f(a)) and (b, f(b)) and enter it as Y2.
  5. Write a function, h(x), that gives the vertical distance between f(x) and the line found in step 3. (Hint: upper curve minus the lower.) Enter this as Y3
  6. Find the x-coordinate local extreme value of h(x). Store this number to c.
  7. Find the slope of the tangent line to f(x) at the value found in step 6.
  8. What do you notice? Compare your result and conclusion with the other in your class. Discuss.

My solution.

Step 1: I choose f\left( x \right)=x+2\sin \left( x \right) and entered this in my calculator as Y1

Step 2: I choose a = 1 and b = 3 and stored them in my calculator.

Step 3: I calculated the slope in my calculator – see first figure.

Step 4: The equation of the line is    y=f\left( a \right)+m\left( {x-a} \right). I entered this as Y2 in my calculator.

Step 5  h\left( x \right)=Y1(x)-Y2(x)=\left( {x+2\sin (x)} \right)-\left( {f\left( a \right)+m\left( {x-a} \right)} \right)

Step 6:  {h}'\left( x \right)=1+2\cos (x)-m

Solve  {h}'\left( x \right)=0 for the value between a and b on your calculator. See second figure.

Step 6 and 7: I stored this value to C in my calculator and computed {f}'(c) on the home screen. See third figure.

Step 8: It is no coincidence that {f}'\left( c \right)=m.

The Mean Value Theorem states that for a function that is continuous on the interval [a, b] and differentiable on the open interval (a, b) there exists a number c in (a, b) such that

\displaystyle {f}'\left( c \right)=\frac{{f\left( b \right)-f\left( a \right)}}{{b-a}}

Additional Exploration:

  1. Can you show why {f}'\left( c \right)=m. ? Hint: Look at the expression for {h}'\left( c \right) in step 5; set it equal to zero. Why must the solution be the value that makes {f}'\left( x \right)=m?
  2. What does this mean graphically?

  1. Pick a different value for a and/or b so that the line between (a, f(a)) and (b, f(b)) intersects f(x) two (or more) times. The derivative will now have two (or more) zeros. Find them and calculate the slope at each one. What do you notice?

Students often confuse the Mean Value Theorem, the Average Rate of Change of a function on an interval, and the Average Value of a function on an interval. This is understandable because of the similarity in their names and the similarity of their results. Be sure to point this out as you teach them and help them learn the meanings of each.

Other posts related to the Mean Value Theorem

Foreshadowing the MVT Other examples using this technique

Existence Theorems

Fermat’s Penultimate Theorem   A lemma for Rolle’s Theorem: Any function extreme value(s) on an open interval must occur where the derivative is zero or undefined.

Rolle’s Theorem   A lemma for the MVT: On an interval if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), there must exist a number c in the open interval (a, b) where f ‘(c) = 0.

Mean Value Theorem I   Proof

Mean Value Theorem II   Graphical Considerations

Darboux’s Theorem   the Intermediate Value Theorem for derivatives.

Mean Tables

The Definite Integral and the FTC

Definition of the Derivative – Unit 2

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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), ex, 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

 

 

 

 

 

2019 CED Unit 5 Analytical Applications of Differentiation

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Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.

Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation Writing on the AP Calculus Exams and its handout

Topics 5.1

Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.

Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval, then it is the absolute (global) maximum or minimum for the function on that interval.

Topic 5.8 Sketching Graphs of Functions and Their Derivatives. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topics 5.10 – 5.11

Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems 

Topic 5.11 Solving Optimization Problems

Topics 5.12

Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.

Timing

Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.

The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context

Previous posts on these topics include:

Then There Is This – Existence Theorems

What’s a Mean Old Average Anyway

Did He, or Didn’t He?   History: how to find extreme values without calculus

Mean Value Theorem

Foreshadowing the MVT

Fermat’s Penultimate Theorem

Rolle’s theorem

The Mean Value Theorem I

The Mean Value Theorem II

Graphing

Concepts Related to Graphs

The Shapes of a Graph

Joining the Pieces of a Graph

Extreme Values

Extremes without Calculus

Concavity

Reading the Derivative’s Graph

Real “Real-life” Graph Reading

Far Out! An exploration

Open or Closed  Should intervals of increasing, decreasing, or concavity be open or closed?

Others

Lin McMullin’s Theorem and More Gold  The Golden Ratio in polynomials

Soda Cans  Optimization video

Optimization – Reflections   

Curves with Extrema?

Good Question 10 – The Cone Problem

Implicit Differentiation of Parametric Equations    BC Topic

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

Seeing the Chain Rule

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Being a believer in the Rule of Four, I have been trying for years to find a good visual (graphical) illustration of why or how the Chain Rule for derivatives works. This very simple example is the best I could come up with.

Consider the function y=\sin \left( x \right),0\le x\le 2\pi . (See figure 1. A tangent segment at \displaystyle \left( {\frac{\pi }{3},\sin \left( {\frac{\pi }{3}} \right)} \right) is drawn.) As you know, this function’s values go smoothly from 0 to 1 to 0 to –1 and back to 0. The slopes of its tangent line, its derivative, appears to go from 1 to 0 to –1 to 0 to 1 as you would expect knowing its derivative is  \frac{{dy}}{{dx}}=\cos \left( x \right). (See figure 2)

Consider the function  y=\sin \left( {3x} \right),0\le x\le 2\pi (See figure 3. A tangent line at \displaystyle \left( {\frac{\pi }{9},\sin \left( {\frac{\pi }{9}} \right)} \right)  is drawn) This takes on all the values of the sine function three times between 0 and  2\pi . It goes through the same values three times as fast and therefore, its rate of change (yeah, the derivative) should be three times as much. Compare the tangent lines in Figures 1 and 3. This agrees with the derivative found by the Chain Rule:  \frac{{dy}}{{dx}}=3\cos \left( {3x} \right). See figure 4)

Next, consider the function y=\sin \left( {\tfrac{1}{2}x} \right),0\le x\le 2\pi (See figure 5. A tangent line at \displaystyle \left( {\frac{{2\pi }}{3},\sin \left( {\frac{{2\pi }}{3}} \right)} \right) is drawn.). This time the function is stretch and only goes through half its period. So, It goes through the same values half as fast as the original and the slope is only half as steep as the original. Compare the tangent lines in Figures 1 and 5.Therefore, the rate of change the derivative, should be only half the original’s. So,  \frac{{dy}}{{dx}}=\frac{1}{2}\cos \left( {\tfrac{1}{2}x} \right) (See figure 6)

I hope this helps your students see what’s happening with the Chain Rule, at least a little bit. I’d be happy to hear and share any ideas you have to illustrate the Chain Rule graphically.

There is a movable Desmos graph here to help illustrate all of this.

Here are links to other posts on the Chain Rule

Foreshadowing the Chain Rule

The Power Rule Implies Chain Rule

The Chain Rule

Derivative Practice – Numbers

Derivative Practice – Graphs

Experimenting with CAS – Chain Rule

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

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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:

Foreshadowing the Chain Rule

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

Differentiability Implies Continuity

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An important theorem concerning derivatives is this:

If a function f is differentiable at x = a, then f is continuous at x = a.

The proof begins with the identity that for all x\ne a

\displaystyle f\left( x \right)-f\left( a \right)=\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}

\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {x-a} \right)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}

\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=0\cdot {f}'\left( a \right)=0

And therefore, \underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)

Since both sides are finite, the function is continuous at x = a.

The converse of this theorem is false: A continuous function is not necessarily differentiable. A counterexample is the absolute value function which is continuous at the origin but not differentiable there. (The slope approaching from the left is not equal to the slope from the right.)

This is a theorem whose contrapositive is used as much as the theorem itself. The contrapositive is,

If a function is not continuous at a point, then it is not differentiable there.

Example 1: A function such as  \displaystyle g\left( x \right)=\frac{{{{x}^{2}}-9}}{{x-3}} has a (removable) discontinuity at x = 3, but no value there.

So, in the limit definition of the derivative, \displaystyle \text{ }\!\!~\!\!\text{ }\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{g\left( {3+h} \right)-g\left( 3 \right)}}{h} there is no value of g(3) to use, and the derivative does not exist.

Example 2:  \displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x\le 1} \\ {{{x}^{2}}+3} & {x>1} \end{array}} \right.. This function has a jump discontinuity at x = 1.

Since the point (1, 1) is on the left part of the graph, if h > 0, f\left( {1+h} \right)-f\left( 1 \right)>3 and the limit  will always be a number greater than 3 divided by zero and will not exist. Therefore, even though the slopes from both side of x =1 approach the same value, namely 2, the derivative does not exist at x = 1.

This also applies to a situation like example 1 if f(3) were some value that did not fill in the hole in the graph.

 

On the AP Calculus exams students are often asked about the derivative of a function like those in the examples, and the lack of continuity should be an immediate clue that the derivative does not exist. See 2008 AB 6 (multiple-choice).

Just as important are questions in which the function is given as differentiable, but the student needs to know about continuity. Just remember: differentiability implies continuity. See 2013 AB 14 in which you must realize the since the function is given as differentiable at x = 1, it must be continuous there to solve the problem.

Continuity of the Derivative

A question that comes up is, if a function is differentiable is its derivative differentiable? The answer is no. While almost always the derivative is also differentiable, there is this counterexample:

\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}\sin \left( {\frac{1}{x}} \right)} & {x\ne 0} \\ 0 & {x=0} \end{array}} \right.

The first line of the function has a removable oscillating discontinuity at x = 0, but since the \displaystyle {{x}^{2}} factor squeezes the function to the origin; the added condition that \displaystyle f\left( 0 \right)=0 makes the function continuous. Differentiating gives

\displaystyle {{f}^{'}}\left( x \right)={{x}^{2}}\cos \left( {\frac{1}{x}} \right)\left( {\frac{{-1}}{{{{x}^{2}}}}} \right)+2x\sin \left( {\frac{1}{x}} \right)=-\cos \left( {\frac{1}{x}} \right)+2x\sin \left( {\frac{1}{x}} \right)

And now there is no way to get around the oscillating discontinuity at x = 0.