Calculus - Korey And Math, Teaching Mathematics · 2020. 7. 16. · Calculus Korey Nishimoto. Math 241 Notes Korey Nishimoto Math Department, Kapiolani Community College January 15,

Cal

culu

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Korey Nishimoto

Math 241Notes

Korey NishimotoMath Department, Kapiolani Community College

January 15, 2020

ContentsMath 241 Page 3

Contents

1 Limits and Their Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Finding Limits Graphically and Numerically . . . . . . . . . . . . . . . 4

1.2.1 Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Limits That Fail to Exist . . . . . . . . . . . . . . . . . . . . . 6

1.3 Evaluating Limits Analytically . . . . . . . . . . . . . . . . . . . . . . 81.3.1 Strategies for Finding Limits . . . . . . . . . . . . . . . . . . . 11

1.4 Continuity and One-Sided Limits . . . . . . . . . . . . . . . . . . . . . 171.4.1 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . 19

1.5 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.6 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.6.1 Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . 262 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.1 The Derivative and the Tangent Line Problem . . . . . . . . . . . . . . 292.1.1 Differentiability and Continuity. . . . . . . . . . . . . . . . . . 32

2.2 Basic Differentiation Rules and Rates of Change . . . . . . . . . . . . . 352.3 Product and Quotient Rules and Higher Order Derivatives . . . . . . . . 41

2.3.1 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . 432.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.5 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 492.6 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3 Applications of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 593.1 Extrema on an Interval . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2 Rolle’s Theorem and The Mean Value Theorem . . . . . . . . . . . . . 643.3 Increasing and Decreasing Functions and the First Derivative Test. . . . 703.4 Concavity and The Second Derivative Test . . . . . . . . . . . . . . . . 753.6 A Summary of Curve Sketching . . . . . . . . . . . . . . . . . . . . . . 823.7 Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 863.8 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Limits andTheir Properties

Math 241 Finding Limits Graphically and Numerically Page 4

1 Limits and Their Properties

1.2 Finding Limits Graphically and Numerically

The limit is asking us if the function (output) f(x) is converging as the input variablex approaches some value. When we thing about the key word in English we get a moreintuitive understanding of what a limit is doing. The word converge is defined by theWebsters dictionary as ”to tend or move toward one point or one another : come together:”. This is exactly what the math definition means as well.

1.2.1 Introduction to Limits

Does the height of the function, or output, approach a certain value as we move to a certainvalue of the input from both sides. This idea will expand as we discuss another version ofthe limit. Let’s see what this means visually. Consider the graph of f(x) = x2.

−4 −3 −2 −1 1 2 3 4

−4

−3

−2

−1

1

2

3

4

x

y

Figure 1: f(x) = x2

What is the limit of the function as x approaches 1? Can we see that as we follow thegraph from the left and the right to where x = 1, the height of the function is height one.This means that the limit of the function f(x) is 1.

If the limit exists we write this mathematically as

limx→c

f(x) = L.

The c variable is some real number and so is L. In the problem above, we would writelimx→1

f(x) = 1. What you should notice about the definition of the limit is the absence of thefunction existing at the point that it converges at. The function does not need to exist forthe limit to exist.

Example: 1.2.1:

Does the function f(x) = x3+8x+2 have a limit as x tends to −2.

f(x) = x3 + 8x+ 2 =

(x+ 2)(x2 − 2x+ 4)x+ 2 = x

2 − 2x+ 4 x 6= −2

This function is undefined at x = −2 and graphically it has a hole at that point.



−4 −3 −2 −1 1 2 3 4

−14−12−10−8−6−4−2

2468

101214

◦

x

y

Figure 2: f(x) = x3+8x+2

If we continue to use the concept that a limit is convergence, or the function isapproaching a particular height as the input is approaching a particular value, theneven thought the graph has a hole at the point (−2, 12) this function has a limit.The function still approaches the height of 12 as we follow the graph from the leftand the right. This means that this function has a limit at x = −2 and the limit is12.

limx→2

f(x) = 12.

Both of these examples have used a graph to determine if the function has a limit, butthis is something that is difficult to do if you do not know what the graph looks like. Thinkabout what we have been saying this whole time. We need the inputs to approach a givenvalue and check to see if the function approaches a given height. This can be done byplugging into the function. Let’s use the functions above to determine if our visual can berepresented numerically.

Example: 1.2.2:

1. Show that f(x) = x4 has a limit as x tends to 1 numerically.We will pick values of x that are approaching x = 1

x = 0 .5 .9 .99 1.01 1.1 1.5 2f(x) = 0 .25 .81 .9801 1.0201 1.21 2.25 4

We can see that the output of the function f(x) is approaching height 1 as thex value approaches 1 from the left and the right hand side.

2. Show that f(x) = x3+8x+2 has a limit as x tends to 2 numerically.We will pick values of x that are approaching x = 1

x = 1 1.5 1.8 1.99 2.01 2.2 2.5 3f(x) = 1 3.35 3.64 3.9801 4.0201 4.44 5.25 7

We can see that the output of the function f(x) is approaching height 4 as thex value approaches 2 from the left and the right hand side even thought thefunction is undefined at x = 2.



1.2.2 Limits That Fail to Exist

There are cases where the limit fails to exist. Let’s consider the description of limits thatwe have discussed earlier. A function that approaches a certain output/height as the inputapproaches a certain value has a limit. This also implies that if the function does notapproach a certain output/height as the input value approaches a certain value does nothave a limit. Let’s look a few examples of this.

Example: 1.2.3:

Determine if the function

f(x) ={x if x < 313x

2 − 3 if x ≥ 3

has a limit as x tends to 3. Calculate limx→3

f(x).

Let’s again look at the graph to determine if the function has a limit as x approaches3.

−6 −5 −4 −3 −2 −1 1 2 3 4 5 6

−6−5−4−3−2−1

123456

◦

• x

y

Figure 3: f(x) = x if x < 3 and f(x) = 13x2 − 3 if x ≥ 3

Notice that the function approaches height 3 from the left while it approaches 0 fromthe right. This means that the function does not have a limit at x = 3. You cannotconverge to 2 different values at the same time.

This example leads us to a theorem about the existence of a limit of a function.Theorem 1.2.1:

A function f(x) has a limit at x = c if and only if f(x) has a left hand limit (convergesfrom the left), a right hand limit (converges from the right, and they are the same.This is written mathematically as

limx→c−

f(x) = limx→c

f(x) = limx→c+

f(x)

Notice that we have been using this the entire time. Moving along the graph of thefunction from the left and the right to see what height the function converges to. We nowknow a formal mathematical theorem that will allow us to determine if the functions has alimit.



Example: 1.2.4:

Determine if the function f(x) = x+1x−2 has a limit as x tends to 2.

−10−9−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8 9 10

−10−9−8−7−6−5−4−3−2−1

123456789

10

x

y

Figure 4: f(x) = x+1x−2

Similarly as the previous example, notice that the function approaches negative in-finity as we approach x = 2 from the left and we approach infinity as we approachx = 2 from the right. This also means that the limit does not exist.

We call functions that approaches the same infinity from both sides as divergent. Thesefunctions still do not have a limit at the divergent point, they just have a special nameattached to them.

Example: 1.2.5:

Does the function f(x) = sin( 1x ) have a limit as x tends to 0.

−1 1

−2

−1

1

2

x

y

Figure 5: f(x) = sin( 1x )

This function is an oscillating function that bounces from −1 to 1. It also get moreerratic as we get closer to 0. Since we would never approach a fixed height value,this function does not converge as x tends to 0. As a side note though, this functionwill converge as x tends to c for all other values of c other than 0.


Math 241 Evaluating Limits Analytically Page 8

1.3 Evaluating Limits Analytically

Limits are easy to calculate when the function is continuous. We will discuss this more in thenext section. For now let’s use an intuitive understanding of continuous. To be continuousmeans that there is no breaks, jumps, or holes. We can see in some of the examples offunctions which were discontinuous. The perk of knowing that a function is continuous isthe following theorem.

Theorem 1.3.1:

If f(x) is continuous at x = c, then

limx→c

f(x) = f(c).

This is a powerful theorem and can be used as a substitute for many of the followingproperties that will be discussed.

Theorem 1.3.2:

Let b and c be real numbers, let n be a positive integer, and let f and g be functionswith limits

limx→c

f(x) = L and limx→c

g(x) = K

1. limx→c

b = b 2. limx→c

x = c 3. limx→c

xn = cn

4. Scalar multiple: limx→c

b · f(x) = b limx→c

f(x) = bL

5. Sum or difference: limx→c

f(x)± g(x) = limx→c

f(x)± limx→c

g(x) = L±K

6. Product: limx→c

f(x) · g(x) = limx→c

f(x) · limx→c

g(x) = LK

7. Quotient: limx→c

f(x)g(x) =

limx→c

f(x)

limx→c

g(x) =LK where K 6= 0.

8. Power: limx→c

(f(x))n = Ln

Note that properties 4-8 are allowed only when the functions f and g have limits inthe first place. You cannot split the limit if either g or f do not have a limit.

Let’s do examples of each of these properties and describe how they are used.Example: 1.3.1:

Find the limit of1. lim

x→−24

limx→−2

4 = 4

This is because the constant function y = 4 is al-ways at height 4. Therefore when we approach x = −2from the left and the right we approach function height 4.

2. limx→7

x

limx→7

x = 7

This is because the function f(x) = x is a continuous function. (It is linear)Hence we can plug in c into x to get f(c) = c. So the limit is f(7) = 7.



3. limx→π

xn

limx→π

xn = πn

Similarly to part 2, xn is a continuous function, so the limit is given by f(π) =πn.

4. limx→ 12

(4x2 + 8)

limx→ 12

(4x2 + 8) = limx→ 12

4(x2 + 2)

= 4 limx→ 12

(x2 + 2)

= 4((

12

)2+ 2)

= 1 + 8

= 9

This uses two properties. The scalar multiple and continuous function proper-ties. First we can factor out a 4 and then pull it out of the limit using ”scalarmultiple”. We then use the fact that x2 +2 is a continuous function, so we plugin to get the limit.

5. limx→5

[(x2 − 5) +√x+ 4 ]

limx→5

[(x2 − 5) +√x+ 4 ] = lim

x→5(x2 − 5) + lim

x→5

√x+ 4

= (52 − 5) +√

5 + 4

= 20 + 3

= 23

This uses the addition property. Since both the limits for limx→5

(x2 − 5) andlimx→5

√x+ 4 exist we can split the limit and then apply the continuity property.

The continuity property however is powerful. If you know that the functionis continuous, then you can plug in directly. This is the case for this function.It is continuous on an open interval of x = 5. So we can plug in x = 5 directlyand we will get the same answer.

6. limx→−3

(x+ 1)(3x2 + 2)



limx→−3

(x+ 1)(3x2 + 2) = limx→−3

(x+ 1) · limx→−3

(3x2 + 2)

= (−3 + 1)(3(−3)2 + 2)

= −2(29)

= −58

Since both limx→−3

(x+ 1) and limx→−3

(3x2 + 2) both exist we can use the productproperty.

7. limx→0

x+12x2+3x−5

limx→0

x+ 12x2 + 3x− 5 = limx→0

x+ 1(2x+ 5)(x− 1)

= −15

In this problem we first have to show that the function is not zero at thepoint being discussed. While we may still be able to determine the limit ifthe function does not exist there, we want to use the properties of limits. Todo this the denominator cannot be 0 and the limits of the numerator and thedenominator have to exist. Both of these are true so we can use the quotientrule.

8. limx→2

(x2 − 3)4

limx→2

(x2− 3)4 = (22− 3)4 = 1 Since the limx→2

(x2− 3) exists we can use the powerproperty.

During the last example we have used many properties. We have even used propertiesthat are much more difficult, probably without you even realizing it. We will now look atthese properties.

Theorem 1.3.3:

Some more advanced properties of limits.

1. If p(x) is a polynomial and c is a real number, then

limx→c

p(x) = p(c).

This is because every polynomial is continuous and we can therefore plug in cto calculate the limit.

2. If r(x) = p(x)q(x) and c is a real number such that q(c) 6= 0, then

limx→c

r(x) = r(c) = p(c)q(c) .

This is also because of continuity. The function r(x) is continuous at all pointof the real line where the denominator is not equal to zero.

3. Let n be a positive integer. The limit below is valid for all values of n when n



is odd. If n is even, then the limit is valid when c is positive. That is

limx→c

n√x = n

√c

4. if f and g are functions such that limx→c

g(x) = L and limx→L

f(x) = f(L), then

limx→c

f(g(x)) = f(

limx→c

g(x))

= f(L).

The only one that we haven’t seen yet is part 4. So let’s look at an example of thisproperty.

Example: 1.3.2:

Find the limit of limx→2

(x2 + 2x+ 1) 12 .

This can be written as a composition of functions where g(x) = x2 + 2x + 3 andf(x) = x 12 . Notice that lim

x→2g(x) = 22 + 2(2) + 1 = 9. Since f(x) is continuous when

x is positive, limx→9

f(x) = f(9) = (9) 12 = 3.So using part 4, we get

limx→2

f(g(x)) = f( limx→2

g(x)) = f(L) = 3

The last part of the properties of limits are the limits of trig functions. We can howeversee what these items will do based on our knowledge of trig functions. We have all seenthat the trig functions are continuous on their domain. That is every point where the trigfunction exists, the function is continuous and we can therefore plug in the value c into thefunction.

Theorem 1.3.4:

Let c be a real number in the domain of the given trigonometric function.

1. limx→c

sin(x) = sin(c) 2. limx→c

cos(x) = cos(c) 3. limx→c

tan(x) = tan(c)

4. limx→c

sec(x) = sec(c) 5. limx→c

csc(x) = csc(c) 6. limx→c

cot(x) = cot(c)

Example: 1.3.3:

Find the limit of

1. limx→π2

sin(x) = sin(π2 ) = 1

2. limx→π

x tan(x) = π tan(π) = π(0) = 0

3. limx→π4

(sec(x))2 = sec2(π4 ) = (√

2)2 = 2

1.3.1 Strategies for Finding Limits

Simplification

One of the most common strategies for finding a limit is simplification. Simplificationis as it sounds, making the problem simpler. If the functions that we use are algebraicallyequivalent up to some restriction on the domain, then the functions graphically are the sameup to some hole or removable discontinuity.



Theorem 1.3.5:

Let c be a real number, and let f(x) = g(x) for all x 6= c in an open interval containingc. If the limit of g(x) as x approaches c exists, then the limit of f(x) also exists and

limx→c

f(x) = limx→c

g(x)

This theorem states that if we cannot determine the limit of a function f , we should tryto simplify the function into a different function g and see if we can calculate the limit of ginstead.

Example: 1.3.4:

Find the limit of the following

1. limx→−5

x2−25x2+2x−15 .

Notice that if we plug in x = −5 we get division by zero. This function is notcontinuous at x = −5 and therefore we cannot plug in. We can however simply.

limx→−5

x2 − 25x2 + 2x− 15 = limx→−5

(x− 5)(x+ 5)(x+ 5)(x− 3)

= limx→−5

x− 5x− 3 Using theorem 1.3.5

= −5− 5−5− 3

= 54

2. limz→8

2z2−17z+88−z

Similarly to the problem above, we cannot plug in z = 8. So let’s simplify andsee what happens.

limz→8

2z2 − 17z + 88− z = limz→8

(2z − 1)(x− 8)−(z − 8)

= limz→8−(2z − 1) Using theorem 1.3.5

= −(2(8)− 1)

= −15

Rationalizing

Another strategy is rationalizing. Rationalizing can again be used to remove some ofthe troubles that we run into when evaluating limits. Let’s look at an example where wecontinue to use theorem 1.3.5 but now with rationalizing instead of cancelation.



Example: 1.3.5:

Find the limit limx→−3

√2x+22−4x+3

Notice again that plugging in x = −3 does not work since we have division byzero. So we need to algebraically manipulate this function. Our previous strategy ofsimplification is not as easy. Let’s rationalize the numerator and see what happens.

limx→−3

√2x+ 22− 4x+ 3 = limx→−3

√2x+ 22− 4x+ 3 ·

√2x+ 22 + 4√2x+ 22 + 4

= limx→−3

2x+ 22− 16(x+ 3)(

√2x+ 22 + 4)

= limx→−3

2(x+ 3)(x+ 3)(

√2x+ 22 + 4)

= limx→−3

2√2x+ 22 + 4

theorem 1.3.5

= 2√2(−3) + 22 + 4

= 28

= 14

Squeeze Theorem

We will use this last strategy to show some special trigonometric limits. Before we cando that we must first know what the Squeeze Theorem is.

Theorem 1.3.6:

Squeeze Theorem: If h(x) ≤ f(x) ≤ g(x) for all x in an open interval containing c,except possibly at c itself, and if

limx→c

h(x) = L = limx→c

g(x),

then limx→c

f(x) = L.We can see this visually.

x

y

Can we see that the green graph is always bigger or equal to the red. Similarly withthe orange and the red but smaller. Both the orange and the green graph converge



to zero as x→ 0. So the red graph must also have a limit of 0 as x→ 0.

This theorem is quite difficult to use but is very useful when evaluating limits that youcan find bounds to.

Theorem 1.3.7:

1. limx→0

sin(x)x = 1 2. limx→0

1−cos(x)x = 0

Proof. 1. There are more complicated and mathematical approaches to showing thissqueeze theorem, but for now let’s look at the graph bellow.

−4 −2 2 4

−2

−1

1

2

x

y

Figure 6: f(x) = sin(x)x , g(x) = cos(x), and h(x) = 1

If you look at the interval from (−4, 4), we have that g(x) ≤ f(x) ≤ h(x). This is theopen interval required to use the squeeze theorem. We can now calculate the limit off by calculating the limit of g and h.

g(x) ≤f(x) ≤ h(x)

cos(x) ≤ sin(x)x≤ 1

limx→0

cos(x) ≤ limx→0

sin(x)x≤ limx→0

1

1 ≤ limx→0

sin(x)x≤ 1

This says limx→0

sin(x)x = 1 by the squeeze theorem.



2. This can be shown using the conjugate.

limx→0

1− cos(x)x

= limx→0

1− cos(x)x

· 1 + cos(x)1 + cos(x)

= limx→0

1− cos2(x)x(1 + cos(x))

= limx→0

sin2(x)x(1 + cos(x))

= limx→0

sin(x)x· sin(x) · 1(1 + cos(x))

= 0(1)(

12

)Using part one

= 0

As always, let’s now apply the ideas that we have just learned.Example: 1.3.6:

Evaluate the limits.

1. limx→0

sin(4x)x

Notice that if you plug in directly you get a fraction of the form 00 . This iscalled an indeterminate form and will need some extra work to solve.

limx→0

sin(4x)x

= limx→0

4 sin(4x)4x

= 4 limx→0

sin(4x)4x

= 4(1) By theorem 1.3.7

There are more indeterminate forms that you will learn later in your calculuscareer.

2. limx→0

sec(x)−1x



limx→0

sec(x)− 1x

= limx→0

1cos(x) − 1

x

= limx→0

1cos(x) −

cos xcos x

x

= limx→0

1−cos(x)cos(x)

x

= limx→0

1− cos(x)x cos(x)

= limx→0

1− cos(x)x

· 1cos(x)

= 0(1)

= 0


Math 241 Continuity and One-Sided Limits Page 17

1.4 Continuity and One-Sided Limits

We have already discussed this topic quite a bit. Intuitively, continuity is represented by afunctions that does not have any, removable discontinuities and no non-removable discon-tinuities. You may have learned this in another class as holes and vertical asymptotes. Inboth of these cases, the holes and the vertical asymptotes are restrictions on the domain. Agood check is to see if a function is continuous by checking the restrictions on the domain.

• If the function is continuous at the c, then we can plug in to evaluate the limit.

• If the function is discontinuous at c, then we need to use one of our strategies toevaluate the limit.

Example: 1.4.1:

Determine if the functions are continuous. What do you do to evaluate the givenlimit?

1. f(x) = 1x and limx→0 f(x)This function is discontinuous at x =0. This means thah we cannot plug 0into f to evaluate the limit. We willhave to try one of the other techniquesto remove the division by zero.

2. g(x) = x2−4x+2 and limx→−2 g(x)This function is discontinuous at x =−2. We again cannot plug in to evalu-ate the limit. Using factoring and sim-plification, we can evaluate this limit.

3.

f(x) =

0 x ≤ 0x 0 < x ≤ 2x2 x > 2

limx→2

f(x)

Notice that this function is continuouseverywhere except at x = 2. We cansee this graphically, or using the factthat when we evaluate x at 2 we get adifferent output that when we use x2.Hence the function doesn’t approachthe same value from the left and theright and has no limit.

4. h(x) = cos(x) limx→0

h(x)

This function is continuous. (Some-thing you learned in trig.) We can thusplug in x = 0 into h to evaluate thelimit.

We have already discussed much of what is needed to evaluate limits. Just as a reminder.

• A function has a limit only if its left hand limit and its right hand limit are the same.

• You may plug into a function to evaluate a limit only when the function is continuous.

• Properties of the limit can only be applied if both the limits exist and the limits arein the domain of the combined functions.

Similar to these properties we have properties of continuity.Theorem 1.4.1:

If b is a real number and f and g are continuous functions at x = c, then the functionslisted below are also continuous.

1. Scalar multiplication: bf 2. Sum or Difference: f ± g



3. Product: fg 4. Quotient: fg where g(c) 6= 0

If g is continuous at c and f is continuous at g(c), then

5. (f ◦ g)(x) = f(g(x)) is continuous at c.

Example: 1.4.2:

Determine if the functions are continuous at the given c value.

1. f(x) = cos(x) + sin(x) c = 2Since both cos(x) and sin(x) are con-tinuous everywhere, their sum is alsocontinuous everywhere and hence con-tinuous at c = 2.

2. f(x) = 4(x2 + 1) c = 3The function x2 + 1 is continuous ev-erywhere. Hence a scalar multiple ofit will also be continuous everywhere.So f is continuous at c = 3.

3. f(x) = sin(x) · (x+ 3)2 c = 6

Both of these functions are continuouseverywhere. Similarly as before thisproduct must be continuous at c = 6.

4. f(x) = sin(x)x2−4 c = 2 Both of these func-tions are continuous everywhere, butdivision has an added restriction thatthe denominator cannot be zero at c.This one is and hence f is not contin-uous.

5. f(x) = sin(x2 + 1) c = πThe function x2 + 1 is continuous at c = π.. We also have that sin(x) iscontinuous at π2 + 1. So the composition is also continous.

Let’s determine where a function is continuous on a few more example that are a bitmore complicated.

Example: 1.4.3:

Where are the function is continuous.

1. f(x) = tan(x)The function f(x) = tan(x) = sin(x)cos(x) is discontinuous when cos(x) = 0. Thishappens when we have . . . ,− 3π2 ,−

π2 ,

π2 ,

3π2 ,

5π2 , . . .. This can be written as

π2 + nπ where n is an integer. This means that the function is continuouseverywhere else or

. . .

(−3π2 ,−

π

2

)∪(−π2 ,

π

2

)∪(π

2 ,3π2

)∪(

3π2 ,

5π2

). . . .

The next problem is a tricky example that I believe warrants some explanation.

f(x) = xsin(x)

We know that this has a restriction on its domain dependent on when sin(x) = 0. Thishappens at multiples of π of nπ where n is an integer. In a previous class you where taughtthat the division by 0 gives you a vertical asymptote. So no limits should exist when xapproaches one of the nπ values. However our assumption was wrong. This is only truein rational functions where the numerator and the denominator are polynomials. You can



check in DESMOS that it does have a limit at x = 0. While it is not continuous at x = 0the limit exists. This is possible and was discussed in the first section. I will leave it up toyou to check that it works.

1.4.1 Intermediate Value Theorem

The Intermediate Value Theorem is a nice theorem which intuitively should make sense. Ifyou have a continuous function on a closed interval where, then the function must hit allthe heights between the outputs of the functions at the end points.

Let’s think about this. The function has no holes, breaks, or jumps. It also hits heightf(a) and f(b). So shouldn’t the function also hit all heights in between? Look at the graphbelow and consider the closed interval [−3, 1].

−4 −2 2 4

−4

−2

2

4

x

y

Figure 7: f(x) = x3 + 3x2

The green lines represent the function heights at f(−3) and f(1). Notice that the redfunction hits all the height values between these two lines. This is the idea of the intermediatevalue theorem.

Formally, this is written asTheorem 1.4.2:

If f is continous on the closed interval [a, b], f(a) 6= f(b), and k is any numberbetween f(a) and f(b), then the is at least one value c in [a, b] such that f(c) = k.

The most common application of the Intermediate Value Theorem is finding roots offunctions. This can be seen in the following example.

Example: 1.4.4:

Determine if the function f(x) = x5+2x4−3x+2 has a root on the interval (0, 2).

To do this we must first determine that the function is continuous on some closedinterval. In this case we can use [0, 2]. We know this because the function is discon-tinuous at x = −2 and no where else. Now look at the function at a = 0 and b = 2.We have

f(a) = f(0) = 05 + 2(0)4 − 3

0 + 2 = −32

We also havef(b) = f(2) = 2

5 + 2(2)4 − 32 + 2 .

We don’t need an exact number. We only need to know that it is positive. Thismeans that the function was negative, then positive. Hence we can find some k valuein between 0 and 2 such that f(k) = 0. (f(a) < 0 < f(b)).

https://www.desmos.com


Math 241 Infinite Limits Page 20

1.5 Infinite Limits

So far we have glanced over situations where the function increases to infinity or decreasesto negative infinity. These limits do not exist even though we write them as

limx→c

f(x) =∞

orlimx→c

f(x) = −∞

This notation is used when the function approaches the same infinity as we approach x = cfrom the left and right hand side. If they approach different infinities, then we say that thelimit does not exist (DNE).

Let’s look at a few examples that we can see visually.Example: 1.5.1:

Determine the limit of the following functions.

1. limx→3

x2+2x+1x−3

−20 20

−20

20

x

y

Figure 8: f(x) = x2+2x+1x−3

Visually, we can see that the function approaches −∞ as we approach x = 3from the left ( lim

x→3−x2+2x+1x−3 ) and ∞ as we approach x = 3 from the right.

( limx→3+

x2+2x+1x−3 ) This tells us that the limit does not exist. If you are unable

to see this visually, you have other means of determining the limit numerically.Plug in numbers that approach x = 3 to the left and the right to see whathappens to the function.

2. limx→3

x+1x2−6x+9



−10 −5 5 10

−10

−5

5

10

x

y

Figure 9: f(x) = x+1x2−6x+9

Notice that as we approach x = 3 from both sides we get ∞. This means thatthe limit does not exist, but we can write it as

limx→3

x+ 1x2 − 6x+ 9 =∞

Can we see that these functions will have a limit which approaches either ∞ or −∞when we have a vertical asymptote at x = c. You have also done many problems in previousmath classes to determine the sign of functions using multiplicity of roots and verticalasymptotes. To use these concepts we must first determine if the division by zero leads toa vertical asymptote.

Theorem 1.5.1:

Let f and g be continuous on an open interval containing c. If f(c) 6= 0, g(c) = 0 andthere exists and open interval where g(x) 6= 0 for all x values in the open intervalother than c, then the function

h(x) = f(x)g(x)

has a vertical asymptote at x = c.

It is really important to note that this theorem requires that f(c) 6= 0 while g(c) = 0. Ifwe had bot equalling zero we have an indeterminate form which can cause problems. Whenthis happens, we must check to see if the function has a vertical asymptote at that point ora hole by checking to see if the right and left hand limits are the same.

Example: 1.5.2:

Determine if the following function has a vertical asymptote at the given x value andfind the limit asked.

f(x) = x+ 1x+ 2 at x = −2. Find limx→−2 f(x).

Notice that x+ 1 is not zero at x = −2 but x+ 2 is. We also have that the functionx+2 exists everywhere, hence we may apply theorem 1.5.1 to conclude that f(x) hasa vertical asymptote at x = −2 by using a sign chart we can determine what infinitythe function approaches.

−9 −8 −7 −6 −5 −4 −3 0 1 2 3 4 5 6 7 8 9−1−2

+ − +

x



We can determine this sign chart by plugging in values in each of the intervals givenby the roots and the vertical asymptotes, or by using the multiplicity rules. (Oddmultiplicity the sign changes, even multiplicity the sign stays the same.) In the end,we only care about the −2 part since that is where we are calculating the limit, butwe needed the x = −1 root to ensure that we plugged in a value that was in thecorrect interval. Notice that the function is positive to the left of −2 and negativebetween −2 and −1. This means that

limx→−2−

f(x) =∞

andlim

x→−2+f(x) = −∞.

Thus the limit does not exist.

You can also see thisthrough the graph. Thisis essentially what we didby creating a sign chart.

−10 −5 5 10

−10

−5

5

10

x

y

Example: 1.5.3:

Determine if the following function has a vertical asymptote at the given x value andfind the limit asked.

f(x) = x cot(x) at x = 0. Find limx→0

f(x).

This function is equivalent to x cos(x)sin(x) . The numerator and the denominator are bothzero at x = 0. This means that we need to find the limit in order to determine if ithad a vertical asymptote or not. In this case theorem 1.5.1 does not apply. We willhave to calculate using a trick. We know from theorem 1.3.7 lim

x→0sin(x)x = 1. We can

use this to determine that

limx→0

x

sin(x) = limx→01

sin(x)x

=limx→0

1

limx→0

sin(x)x

= 11

= 1

We can now apply this as such.

limx→0

x cos(x)sin(x) = limx→0

x

sin(x) · cos(x)

= limx→0

x

sin(x) · limx→0 cos(x)

= 1(1)

= 1

Since this limit exists and is equal to 1, x = 0 is not a vertical asymptote. It is ahole and is not part of the domain.

Similarly to all the sections previously, we would like to know some rules that allow usto calculate the limit easier. We have the following properties of infinite limits.



Theorem 1.5.2:

Let c and L be real numbers, and let f and g be continuous functions such thatlimx→c

f(x) =∞ and limx→c

g(x) = L, then

1. limx→c

[f ± g] =∞

2. limx→c

f(x)g(x) =∞ L > 0

limx→c

f(x)g(x) = −∞ L < 0

3. limx→c

g(x)f(x) = 0 Similar properties hold for one sided limits and for functions that

approach −∞ as x→ c.


Math 241 Limits at Infinity Page 24

1.6 Limits at Infinity

Remember, the goal of finding the limits is to determine what the function height is ap-proaching as we approach the given input value. This is still true in the next topic with aslight modification with respects to the left and right hand limit.Definition 1.6.1: Definition of a limit at infinity: Let L be a real number.

1. For every � > 0 there exists an M > 0 such that |f(x)−L| < � for x > M implies thatlimx→∞

f(x) exists andlimx→∞

f(x) = L

2. For every � > 0 there exists an N < 0 such that |f(x)−L| < � for x < N implies thatlimx→∞

f(x) exists andlim

x→−∞f(x) = L

This means that the function height converges to L as we move further out to the leftor the right hand side. Notice that since we cannot approach ∞ from the right we cannothave a right hand limit and conversely for −∞. We use these to determine the action of thefunction in its tail ends.

Think about what is happening in the two situations. Either the input value is gettingvery large, or very small. We can see what happens graphically and numerically so get abetter understanding of the concept.

Example: 1.6.1:

Calculate the limit limx→∞

1x and limx→−∞

1x .

Notice that we can see what the limit will be in both cases graphically. In case youforgot what the graph of 1x looks like

−10−9−8−7−6−5−4−3−2−1 1 2 3 4 5 6 7 8 9 10

−10−9−8−7−6−5−4−3−2−1

123456789

10

x

y

We can see that both of these limits converge to zero as we approach arbitrarily largeor small values. Numerically this looks like,

x = 10 100 1000 10000 105 106 107 108

f(x) = 1101

1001

10001

100001

1051

1061

1071

108

and

x = -10 -100 -1000 -10000 −105 −106 −107 −108

f(x) = − 110 −1

100 −1

1000 −1

10000 −1

105 −1

106 −1

107 −1

108

This tells us thatlimx→∞

1x

= limx→−∞

1x

= 0



Using this information we can understand the following common limits at infinity.Theorem 1.6.1:

1. If r is a positive rational number and c is any real number, then

limx→∞

1xr

= limx→−∞

1xr

= 0

2. If m and n are positive rational number and m > n, then

limx→∞

anxn + an−1xn−1 + . . .+ a2x2 + a1x+ a0

amxm + am−1xm−1 + . . .+ a2x2 + a1x+ a0= 0

andlim

x→−∞

anxn + an−1xn−1 + . . .+ a2x2 + a1x+ a0

amxm + am−1xm−1 + . . .+ a2x2 + a1x+ a0= 0

If the degree of the denominator is larger than the degree of the numerator,then the limit is 0. Notice, this is the general concept of part one.

3. If m and n are positive rational number and m = n, then

limx→∞

anxn + an−1xn−1 + . . .+ a2x2 + a1x+ a0

amxm + am−1xm−1 + . . .+ a2x2 + a1x+ a0= anam

andlim

x→−∞

anxn + an−1xn−1 + . . .+ a2x2 + a1x+ a0

amxm + am−1xm−1 + . . .+ a2x2 + a1x+ a0= anam

In words, If the degree of the numerator is the same as the degree of thedenominator, then the limit as x → ∞ or x → −∞ is the ratio of the leadingcoefficients.

To show the last two parts we can use part one. We will do this by example. (Note thatproofs cannot be done by example)

Example: 1.6.2:

Find the following limits.

1. limx→∞

−3x2+2x−57x4+3x2

Notice that if we evaluate each of the limits seperately they both equal∞. Thiswould give us ∞∞ which is another indeterminate form. We only have the toollimx→∞

1xr = 0. So let’s divide the numerator and the denominator by x4.

limx→∞

−3x2 + 2x− 57x4 + 3x2 = limx→∞

−3x2x4 +

2xx4 −

5x4

7x4x4 +

3x2x4

= limx→∞

−3x2 +

2x3 −

5x4

7 + 3x2

= 0 + 0− 07 + 0

= 0



2. limx→−∞

x7−2x2+5−6x7−3x+4

Using the same trick as above,

limx→−∞

x7 − 2x2 + 5−6x7 − 3x+ 4 = limx→−∞

x7

x7 −2x2x7 +

5x7

− 6x7x7 −3xx7 +

4x7

= limx→−∞

1− 2x5 +5x7

−6− 3x6 +4x7

= limx→−∞

1− 0 + 0−6− 0 + 0

= −16

1.6.1 Horizontal Asymptotes

These cases of limits to infinity or minus infinity are something that you have probablydiscussed in a previous class. What does it mean to have the graph of a function approacha value far out to the right and far out to the left? The example for f(x) = 1x we had alimx→∞

1x = 0. What is y = 0 in for this function? This is called a horizontal asymptote.

Theorem 1.6.2:

If limx→∞

f(x) = L1 and limx→−∞

f(x) = L2, then y = L1 and y = L2 are horizontalasymptotes of f(x).

• If L1 = L2, then the function has one horizontal asymptote.

• If L1 6= L2, then the function has two horizontal asymptote.

All of the examples above have had one horizontal asymptote because in each case

limx→∞

f(x) = limx→−∞

f(x)

You can double check if you like.Let’s consider an example that has two horizontal asymptotes.Example: 1.6.3:

Find the limit of limx→∞

−2x−3√4x2+1 and limx→−∞

−2x−3√4x2+1

This function is given by the graph

−20−16−12 −8 −4 4 8 12 16 20

−5−4−3−2−1

12345

x

y



I understand that we will not be able to graph things to see our solution before hand,but this will help you understand what is happening before we do the calculations.

Since x > 0 in this limit x =√x2

limx→∞

−2x− 3√4x2 + 1

= limx→∞

(−2x− 3) 1x(√

4x2 + 1) 1x

= limx→∞

(−2x− 3) 1x(√

4x2 + 1) 1√x2

= limx→∞

−2xx −

3x√

4x2x2 +

1x2

= limx→∞

−2− 3x√4 + 1x2

= − 2√4

= −1

Since x < 0 in this limit x = −√x2

limx→−∞

−2x− 3√4x2 + 1

= limx→−∞

(−2x− 3) 1x(√

4x2 + 1) 1x

= limx→−∞

(−2x− 3) 1x(√

4x2 + 1) 1−√x2

= limx→−∞

−−2xx −

3x√

4x2x2 +

1x2

= limx→−∞

−−2− 3x√

4 + 1x2

= 2√4

= 1

If you have noticed, the rational functions that we have been working on have been fairlystraight forward to evaluate. The one that was a bit more complicated was the function thathad a fraction but was not a rational function. Another example of these are trig functions.Let’s consider the functions below.

Example: 1.6.4:

Calculate the limit of the following.

1. limx→∞

cos(x)

This function is an oscillating function that bounces between −1 and 1. Thismeans that it will never approach a value as x→∞.

2. limx→∞

cos(x)x

This limit can be calculated using the squeeze theorem, but before we do thatlet’s think about it intuitively. The cos(x) part also oscillates from −1 to 1.This means that it never gets arbitrarily large or small. This is not the samefor the denominator. The x increases without bound. If the denominator of afraction increases but the numerator is relatively constant, then the functionwill shrink to zero. Let’s show this using the squeeze theorem.Since −1 ≤ cos(x) ≤ 1 and x > 0 we have

−1x≤cos(x)

x≤ 1x

limx→∞

−1x≤ limx→∞

cos(x)x

≤ limx→∞

1x

0 ≤ limx→∞

cos(x)x

≤ 0

=⇒ limx→∞

cos(x)x

= 0

We can also see this graphically as such.



5 10 15 20 25 30

−2

−1.5

−1

−0.5

0.5

1

1.5

2

x

y

This last part can be a little tricky because you need to consider if the function willapproach ∞ or −∞. This can be shown quickly using a few simple examples.

limx→∞

x2 =∞ limx→−∞

x2 =∞

and

limx→∞

x3 =∞ limx→−∞

x3 = −∞

Notice that in the x2 question the limits are both ∞. This is because when we plug ineither positive or negative number the square outputs a positive number, hence the limit is∞. In the x3 case the limit for x → −∞ will give us −∞ because negative numbers in x3output negative numbers. You must consider these items when determining that the limitdoes not exist, but approaches either ∞ or −∞.

Theorem 1.6.3:

Let m and n are positive rational numbers, n > m, f be a polynomial of degree n,and g be a polynomial of degree m. The limit of f(x)g(x) as x→∞ is either ∞ or −∞.Similarly for x→ −∞.

In the following example I will show two ways to do problems like these.Example: 1.6.5:

Find the limit of limx→∞

−4x2+32x+5 and limx→−∞

−4x2+32x+5

1. I will first demonstrate how to solve this algebraically. This requires longdevision.

limx→∞

−4x2 + 32x+ 5 = limx→∞−2x+ 5−

222x+ 5

= −∞

We can see that this is negative as we plug in large values of x, hence the limit

DifferentiationMath 241 The Derivative and the Tangent Line Problem Page 29

is −∞.

limx→−∞

−4x2 + 32x+ 5 = limx→−∞−2x+ 5−

222x+ 5

=∞

A similar argument hold for this limit.

2. Let’s now look at this intuitively. The function −4x2+32x+5 has an x2 in thenumerator and an x in the denominator. If we plug in large positivenumbers into x, the x2 is alway positive and so is the x. This means thatthe sign of the −4 and the two are what dictate the sign of the limit, hence −∞.

For the other limit as x→ −∞. If we plug large negative numbers into x2 westill stay positive, however when we plug into x we get a negative number. Thismeans that the −4x2 is negative and the 2x is also negative. When we dividetwo negative numbers the result is positive, hence ∞ is the limit.

You may be wondering why we disregarded the 3 and the 5. This is because the numbersthat we are dealing with are so large, that the addition of constants have no affect on thelimit. Further more they have no affect on the sign of the limit as addition or subtractionof constants only affect the sign when they are larger that the other term.

2 Differentiation

2.1 The Derivative and the Tangent Line Problem

The first thing that we need to discuss is the idea behind what a derivative is. This can beseen visually as a tangent line. We can think of a tangent line as a line created by connectingtwo points on a graph where the points are converging to the point that you would like todiscuss the derivative. The tangent line is the last line draw of the convergent sequence oflines. Let’s look at this visually.

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•

•

∆x

∆y

x

y

−4 −3 −2 −1 2 3 4

−4−3−2−1

1234

•

•

∆x

∆y

x

y

−4 −3 −2 −1 2 3 4

−4−3−2−1

1234

•

•

∆x

∆yx

y


−4 −3 −2 −1 2 3 4

−4−3−2−1

1234

••

∆x∆y x

y

−4 −3 −2 −1 2 3 4

−4−3−2−1

1234

• •∆x

∆y x

y

−4 −3 −2 −1 2 3 4

−4−3−2−1

1234

• x

y

In all of these cases we can write the slope of these lines fairly easily. Let’s discuss thisgiven the two points on the line. The first point is given by (x1, y1) and the second point isgiven by (x2, y2). If we want to find the slope of the tangent line at a given point, let’s sayx = c, then the first point can be written as (c, f(c)) and the second point can be writtenas (c+ ∆x, f(c+ ∆x)). Using these two points and plugging into our slope formula, we get

m = y2 − y1x2 − x1

= f(c+ ∆x)− f(c)(c+ ∆x)− c

= f(c+ ∆x)− f(c)∆x

In the visual above, we can see that ∆x converges to zero because we wanted the points toconverge to the point that we wanted to discuss the derivative. This means that the ∆x hasshrink till it is zero. Using this idea, the slope of the tangent line is the limit as ∆x→ 0 ofthe slopes of the secant lines.Definition 2.1.1: If f is defined on an open interval containing c, and if the limit

lim∆x→0

∆y∆x = lim∆x→0

f(c+ ∆x)− f(c)∆x = m

or equivalently

lim∆x→c

f(x)− f(c)x− c

= m

exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graphof f at the point (c, f(c)).

Iflim

∆x→0

f(c+ ∆x)− f(c)∆x =∞ or lim∆x→0

f(c+ ∆x)− f(c)∆x = −∞,


then the function has a vertical tangent line at x = c.Let’s now find the slope of tangent lines to given points of different functions.Example: 2.1.1:

Find the slope of tangent line at the given point using the definition of the tangentlines slope.

1. f(x) = 23x+ 4 at x = 3.

lim∆x→0

f(c+ ∆x)− f(c)∆x = lim∆x→0

23 (c+ ∆x) + 4− (

23c+ 4)

∆x

= lim∆x→0

23c+

23∆x+ 4−

23c− 4

∆x

= lim∆x→0

23∆x∆x

= lim∆x→0

23

= 23

This should make sense that the slope of the tangent line of aline is the slope of the actual line since a line that goes throughtwo points is unique. It is true that a the slope of a tan-gent line of a linear equation is always the slope of the line itself.

2. f(x) = −x2 + 5 at x = 3.

lim∆x→0

f(c+ ∆x)− f(c)∆x = lim∆x→0

−(3 + ∆x)2 + 5− (−32 + 5)∆x

= lim∆x→0

−(∆x)2 − 6∆x− 9 + 9∆x

= lim∆x→0

−∆x− 6

= lim∆x→0

−6

3. f(x) = 3√x at x = 0.

lim∆x→0

f(c+ ∆x)− f(c)∆x = lim∆x→0

3√

0 + ∆x− 3√

0∆x

= lim∆x→0

3√

∆x∆x

= DNE


Sincelim

∆x→0−

3√

∆x∆x = −∞

andlim

∆x→0+

3√

∆x∆x =∞

This function has a vertical asymptote at x = 0.

2.1.1 Differentiability and Continuity

A function has a derivative (is differentiable) when definition 2.1.1 exists. There are manynames for the derivative of a function,

• Derivative

• Instantaneous rate of change

• Velocity vector

• Slope of the tangent line

These different versions of the word are useful to describe different situations. In all casesthe limit must exist in order for the function to have a derivative. We do have a theoremthat can test if a function is differentiable. However, it only works on very specific functionat very specific points.

Theorem 2.1.1:

If f is differentiable at x = c, then f is continuous at x = c.

Proof. Assume that f is differentiable at x = c. This means that

limx→c

[f(x)− f(c)] = limx→c

[(x− c)f(x)− f(c)x− c

]

= limx→c

(x− c) limx→c

f(x)− f(c)x− c

]

= 0(f ′(c)) Where f ′(c)=derivative of f at c.

= 0

Hence the function height f(x)− f(c) shrinks as x→ c, this function is also continuous.

This theorem allows us to use an equivalent version of it called the contrapositive. If afunction f is not continuous at x = c, then it not differentiable at x = c. Let’s look at someexamples where a function is not differentiable at a given point.

Example: 2.1.2:

Determine if the following functions are differentiable at the given point.

1.

f(x) ={x2 x > 0x+ 1 x ≤ 0

at x = 0.

Notice that this function looks like x2 as x → 0+ and it looks like x + 1 as


x→ 0−. So we can write the left and right hand limits as such.

limx→c+

f(x)− f(c)x− c

= limx→0+

x2 − 0x− 0

= limx→0+

x2

x

= limx→0+

x

= 0

and

limx→c−

f(x)− f(c)x− c

= limx→0−

x+ 1− (0 + 1)x− 0

= limx→0−

x

x

= limx→0−

1

= 1

This limit does not converge and therefore the derivative does not exist.We could however have used theorem 2.1.1. Notice that this function is notcontinuous at x = 0 since the outputs are different. Using theorem 2.1.1, thisfunction is not differentiable at x = 0.

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•◦ x

y

2. f(x) = 2|x+ 1| − 3 at x = −1Notice that this functions graph looks like a ”v”. The tip of the point is atx = −1. if we calculate the limits we get

limx→c+

f(x)− f(c)x− c

= limx→−1+

2(x+ 1)− 3− (2(−1 + 1)− 3)x− (−1)

= limx→−1+

2x+ 2x− (−1)

= limx→−1+

2(x+ 1)x+ 1

= 2


and

limx→c−

f(x)− f(c)x− c

= limx→−1−

−2(x+ 1)− 3− (−2(−1 + 1)− 3)x− (−1)

= limx→−1−

−2x− 2x− (−1)

= limx→−1−

−2(x+ 1)x+ 1

= −2

These limits are not the same and therefore the function is not differentiable.

3. I won’t write out the last example because we have already done one like it. Itrepresents a function that has limit of infinity. This does not exist and thereforedoes not have a derivative at x = c. (Previous example part 3)

DifferentiationMath 241 Basic Differentiation Rules and Rates of Change Page 35

2.2 Basic Differentiation Rules and Rates of Change

When discussing the rules of mathematics I believe that it is a disservice to teach them inparts. Most people can remember the rule, but applying them at the correct time is difficult.I will state the differentiation rules and we will do random problems.

Theorem 2.2.1:

If we know that f and g are differentiable functions, then the derivative rules aregiven by the following.

1. Constant rule: ddxc = 0 where c is a real number.

2. Constant Multiple Rule: ddxcf(x) = xddxf(x)

3. Power Rule: ddxxn = nxn−1

4. Sum and Difference Rule: ddx [f(x)± g(x)] =ddxf(x)±

ddxg(x)

5. Product Rule: ddx [f(x)g(x)] =ddx [f(x)] · g(x) +

ddx [g(x)] · f(x)

6. Quotient Rule: ddxf(x)g(x) =

ddx [f(x)]·g(x)−

ddx [g(x)]·f(x)

(g(x))2

We also write derivatives of functions as f ′(x) or dfdx instead ofddxf(x). If the function

is written as y = we use the notation dydx .

Along with these rules of differentiation we have derivative of trig functions. We won’tprove these derivatives but will state them.

Theorem 2.2.2:

The derivative of the trigonometric functions are as follows

1. ddx sin(x) = cos(x)

2. ddx cos(x) = − sin(x)

3. ddx tan(x) = sec2(x)

4. ddx cot(x) = − csc2(x)

5. ddx sec(x) = sec(x) tan(x)

6. ddx csc(x) = − csc(x) cot(x)

In this section we will do random problems from parts 1-4 of the rules of differentiation.Example: 2.2.1:

Find the derivative of the following functions.

1. f(x) = x5This is the exponent rule. So

f ′(x) = 5x5−1 = 5x4

.

2. f(x) = πThe number π is a constant, so we willuse part 1 to get

df

dx= 0.


3. g(x) = x22This problem is a blend of two rules.We will first use the constant multiplerule, then the exponent rule.

d

dxf(x) = 12

d

dxx2 = 12 · 2x

1 = 2x

4. g(x) = x5 + 3x3 − 2x2 + 4We will use the sum and difference rulehere to take the derivative of each partseperately.

g′(x) = 5x4 + 9x2 − 4x+ 0

5. f(x) = x−2 − 23x−1 − 5We can use the sum rule and the ex-ponent rule to get

f ′(x) = −2x−3 + 23x−2.

6. f(x) = 1x2 +1x3

Similarly in this problem but we firstmust write the variables as negativeexponents. f(x) = x−2 + x−3 Thisallows us to use the same rules as theprevious problem.

f ′(x) = −2x−2 − 3x−4.

7. f(x) = sin(x) + csc(x)

f ′(x) = cos(x)− csc(x) cot(x)

8. f(x) = tan(x) + cot(x)

f ′(x) = sec2(x)− csc2(x)

Before we continue, we should ask ourselves what is a derivative? We have seen that thederivative gives us the slope of a tangent line. But as we have seen in the previous examples,the derivative is a function. For example f ′(x) = 2x. This means that as we change theinput value the derivative also changes. This implies that the slope changes? Does this makesense? YES it does. This is because the slope of the tangent line changes as we change thevalues of the input. We can see this below.

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•

x

y

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•x

y

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•x

y

−4 −3 −2 −1 1 2 3 4

−4−3−2−1

1234

•

x

y


This says that the derivative is a function which will tell us the slope of the tangent lineat any point you want on the graph. All we have to do is plug in a value for the input andout comes the slope of the tangent line at that input value. This also means that we canfind the equation of the tangent line. This is because we have the slope and a point on theline, that is (x, f(x)) since the tangent line intersects the graph at a single point on someopen interval. Let’s see if we can find the equation of the tangent lines.

Example: 2.2.2:

Find the equation of the tangent line to the function f(x) at the given x = c point.

1. f(x) = −3x2 + 2x− 5 at x = −2

We know that the slope of the tangent line is given by the derivative at x = −2.

f ′(x) = −6x+ 2

andf ′(−2) = 14.

We also have the point (−2, f(−2)) = (−2,−3(−2)2 + 2(−2)− 5) = (−2,−21).So we can find the equation of the line

y = mx+ b

−21 = 14(−2) + b

b = 7

=⇒ y = 14x+ 7

−5 −4 −3 −2 −1 1 2 3 4 5

−30

−25

−20

−15

−10

−5

•

xy

Figure 10: f(x) = −3x2 + 2x− 5 and y = 14x+ 7

2. What if we changed the value of x to 3 in part 1? What is the equation of thetangent line now?

The slope is given by f ′(3) = −6(3) + 2 = −16. Now we need the point ofintersection which is given by (3, f(3)) = (3,−3(3)2 + 2(3) − 5) = (3,−26).


Lastly the equation of the line is given by

y = mx+ b

−26 = −16(3) + b

b = 22

=⇒ y = −16x+ 22

−5 −4 −3 −2 −1 1 2 3 4 5

−30

−25

−20

−15

−10

−5

•

xy

Figure 11: f(x) = −3x2 + 2x− 5 and y = −16x+ 22

Again, the derivative function gives us the slope of the tangent line at any inputvalue.

Rates of Change

There will be two different rates of changes that will be discussed in this section. Thefirst is called the average rate of change. This can be discussed as it sounds, like an average.The average that we will be considering is the change in the output relative to the input.This is written as a fraction as such

Average rate of change = Change in outputChange in input =∆y∆x

Notice though that the change in output is the same as y2− y1 and the change in the inputis the same as x2 − x1. This means that the average rate of change over an interval is givenby the slope of the secant line draw through the two end points of the said interval. In thecase of velocity we want to calculate the change in distance over the change in time. Wewill write this as

∆s∆t

Example: 2.2.3:

You are performing an experiment where you drop a ball of the top of your schoolsbuilding. You drop the ball from a height of 150 ft. The position function of the ballis given by

s = −16t2 + 150

where s is in feet and t is in seconds. Find the average velocity over each timeinterval.

1. [0, 2] 2. [1, 2] 3. [.5, 1.5]


1. We know that the interval is [0, 2]. Plugging this in we get s(0) = −16(0)2 +150 = 150 and s(2) = −16(2)2 + 150 = 86. Hence the average rate of changeon this interval is

150− 860− 2 =

64−2 = −32

This means that the ball is dropping at a rate of 32 feet per second.

2. We know that the interval is [1, 2]. Plugging this in we get s(0) = −16(1)2 +150 = 134 and s(2) = −16(2)2 + 150 = 86. Hence the average rate of changeon this interval is

134− 861− 2 =

48−1 = −48

This means that the ball is dropping at a rate of 48 feet per second. Thisshould make sense because we are calculating the tail end of the first interval.Gravity increases the velocity of the ball the longer it acts on it.

3. We know that the interval is [.5, 1.5]. Plugging this in we get s(0) = −16(.5)2 +150 = 146 and s(2) = −16(1.5)2 +150 = 114. Hence the average rate of changeon this interval is

146− 114.5− 1.5 =

32−1 = −16

This means that the ball is dropping at a rate of 16 feet per second

Now that we have calculated the average velocity (average rate of change) what wouldhappen if we took the limit as the input variable tended to zero (∆t→ 0). We would get

v(t) = lim∆t→0

∆s∆t =

s(t+ ∆t)− s(t)∆t = s

′(t)

This is called the velocity function. We have already discussed that the derivative canalso describe velocity. This is how. This is also called the instantaneous velocity. Hopefullythis makes sense already. If the change in time tends to zero, we are no longer talking aboutthe rate of change over an interval. Rather, we are discussing the rate of change at a singlepoint, or the instance where we would like to discuss velocity. If we take the absolute valueof the derivative, we will have the speed.

In this section we will discuss the instantaneous velocity using the equation of a freefalling object. This formula is given by

s(t) = −12gt2 + v0t+ s0

where s0 is the initial height, v0 is the initial velocity, and g is gravity which is 32 feet persecond per second on earth or 9.8 meters per second per second.

Example: 2.2.4:

A rock is dropped from the edge of a cliff that is 214 meters above water.

1. Determine the position and velocityfunctions for the rock.

2. Find the instantaneous velocity whent = 2 and t = 5. Where t is in seconds.

3. Find the time required for the rock toreach the surface of the water

4. Find the velocity of the rock at impact.


1. The position function is given by

s(t) = −12(9.8)t2 + 0(t) + 214 = −4.9t2 + 214

and the velocity functions′(t) = −4.9t

2. The instantaneous velocity when t = 2 is

s′(2) = −9.8

and when t = 5 iss′(5) = −24.5

At 2 seconds the rock is falling at a rate of 9.8 meters per second, while at 5seconds the rock is falling at a rate of 24.5 meters per second.

3. It will take

−9.8t2 + 214 = 0

−9.8t2 = −214

t2 = 2149.8 ≈ 21.836734

t = ±√

2149.8

t ≈ ±4.67298

Since time cannot be negative, t = 4.67298.

4. The rocks velocity when it hits the water will be determined by the derivativeat t = 4.67298 or

s′(4.67298) = −9.8(4.67298) = −45.795

Hence the the rock was falling at a rate of about 45.795 meters per second whenit hit the water. Notice that this is discussing the fall since we have a negativevelocity. If we take the absolute value of the derivative we would have 45.795which gives us the speed at which the rock hits the water.

DifferentiationMath 241 Product and Quotient Rules and Higher Order Derivatives Page 41

2.3 Product and Quotient Rules and Higher Order Derivatives

We have already looked at the rule for the product and the quotient in theorem 2.3.1. Inthis section we will discuss them in more detail and demonstrate how they work. In caseyou forgot, we will restate the product and quotient rule.

Theorem 2.3.1:

If we know that f and g are differentiable functions, then the derivative rules aregiven by the following.

1. Product Rule: ddx [f(x)g(x)] =ddx [f(x)] · g(x) +

ddx [g(x)] · f(x)

2. Quotient Rule: ddxf(x)g(x) =

ddx [f(x)]·g(x)−

ddx [g(x)]·f(x)

(g(x))2

We will now do an abundance of problems that involve the all the rules of derivatives.Example: 2.3.1:

Find the derivatives of the following functions.

1. Prove that the derivative of tan(x) is sec2(x).

d

dxtan(x) = d

dx

sin(x)cos(x)

= (cos(x))(cos(x))− (− sin(x))(sin(x))cos2(x)

= cos2(x) + sin2(x)

cos2(x)

= 1cos2(x)

= sec2(x)

2. f(x) = tan(x)−sin(x)−3x2+2x−7

d

dx

tan(x)− sin(x)−3x2 + 2x− 7

=ddx [tan(x)− sin(x)](−3x2 + 2x− 7)−

ddx [−3x2 + 2x− 7](tan(x)− sin(x))

(−3x2 + 2x− 7)2

= (sec2(x)− cos(x))(−3x2 + 2x− 7)− (−6x+ 2)(tan(x)− sin(x))

(−3x2 + 2x− 7)2

3. f(x) = x2 cos(x)

(x+1)(csc(x))


d

dx

x2 cos(x)(x+ 1)(csc(x))

=ddx [x2 cos(x)](x+ 1)(csc(x))−

ddx [(x+ 1)(csc(x))]x2 cos(x)

[(x+ 1)(csc(x))]2

= [2x cos(x) + x2 sin(x)](x+ 1)(csc(x))− [csc(x)− (x+ 1) csc(x) cot(x)]x2 cos(x)

[(x+ 1)(csc(x))]2

We can see this done in parts using the product rule.

d

dx[x2 cos(x)](x+ 1)(csc(x))

= [ ddx

[x2] cos(x) + ddx

[cos(x)]x2](x+ 1)(csc(x))

= ([2x] cos(x) + [sin(x)]x2)(x+ 1)(csc(x))

− ddx

[(x+ 1)(csc(x))]x2 cos(x)

= −[ ddx

[(x+ 1)] csc(x) + ddx

[csc(x)](x+ 1)]x2 cos(x)

= −(csc(x)− csc(x) cot(x)(x+ 1))x2 cos(x)

4. f(x) = (x+ 1) sin(x) cos(x)This function is a product of three functions. We will need to apply the productrule.

d

dx(x+ 1) sin(x) cos(x)

= ddx

[x+ 1] sin(x) cos(x) + ddx

[sin(x) cos(x)](x+ 1)

= sin(x) cos(x) + [ ddx

[sin(x)] cos(x) + ddx

[cos(x)] sin(x)](x+ 1)

= sin(x) cos(x) + [cos(x) cos(x)− sin(x) sin(x)](x+ 1)

= sin(x) cos(x) + [cos2(x)− sin2(x)](x+ 1)

5. f(x) = 3 cot(x)−5 tan(x)x−2


Let’s do this problem in two ways. First using the quotient rule

d

dx

3 cot(x)− 5 tan(x)x−2

=ddx [3 cot(x)− 5 tan(x)]x−2 −

ddx [x−2](3 cot(x)− 5 tan(x))

(x−2)2

= (−3 csc2(x)− 5 sec2(x))x−2 + 2x−3(3 cot(x)− 5 tan(x))

x−4

= [(−3 csc2(x)− 5 sec2(x))x−2 + 2x−3(3 cot(x)− 5 tan(x))]x4

= (−3 csc2(x)− 5 sec2(x))x2 + 2x(3 cot(x)− 5 tan(x))

Now let’s do this using the product rule. To do this we must first write thefunction as a product.

d

dx

3 cot(x)− 5 tan(x)x−2

= ddx

[(3 cot(x)− 5 tan(x))x2]

= ddx

[3 cot(x)− 5 tan(x)]x2 + ddx

[x2](3 cot(x)− 5 tan(x))

= (−3 csc2(x)− 5 sec2(x))x2 + 2x(3 cot(x)− 5 tan(x))

Notice that these both have the same answer. We may find derivatives inmultiple ways. Unless stated specifically, find the derivative in what ever wayworks best for you.

2.3.1 Higher Order Derivatives

We will now discuss what a higher order derivative is. We already understand some of thefunctions.

• s(t) Position function.

• s′(t) = v(t) Velocity function.

If we have the second derivative or

s′′(t) = v′(t) = a(t),

we have the acceleration function. Let’s think about why this is. The acceleration functionis the derivative of the velocity function. What does it tell us when we take a derivativeof a function? The derivative is the rate of change, so when we take a derivative of thevelocity function we are asking what is the rate of change of velocity. How fast the velocityis changing is given by acceleration.

While we will discuss at most the second derivative, we way continue to take derivativesas many times as possible for the given function. We label this as the nth derivative and iswritten as f (n)(x).


Example: 2.3.2:

It has been shown that the removal of air resistance allows objects to fall at the samerate. This is even true on the moon. Given the position function

s(t) = −0.81t2 + 2

for free falling objects, where t is time in seconds and s(t) is height in meters,determine the gravitational force of the moon.

We know the position function but we need to know the acceleration function. Thismeans that we need to find the second derivative of s(t)

• s(t) = −0.81t2 + 2

• s′(t) = −1.62t+ 2

• s′′(t) = −1.62

This tells us that the moon has a gravitational pull of 1.62 meters per second.If we compare the Earths gravitational pull to the moons, we get 9.81.2 ≈ 6. This meansthat the earths gravitational pull is 6 times stronger than the moons.

We may also use the acceleration function without a real life scenario. We would have afunction and determine is velocity and/or acceleration using the derivative.

DifferentiationMath 241 The Chain Rule Page 45

2.4 The Chain Rule

This section, in my opinion, is by far the most important section on derivatives. It is alsoby far the most difficult. The chain rule however, allows us to calculate the derivative offunctions that have compositions of multiple functions that can be quite involved.

Theorem 2.4.1:

If f(u) is a differentiable of u and u = g(x) is a differentiable function of x, thenf(g(x)) is a differentiable of x and

d

dxf(g(x)) = d

dx[f ] ◦ g(x) · d

dxg(x) = f ′(g(x)) · g′(x)

We will first do some simple examples to get a grasp of the idea of the chain rule.Example: 2.4.1:

Determine if the function requires the chain rule. If it does, find the derivative.

1. f(x) = (−2x2 + 3x− 5)5

This is a function that requires thechain rule. This is because we arecomposing g(x) = −2x2 + 3x− 5 andh(x) = x5 as h(g(x)) = f(x). Thismeans that we must use the chainrule as follows.

f ′(x) = h′(g(x)) · g′(x)

= 5(g(x))4 · (−4x+ 3)

= 5(−2x2 + 3x− 5)4(−4x+ 3)

2. f(x) = sin(x) tan(x)

This function does not require a chainrule. There is no composition in-volved. Even thought the instruc-tions state to find the derivative if weuse the chain rule, let’s calculate itanyway.

f ′(x)

= ddx

[sin(x)] tan(x) + ddx

[tan(x)] sin(x)

= cos(x) tan(x) + sec2(x) sin(x)

3. f(x) = sin(x2 + 1)

This does require the chain rulewhere h(x) = sin(x) and g(x) =x2 + 1, where h(g(x)) = f(x). So

f ′(x) = h′(g(x)) · g′(x)

= cos(g(x)) · 2x

= 2x cos(x2 + 1)

4. f(x) = sin(cos(x))

This does require the chain rulewhere h(x) = sin(x) and g(x) =cos(x), where h(g(x)) = f(x). So

f ′(x) = h′(g(x)) · g′(x)

= cos(g(x)) · (− sin(x))

= − sin(x) cos(cos(x))

Now that we have an understanding of the chain rule we need to do more complicatedproblems. We will first find the derivative of functions that require more than one derivativerule. We will then learn how to find the derivative of a function that has multiple chainrules.


Example: 2.4.2:

Find the derivative of the following functions

1. f(x) = 3√

(x2 − 4)2

We have a composition of h(g(x)) where h(x) = x 23 and g(x) = x2 − 4.

f(x) = (x2 − 4) 23

=⇒ f ′(x) = 23(g(x))− 13 · (2x)

= 4x3 (x2 − 4)− 13

= 4x3 3√

(x2 − 4)

Notice that this function has a horizontal tangent line at x = 0 and no deriva-tives at x = −2, 2.

2. f(x) = sin(−5x2+3)

(x+1)2

=⇒ f ′(x) =ddx [sin(−5x2 + 3)](x+ 1)2 −

ddx [(x+ 1)2] sin(−5x2 + 3)

(x+ 1)4

We now have two parts with chain rule

h(x) = sin(x)

g(x) = −5x2 + 3

p(x) = x2

q(x) = x+ 1

Where h(g(x)) = sin(−5x2 + 3) and p(q(x)) = (x+ 1)2

=⇒ f ′(x) =ddx [sin(−5x2 + 3)](x+ 1)2 −

ddx [(x+ 1)2] sin(−5x2 + 3)

(x+ 1)4

= cos(g(x))(−10x)(x+ 1)2 − 2(q(x))(1) sin(−5x2 + 3)

(x+ 1)4

= cos(−5x2 + 3)(−10x)(x+ 1)2 − 2(x+ 1) sin(−5x2 + 3)

(x+ 1)4

= cos(−5x2 + 3)(−10x)(x+ 1)−2 − 2(x+ 1)−3 sin(−5x2 + 3)

3. f(x) = tan(x2) sec(sin(x))


f ′(x) = ddx

[tan(x2)] sec(sin(x)) + ddx

[sec(sin(x))] tan(x2)

We again have two parts with chain rule

h(x) = tan(x)

g(x) = x2

p(x) = sec(x)

q(x) = sin(x)

Where h(g(x)) = tan(x2) and p(q(x)) = sec(sin(x))

f ′(x) = ddx

[tan(x2)] sec(sin(x)) + ddx

[sec(sin(x))] tan(x2)

= sec(g(x))(2x) sec(sin(x)) + sec(q(x)) tan(q(x)) cos(x) tan(x2)

= sec(x2)(2x) sec(sin(x)) + sec(sin(x)) tan(sin(x)) cos(x) tan(x2)

We now have the ability to find the derivatives of complicated functions. We can evendo functions that have more than one composition.

Example: 2.4.3:

Find the derivative of the following function.

1. f(x) = sin(cos(tan(x)))

This uses the chain rule two times because it is the composition of h(x) = sin(x),g(x) = cos(x), and p(x) = tan(x) where h(g(p(x))) = sin(cos(tan(x)))

f ′(x) = ddx

[sin(cos(tan(x)))]

= cos(cos(tan(x))) ddx

[cos(tan(x))]

= cos(cos(tan(x))) · sin(tan(x)) ddx

[tan(x)]

= cos(cos(tan(x))) · sin(tan(x)) · sec2(x)

2. f(x) = tan2(sin(x) cos(x))

There are two chain rules here that need to be applied at different times.


f ′(x) = ddx

[tan2(sin(x) cos(x))]

= 2 tan(sin(x) cos(x)) ddx

[tan(sin(x) cos(x))]

= 2 tan(sin(x) cos(x)) · sec2(sin(x) cos(x)) ddx

[sin(x) cos(x)]

= 2 tan(sin(x) cos(x)) · sec2(sin(x) cos(x)) · (cos(x) cos(x)− sin(x) sin(x))

= 2 tan(sin(x) cos(x)) · sec2(sin(x) cos(x)) · (cos2(x)− sin2(x))

DifferentiationMath 241 Implicit Differentiation Page 49

2.5 Implicit Differentiation

We have been differentiating functions that were in explicit form. This means that thefunctions can be solved for y and the y is equal to an expression of the x variable. Afunction that is in implicit for is a function that cannot be solved for y. An example of thisis

x2 + 3y3 − 2y = −7

We cannot get the y variable by itself. We can however still differentiate this functionby using implicit differentiation. We must first understand that the y variable is a functionof x, so when we take the derivative of y we treat it like a function of x and apply the chainrule.

• ddxx3 = 3x2.

since dx and x have the same variable.

• ddxy3 = 3y2ddxy.

This is the chain rule.

• ddx [x3 + y3] = 3x2 + 3y2dydx

Notice that we apply the same rules as before but we now add the dydx for the chainrule.

Instructions: 2.5.1:To apply implicit differentiation.

1. Check that you cannot solve explicitly for y.

2. Different both sides with respect to x.

3. Collect all terms with dydx on the left hand side of the equation and move the otheritems to the right.

4. Factor out the dydx .

5. Solve for dydx .

Example: 2.5.1:

Find dydx of the following functions.

1. 2y3 + 4x2 − y = x6

(a) We cannot solve for y.(b)

d

dx[2y3 + 4x2 − y] = d

dx[x6]

6y2 dydx

+ 8x− dydx

= 6x5

(c) Done(d) dydx (6y2 − 1) = 6x5 − 8x


(e) dydx =6x5−8x6y2−1

2. cos(x2 + 2y) + y2 = 1

d

dxcos(x2 + 2y) + y2 = d

dx1

− sin(x2 + 2y) ·(

2x+ 2dydx

)+ 2y dy

dx= 0

−2x sin(x2 + 2y) + 2 sin(x2 + 2y)dydx

+ 2y dydx

= 0

dy

dx

(2 sin(x2 + 2y) + 2y

)= 2x sin(x2 + 2y)

dy

dx= 2x sin(x

2 + 2y)2 sin(x2 + 2y) + 2y

= x sin(x2 + 2y)

sin(x2 + 2y) + y

3. tan(x2y4) = 3x+ y2

d

dxtan(x2y4) = d

dx[3x+ y2]

sec2(x2y4) ·(

2xy4 + 4y3 dydxx2)

= 3 + 2y dydx

sec2(x2y4)2xy4 + sec2(x2y4)4y3 dydxx2 = 3 + 2y dy

dx

sec2(x2y4)4y3 dydxx2 − 2y dy

dx= 3− sec2(x2y4)2xy4

dy

dx(4x2y3 sec2(x2y4)− 2y) = 3− 2xy4 sec2(x2y4)

dy

dx= 3− 2xy

4 sec2(x2y4)4x2y3 sec2(x2y4)− 2y

Like all derivatives, this type of differentiation can be thought of as the slope of thetangent line that intersects the graph at a given point. Let’s use this to find the equationof the tangent line for the following problems.


Example: 2.5.2:

Find the equation of the line tangent to the function at the given point.

1. x2 + (y − x)3 = 9 at x = 1.

We need a slope which is given by the derivative at x = 1.

d

dx[x2 + (y − x)3] = d

dx[9]

2x+ 3(y − x)2 ·(dy

dx− 1)

= 0

2x+ dydx

3(y − x)2 − 3(y − x)2 = 0

dy

dx3(y − x)2 = −2x+ 3(y − x)2

dy

dx= −2x+ 3(y − x)

2

3(y − x)2

Now that we have the slope of the tangent line, we also need a point. This isgiven by plugging in x = 1 into the original.

12 + (y − 1)3 = 9

(y − 1)3 = 8

y − 1 = 2

y = 3

Using this we can get the slope of

−2(1) + 3(3− 1)23(3− 1)2 =

−2 + 1212 =

1012 =

56 .

Lastly, find the equation by plugging (1, 3) into y = 56x+ b to get

3 = 56(1) + b

b = 136

Hence the equation of the line that is tangent to x2 + (y − x)3 = 9 at x = 1 isy = 56x+

136 .

2. (x2 + y2)3 = 8x2y2 at (−1, 1).

Taking the derivative to find the slope we get


d

dx[(x2 + y2)3] = d

dx[8x2y2]

3(x2 + y2)2 ·(

2x+ 2y dydx

)= 16xy2 + 16x2y dy

dx

6x(x2 + y2)2 + 6y(x2 + y2)2 dydx

= 16xy2 + 16x2y dydx

6y(x2 + y2)2 dydx− 16x2y dy

dx= 16xy2 − 6x(x2 + y2)2

dy

dx(6y(x2 + y2)2 − 16x2y) = 16xy2 − 6x(x2 + y2)2

dy

dx= 16xy

2 − 6x(x2 + y2)26y(x2 + y2)2 − 16x2y

This gives us a slope of

16(−1)(1)2 − 6(−1)((−1)2 + (1)2)26(1)((−1)2 + (1)2)2 − 16(−1)2(1) =

−16 + 6(4)6(4)− 16 = 1

Now plugging into y = x+ b we get

1 = −1 + b

b = 2

Hence y = x+2 is the equation of the line that is tangent to (x2 +y2)3 = 8x2y2at (−1, 1).

The last item which needs to be discussed is the higher order derivatives using implicitdifferentiations. This is exactly the same as before. We only change the notation of thederivatives from f ′(x) and f ′′(x) to dydx and

d2ydx2 .

Example: 2.5.3:

Find the second derivative of 3xy − 4 cos(x) = −6.

d

dx[3xy − 4 cos(x)] = d

dx[−6]

3y + 3xdydx

+ 4 sin(x) = 0

3xdydx

= −3y − 4 sin(x)

dy

dx= −3y − 4 sin(x)3x


Taking the derivative again we get

d2y

dx2= dydx

[−3y − 4 sin(x)

3x

]d2y

dx2= dydx

[(−3y − 4 sin(x))(3x)−1]

d2y

dx2= (−3dy

dx− 4 cos(x))(3x)−1 − 3x−2(−3y − 4 sin(x))

d2y

dx2=(−3−3y − 4 sin(x)3x − 4 cos(x)

)(3x)−1 − 3x−2(−3y − 4 sin(x))

d2y

dx2=(

3y + 4 sin(x)x

− 4 cos(x))

(3x)−1 − 3x−2(−3y − 4 sin(x))

DifferentiationMath 241 Related Rates Page 54

2.6 Related Rates

A related rate is exactly as it sounds, rates that are related. In particular, we are talkingabout the rate of change or the derivative.

For example;Suppose an object is moving along a path described by y = x2 where y and x are functions oft, that is, it is moving on a parabolic path. At a particular time, say t = 5, the x coordinateis 6 and we measure the speed at which the x coordinate of the object is changing and findthat dxdt = 3. At the same time, how fast is the y coordinate changing?

If we take the derivative on both sides with respects to t we get

dy

dt= 2xdx

dt.

We took the derivative with respects to t because we already had information about dxdt .This will allow us to plug in as follows.

dy

dt= 2xdx

dt

dy

dt= 2x(3)

dy

dt= 6x

We also know at time t = 5 we have x = 6 this means that y changes at a rate of 2(6)(3) = 36when t = 5.

Instructions: 2.6.1:To solve related rates problems,

1. Write all the relavent information down. (What we know)

2. Determine what are we trying to find.

3. Write an equation using the known information.

4. Differentiate with respects to t.

5. Plug in known information.

6. Solve.

Example: 2.6.1:

Solve the following related rates problems.

1. A plane is flying directly away from you at 500 mph at an altitude of 3 miles.How fast is the plane’s distance from you increasing at the moment when theplane is flying over a point on the ground 4 miles from you?

plane

you

x

3y


(a) • The plane is 3 miles high. • dxdt = 500

(b) Find the dydt when x = 4.(c) We can use the pythagorean theorem to get

x2 + 32 = y2.

(d)

d

dt[x2 + 32] = d

dt[y2]

2xdxdt

= 2y dydt

(e) We know that x = 4 and y2 = 42 + 32 =⇒ y = 5. Hence

2(4)(500)10 =

dy

dt

So dydt = 400. This means that when the plane is 4 ground miles awayfrom me, the distance between me and the plane will be changing at arate of 400 miles per hour.

2. You are inflating a spherical balloon at the rate of 7 cm3/ sec. How fast is itsradius increasing when the radius is 4 cm?

r

(a) dvdt = 7 This is because we are given 7 cm3/ sec. This implies that this isrelating volume.

(b) Find drdt when r = 4.(c) v = 43πr3

(d)

d

dt[v] = d

dt[ 43πr

3]

dv

dt= 43π(3r

2)drdt

dv

dt= 4πr2 dr

dt


(e)

7 = 4π(4)2 drdt

dr

dt= 764π

So the radius is increasing at a rate of 764π centimeters per second whenthe radius is 4cm.

3. A cylindrical tank standing upright (with one circular base on the ground) hasradius 20 cm. How fast does the water level in the tank drop when the wateris being drained at 25 cm3/ sec.

r

(a) • r = 20 • dvdt = 25 cm3/ sec. Similarly,we know this because it is acube.

(b) We want to find the rate of change in the height of the water or dhdt .(c) v = πr2h = π(20)2h = 400πh(d)

d

dt[v] = d

dt[400πh]

dv

dt= 400πdh

dt

(e)

dv

dt= 400πdh

dt

dh

dt= 25400π

dh

dt= 116π

The height is changing at a rate of 116π cm/s when the cylindrical tankhas a fixed radius of 20cm.


4. A ladder 13 meters long rests on horizontal ground and leans against a verticalwall. The top of the ladder is being pulled up the wall at 0.1 meters per second.How fast is the foot of the ladder approaching the wall when the foot of theladder is 5 m from the wall?

top

footx

h13

(a) • Ladder is 13 meters • dhdt = 0.1 meters per second.

(b) Find dxdt when x = 5.(c) x2 + h2 = 132 =⇒ h =

√169− x2.

(d)

d

dt[x2 + h2] = d

dt132

2xdxdt

+ 2hdhdt

= 0

(e)

2xdxdt

= −2hdhdt

2xdxdt

= −2√

169− x2 dhdt

2(5)dxdt

= −2√

169− (5)2(0.1)

2(5)dxdt

= −2√

144(0.1)

2(5)dxdt

= −2410

2(5)dxdt

= −125

dx

dt= −1250

dx

dt= − 625


This should make sense since the distance between the foot of the ladderand the wall is shrinking as we pull the ladder up the wall. The foot isapproaching the wall at a rate of 625 meters per second when the foot is 5meters away from the wall.

5. Sand is poured onto a surface at 15 cm3/ sec, forming a conical pile whosebase diameter is always equal to its altitude. How fast is the altitude of thepile increasing when the pile is 3 cm high?

D

h

(a) • dvdt = 15 cm3/sec • D = h

(b) Find dhdt when h = 3.(c) v = π3 r2h where r =

12D =

12h

(d)

d

dtv = d

dt

π

3 r2h

d

dtv = d

dt

π

3

(12h)2

h

dv

dt= π123h

2 dh

dt

dv

dt= π4 h

2 dh

dt

(e)

dv

dt= π4 h

2 dh

dt

15π4 (3)2

= dhdt

5π4 3

= dhdt

203π =

dh

dt

Applications ofDifferentiation

Math 241 Extrema on an Interval Page 59

This means that the height is increasing at a rate of 203π cm / sec when theheight is 3 cm.

Related rates are a complicated section but will be easier if you write out the things thatyou know and the objective of the problem. Do not do math in your head. Keeping all theinformation in your head has a tendency to get mixed up. It is easier to write it down.

More Problems

3 Applications of Differentiation

3.1 Extrema on an Interval

Other than graphing the function with thousands and thousands of points, how can wedetermine if a function has a minimum or a maximum? A minimum and maximum of afunction is defined asDefinition 3.1.1: Let f be defined on an interval D and c is in D.

1. f(c) is the global or absolute minimum minimum of f on D where f(c) ≤ f(x)for all x in D.

2. f(c) is the global or absolute maximum of f on D where f(c) ≥ f(x) for all xin D.

Definition 3.1.2: Let f have domain D and c is in D.

1. If there exists a sub interval I such that f(c) ≤ f(x) for all x in I, then f(c) is calledthe local or relative minimum.

2. If there exists a sub interval I such that f(c) ≥ f(x) for all x in I, then f(c) is calledthe local or relative maximum.

These are called extreme values. You have probably done a simple example of this beforein an alg

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Calculus - Korey And Math, Teaching Mathematics · 2020. 7. 16. · Calculus Korey Nishimoto. Math 241 Notes Korey Nishimoto Math Department, Kapiolani Community College January 15,