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QUICK AND EASY

DIFFERENTIAL CALCULUS

WORKBOOK

Leo Jonker

Queens UniversityFall 2007-2008


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Foreword

Most of the students entering in first year engineering at Queens Univer-sity have taken differential Calculus in High School. This applies to stu-dents who have graduated from Ontario high schools, those who have hadsome CEGEP education in Quebec, all Alberta Students (assuming they tookmathematics 31 in their final high school year), and some British Columbiaand Saskatchewan students, . Most of the students coming from outsideCanada have also had some Calculus before they get here. A smaller number,including most of those who come from schools in the Maritime provinces,and those coming from Manitoba have not had any significant amount ofcalculus. This should not worry you. You are probably better prepared inother ways, but it does meant that you will need some extra help at thebeginning of first year.

We would like our first year calculus course to make use of the more advancedpreparation of some students, without creating a situation that is impossiblefor those students for whom Calculus is a completely new experience. Thesenotes are intended to prepare all students for APSC 171, Calculus I.

In them we will introduce students to differential calculus in a highly intu-itive fashion, one that will, we think, fit well with the tenor of the rest of thecourse. We will introduce the idea of the derivative, discover formulas forthe derivatives of polynomial functions, the exponential functions, and loga-rithmic functions, and we will discuss basic rules for calculating derivativesof combinations of functions (the Product Rule, Quotient Rule and ChainRule).

This is a set of (deliberately) incomplete notes prepared for your use. Thenotes reflect the way in which the material should be learned. The intro-

i


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ii

duction to each topic is concise, but complete. You should be able to read

and understand it. At certain points the flow of thought in interrupted bya question for which the answer is not provided. If the notes are used ina class, this is where the instructor will ask you to become involved in thecompletion of the argument. If the notes are used as self-study, then this iswhere you try to make sense of the question. When you get stuck you canconsult James Stewart, Calculus: Early Transcendentals, the text for thecourse, or any Calculus book that may be available to you. At other times,the gaps in the notes follow problems in which the theory you have learnedcan be applied. These, too, are for you to fill in. Some of these problems areindicated as Concept Questions. These are multiple-choice questions thatsignal points at which key concepts are often misunderstood or misapplied.

They have short answers, but do require some thought.

Computer assisted learning

There are also some computer-based learning objects designed specificallyfor the course, and written a few years ago by Thomas Norman, a formerstudent in Mathematics and Engineering. These are small interactive com-puter programmes to help you learn the concepts behind a topic. Whenevera topic has one of these learning objects associated with it, this is indicatedin the interactive notes as follows:

Computer help for this topic is available on the course website

To find these learning objects, go to the course website and click on Com-puter assistance in the column on the left.


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Contents

What is a derivative really? . . . . . . . . . . . . . . . . . . . . . . 1

Rates of Change(Sections 2.7 and 2.9 in Stewart) . . . . . . . . . . . . . . . . 10

Derivatives of Exponential Functions(Section 3.1 in Stewart) . . . . . . . . . . . . . . . . . . . . . 16

Derivatives of Combinations of Functions(Sections 3.1 and 3.2 in Stewart) . . . . . . . . . . . . . . . . 19

The Chain Rule(Section 3.4 in Stewart) . . . . . . . . . . . . . . . . . . . . . 30

Inverse Functions(Section 1.6 in Stewart) . . . . . . . . . . . . . . . . . . . . . 35

Logarithmic Functions(Section 1.6 in Stewart) . . . . . . . . . . . . . . . . . . . . . 43

Derivatives of Log Functions(Section 3.6 in Stewart) . . . . . . . . . . . . . . . . . . . . . 48

iii


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iv CONTENTS


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WHAT IS A DERIVATIVE REALLY? 1

What is a derivative really?

There is just one big secret at the heart of Calculus:

If you consider a function over a very small interval of input values italways looks like a linear function.

This means that if you try successive input values, equally spaced and veryclose to each other, the output values seem to go up at a constant rate. Hereare two examples:

Consider the function f(x) = x2. We know this function is not linear (itsgraph is not a straight line), and yet, look at the following table of (a few)values of this function:

x x2

3.000 9.0003.001 9.0063.002 9.0123.003 9.018

It sure looks linear on this small scale! So does the following table of somevalues of the function f(x) = ex:

x ex

1.000 2.718281.001 2.721001.002 2.723721.003 2.72644

In each case the output goes up by 0.00272 when the input goes up by 0.001;so once again, this function looks linear when you get close.

This property of looking linear when you zoom in to a point becomesvisual when you apply it to the graph of a function. Here are three pictures,produced by Maple, of the graph of the exponential function ex at smallerand smaller scales around the point x = 1:


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2 CONTENTS

x

y

8

6

2

4

1.5

2

010.50

x

y

3.6

3.2

1.2

2.8

1.1

2.4

10.90.8

x

1.11.0510.950.9

y

3.2

3.1

3

2.9

2.8

2.7

2.6

2.5

This property that makes graphs look linear when observed on a small scale,is closely related to the fact that when we look around us, the earth looks

more or less flat. We can, perhaps, forgive our distant ancestors for thinkingthat it really was flat.

Of course, as with all simplifications, there is a caveat. The secret is notquite true of all functions. If the graph of your function has a corner at somepoint, as in the case of f(x) = |x|, no amount of zooming in is going to makethat corner go away. A function that looks linear when you zoom in to apoint on its graph is called a differentiable function. Fortunately mostof the functions we deal with are differentiable, or else are differentiable atmost points.

Eventually we should turn our observation about the local linearity of func-tions into something we can do mathematics with. In particular, the insightshould help us do certain kinds of calculations.

Before we get to that, however, there is a second secret at the heart ofCalculus, which has a lot to do with the one we revealed at the start of thischapter, even though it may seem quite unlike it:

If a quantity is very small, then its square 2 is negligible by comparison.

When we say something is negligible relative to something else, we meanthat its size is a small fraction of the something else. For example, consider = 0.001 then 2 = 0.000001. If we were to subtract 0.000001 (or addit) to 0.001, it would make a negligible difference to the total. It wouldadd only one-thousandth of its value to the total. The difference becomes


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even more pronounced if = 0.000001; for then 2 = 0.000000000001 and

subtracting or adding this to would change it by only one-millionth of itsinitial magnitude.

Concept Question 1. If is very small (say around 1/1000), which of thefollowing is a reasonable approximation? ( means is approximately equalto)

1.5 +

3 5

3A. Equation 3.

2. + 2

2 B. Equation 5.3. 5 + 5 C. Equations 1 and 3.

4.7 + 2

7

D. Equations 1, 3, and 4.

5.13

13 E. All of the equations.

Click here for help

The connection between the two secrets

So how is the secret on page 2 connected to the secret on page 1? Suppose wehave a simple function, say f(x) = x2. We want to use the second secret to
http://quickneasycalculus.blogspot.com/2007/07/concept-question-1.htmlhttp://quickneasycalculus.blogspot.com/2007/07/concept-question-1.html


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4 CONTENTS

explain why this function will seem linear if we stay very close to any given

initial point. Say we start at x = a, and examine what happens to f(x) whenx is replaced by a + x, where x represents a very small number. That is,we want to see how the value of this function changes as x changes, but iskept very small. Then

f(a + x) = (a + x)2 = a2 + 2ax + (x)2 .

By the secret on page 2, we can ignore (x)2 for all practical purposes, aslong as x stays very small. Thus

f(a + x) a2 + 2ax .

Notice that if we think of x as a variable (and keep a constant), the rightside is a linear function g(a + x) = a2 + 2ax. Notice that we do not wantto say

f(a + x) a2 + 2ax a2 ,which is also true, though with a rougher approximation. The reason is, wewant to see what linear function f(a + x) resembles, so if there is a term offirst degree in x we do not want to remove it, for it indicates the mannerin which the value of f(a + x) changes.

If we go back to writing x for a + x we get

g(x) = a2 + 2a(x a) ,

where, as we said already, f(x) and g(x) are extremely close to each other,as long as x is very close to a. Notice that if we let x = a the two functionare precisely equal to each other.

Example 1. If we write down the equation of the graph of g we get y =a2 + 2a(x a). Since a is kept constant, this is clearly the equation of aline. Zooming in to the graph of f around the point (a, f(a)) is the same asrestricting x to values close to a. What do we call the line given by the graph

of g?

Click here for help
http://quickneasycalculus.blogspot.com/2007/07/example-1.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-1.html


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Notice, from the equation of the graph of g, that the tangent line to the

graph of f at (a, f(a)) has slope 2a. This slope is defined as the derivativeof f(x) = x2 at a.

The derivative of a function f at an input value a is the slope of thetangent line to the graph of the function at the corresponding point(a, f(a)) on the graph.

We will give a more important characterization of the derivative in a moment,but for now let us note what we have learned about the derivatives of thefunction f(x) = x2 at various points:

The derivative of f(x) = x2 at input value x is equal to 2x.

Thus its derivative at x = 2 is equal to 4. At x = 0 it comes to 0. At x = it comes to 2.

The process that associates the derivative at x to the input x can be thoughtof as a new function constructed from the original function f. It is called thederivative function, or simply the derivative, and is denoted f. Thus

we have learned that if f(x) = x2

then f

(x) = 2x. It has suddenly become asimple matter to calculate the slope of the parabola y = x2! This process ofcalculating the derivative function from the original is called differentiation.

Notice that since f(a) is the slope of the tangent line from which the graphoff becomes indistinguishable when we zoom in to a, we are saying, in effect,that near a,

f(a + x) f(a) + f(a)x .In other words, the derivative at a is the coefficient of the linear termin the linear function that best approximates f near a.

As we said on page 2, all reasonable functions are differentiable with theexception of isolated points at which the graph of the function has corners.Even when you do not have a formula for a function, but are given its graph,you can say something about its derivative:


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6 CONTENTS

Concept Question 2. Given the graph of f, which of the graphs below it

is the graph of f

? Is it the solid graph A, the dashed graph B or the dottedgraph C?

f

A

C

B

Click here for help


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We would like to be able to calculate a formula for the derivative of anypolynomial. Lets begin with other power functions.

Theorem Iff(x) = xn, where n is a positive integer, then f(x) = nxn1.

Example 2. Prove that this theorem is true, by expanding f(a + x) andthen applying the secret revealed on page 2 to the resulting expression .

Click here for help


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8 CONTENTS

A polynomial function is just a sum of constant multiples of power func-

tions. Suppose we have a polynomial

f(x) = c0 + c1x + c2x2 + + cnxn .

Then to find the derivative f we could apply the same idea to all of theterms at once: replace x by a + x. Then

f(a + x) = c0 + c1(a + x) + c2(a + x)2 + + cn(a + x)n

c0 + c1(a + x) + c2(a2 + 2ax + ) + + cn(an + nan1x + ) ,where the dots indicate terms that involve x to power 2 or higher, and

which can therefore be ignored relative to the lower order terms. Thus

f(a + x) c0 + c1(a + x) + c2(a2 + 2ax) + + cn(an + nan1x)

= f(a) + (c1 + 2c2a + 3c3a2 ncnan1)x .

If we extract the coefficient in front of x, and then replace a by x we getthe derivative (the slope):

Theorem: The derivative of a polynomial

f(x) = c0 + c1x + c2x2 + cnxn

isf(x) = c1 + 2c2x + 3c3x

2 + + ncnxn1 .

Notice that this says that to calculate the derivative of a polynomial youshould calculate the derivatives of the power functions in the terms, and thencombine the results in the way they were combined in the original polynomial.Notice also that the derivative of a constant function is always 0 (see whatyou get if all the coefficients except c0 are equal to zero). The reason is thatthe graph of a constant function is a horizontal line, and therefore the slopeis 0 at each point.


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Example 3. What is the slope of the graph y = 3x2 5x + 7 at (1, 5)?

Click here for help

Example 4. For what values ofx does the graph of f(x) = x4

8x3 + 22x2

24x + 3 have positive slope and for which values does it have negative slope?

Click here for help
http://quickneasycalculus.blogspot.com/2007/07/example-3_27.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-4.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-4.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-3_27.html


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10 CONTENTS

Rates of Change

(Sections 2.7 and 2.9 in Stewart)

NOTE: James Stewart: Calculus, Early Transcendentals 6e, is the text-book we will be using in the first year Applied Science calculuscourses. However, you should be able to read these notes withoutany reference text so there is no need to buy the book before youarrive at Queens.

The most important characterization of a derivative is its ability to measureinstantaneous rate of change. Most of you have discussed rates of change

in high school, especially in cases when the function expresses distance (itsoutput) in terms of time (its input). Iff is the function whose rate of changewe want to discuss, say at an input value a, then we might begin by takinga value not too far from a, say a + x, and comparing the function values atthese inputs by subtracting f(a + x) f(a). This expresses a change inthe function value, but not yet a rate of change. To get that, you have tocompare this change in the output value of the function to the correspondingchange in the input. This comparison is achieved by division: The averagerate of change of the function f between inputs a and a + x is

Average rate of change =f(a + x) f(a)

x.

Sometimes it is convenient to give a name to the output of the function, sayy = f(x). Then the numerator in the above expression represents a changein the y-value and can be denoted as y. That is, the average rate of changeis the ratio

y

x.

a a + x

x

yf


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RATES OF CHANGE 11

In the diagram, this average rate is equal to the slope of the line segment

joining the points (a, f(a)) and (a + x, f(a + x)). The line extending thissegment is sometimes referred to as a secant line. Now suppose we let xget smaller and smaller. Then the point (a + x, f(a + x)) will graduallyslide (along the graph) towards (a, f(a)), and in the process the secant linewill turn (a little) and gradually converge to the tangent line, as in the nextpicture:

a

f

On the one hand, the slope of this tangent line is what we mean by thederivative f(a). On the other hand, this slope is what the slopes of thesecant lines get closer and closer to as we let x get smaller and smaller.This idea is expressed as a limit:

f(a) = limx0

f(a + x) f(a)x

.

This means that if, in the fractional expression on the right, you let x getcloser and closer to 0, then the value of that expression gets closer and closerto f(a). In general, whenever we have any kind of expression E(u) thatdepends on a variable u, then by the limit of E(u) as u c we mean thequantity L that E(u) gets closer and closer to when u is allowed to get closerand closer to c. The notation for this is

L = limuc

E(u) .

Returning to the discussion of the derivative, and writing x in place ofa+x,we get


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12 CONTENTS

f

(a) = limxaf(x)

f(a)

x a .

This is the definition of the derivative you will usually see in atextbook .

Because of this characterization of the derivative as the limit of a ratio ofchange in output to change in input, several alternative notations for a deriva-tive reflect this understanding. If the variable name y is used for the outputof a function f, as in y = f(x), then we also write any of the following:

f(x) = dydx = ddx f(x) .

Let us see how this limit approach might work for the function f(x) = x2:The slope of the secant between (a, a2) and (a + x, (a + x)2) is

(a + x)2 a2x

=a2 + 2ax + (x)2 a2

x=

2ax + (x)2

x.

At this point we can either notice that we can disregard (x)2 because it isso small in comparison to the other terms in the numerator, in which case

we concludef(a) =

2ax + (x)2

x= 2a ,

or we can say, slightly more formally, that

f(a) = limx0

2ax + (x)2

x= lim

x0(2a + x) = 2a .

Either way, we have confirmed the derivative calculation we did much earlier.

We saw earlier that the derivative of xn is nxn1, for any positive integer n.In fact this formula is valid even when n is not a positive integer, but any

fixed real number. We do not have time to do all the proofs necessary toshow this, but we do have time to show it for two instances.


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RATES OF CHANGE 13

Concept Question 3. To find a formula for the derivative of the function

f(x) = x1

= 1/x, at a point a, we have to find the limit

f(a) = limx0

f(a + x) f(a)x

.

What does the expression on the right look like for this function?

A. limx0

1a + x 1a

x

B. limx0

1a+x

1a

x

C. limx0

x

(a + x) a

Click here for help

Example 5. Calculate this limit and thus find the derivative of x1.

Click here for help
http://quickneasycalculus.blogspot.com/2007/07/concept-question-3.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-5.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-5.htmlhttp://quickneasycalculus.blogspot.com/2007/07/concept-question-3.html


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For a second example, we will find the derivative of

x = x1/2.

Concept Question 4. If we want to calculate the derivative of

x, say ata point a, we should begin with the expression for the average rate of changebetween a and a + x. Which of the following is the correct expression forthat?

A.

a + x a

x

B.

(a + x) a

x

C.(

a +

x) ax

Click here for help


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RATES OF CHANGE 15

We must take the limit of this expression as x 0. This does not looksimple, and may require some algebra to transform the expression:

Example 6. Find the limit of the expression for the average rate of changeat a for the function f(x) =

x.

Click here for help

A combination of similar techniques will show that the theorem for derivativesof power functions,

d

dxxn = nxn1 ,


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16 CONTENTS

is true for any fixed rational power n. In fact,as already pointed out, the

theorem is true for any fixed real power whatsoever; but proving that requiresother methods.

Derivatives of Exponential Functions(Section 3.1 in Stewart)

We will now turn to a calculation of the derivative of the exponential func-tion ex. There are several ways to introduce the number e. The one we willuse defines e as that number which is just right so that the derivative of ex

at x = 0 is equal to 1. In other words, among the exponential functions bx

for b = 0.5, 1, 1.5, 2, 3,etc there is one base b = e just right that the slopeof its graph, where it crosses the vertical axis, is precisely equal to 1. (Youmay remember from your high school course that this results in an irrationalnumber, like . The value of e is approximately 2.718 .)

(0.5)x

2x

(1.5)x

1x

ex

This way of introducing the number e immediately tells us that if f(x) = ex

then f(0) = 1.


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DERIVATIVES OF EXPONENTIAL FUNCTIONS 17

Concept Question 5. Which of the following limits expresses this fact?

A. limx0

ea+x eax

= 1

B. limx0

(e0 + ex) e0x

= 1

C. limx0

ex 1x

= 1

Click here for help

We will now use this information to prove

Theorem:d

dx(ex) = ex

In other words, at every point, the value of the function ex is preciselyequal to its rate of change.

If you have had some study of exponential growth in high school, this willsound very familiar. When something grows exponentially, it means that the

rate at which a function increases is proportional to is value.


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18 CONTENTS

Example 7. Use the definition of the derivative as a limit of average rates

of increase to prove this theorem.

Click here for help


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DERIVATIVES OF COMBINATIONS OF FUNCTIONS 19

Derivatives of Combinations of Functions

(Sections 3.1 and 3.2 in Stewart)

In order to differentiate effectively, we should develop a kind of arithmeticof differentiation, which includes not only the derivatives of certain key func-tions, but which also includes ways to calculate derivatives of combinationsof functions whose derivatives we already know. Here is the first such theo-rem:

The Constant Multiple Rule: If c is a constant and f is a differen-

tiable function, then so is cf, and

d

dx[cf(x)] = c

d

dxf(x)

or, equivalently,(cf) = cf

Note that when we speak of the function cf we mean the function whose value

at an input x is equal to the product cf(x). If we evaluate this function ata we get cf(a); if we evaluate it at a + x we get cf(a + x). Thus theaverage rate of change of cf between a and a + x is

cf(a + x) cf(a)x

.

This factors immediately to become

cf(a + x) f(a)

x.

To find the derivative of cf we have to take the limit of this as x 0. Butsince c is constant we can take the limit of the fraction and multiply by cafterwards - it will come to the same thing (think about this!) Therefore

(cf)(a) = c limx0

f(a + x) f(a)x

= cf(a) .

If now we replace a by x we have a proof of the theorem.


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20 CONTENTS

The next theorem tells us how to differentiate the sum of two functions whose

derivatives we know already. Notice that it tells us to go ahead, differentiateeach function, and add the results afterward.

The Sum Rule: If the functions f and g are both differential, then sois f + g, and

d

dx[f(x) + g(x)] =

d

dxf(x) +

d

dxg(x)

or, equivalently,(f + g) = f + g

Notice that we have already seen instances illustrating the sum rule, for whenwe differentiated polynomials we found that it amounted to differentiatingeach of the polynomials terms separately and then adding (or subtracting)them.

Concept Question 6. What is the correct expression for the average rateof increase of the function f(x) + g(x) between a and a + x?

A.(f(a) + g(a)) + x

(f(a) + g(a))

x

B.f(a + x) f(a)

x+

g(a + x) g(a)x

C.(f(a + x) + g(a + x)) (f(a) + g(a))

x

Click here for help


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We are now ready to prove the Sum Rule, by taking the limit of this expres-sion as x 0:

(f + g)(a) = limx0

(f(a + x) + g(a + x)) (f(a) + g(a))x

= limx0

f(a + x) f(a)

x+

g(a + x) g(a)x

= limx0

f(a + x) f(a)x

+ limx0

g(a + x) g(a)x

= f(a) + g(a) .

Example 8. Calculate the derivative of 5ex

+ 3x2

+ 5x7

.

Click here for help


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22 CONTENTS

The following rule for finding derivatives of combined functions is proved in

the same way as the Sum Rule:

The Difference Rule: If the functions f and g are both differential,then so is f g, and

d

dx[f(x) g(x)] = d

dxf(x) d

dxg(x)

or, equivalently,(f g) = f g

Example 9. Find the slope of the graph of the function g(x) = + 7x 3x8 + 0.5x3 2ex at the point (0, 2)

Click here for help

Now that we know how to differentiate the sum and the difference of twofunctions whose derivatives we already know, it is time to turn to their prod-uct. Here the product off and g is thought of as a new function whose value,at input x, is given by f(x)g(x). In other words,

(f g)(x) = f(x)g(x) .


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The Product Rule: If f and g are both differentiable functions, then

so is f g, and

d

dx[f(x)g(x)] = f(x)

d

dx[g(x)] + g(x)

d

dx[f(x)]

or, equivalently,(f g) = f g + gf

You are probably wondering why the formula is so complicated. This will

become clear when we look at the proof:

Example 10. Prove the Product Rule.

Click here for help


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24 CONTENTS

At the beginning of this short course on differentiation we introduced the

derivative as the rate of change of the linear function that best approximatesthe function near a particular point a. Later we moved to a more formaldefinition of the derivative as a limit of average rates of change, and usedthis more formal approach to prove our theorems. To indicate how the earlier,less formal, approach can help us form good intuition in many situations, wewill now give an alternative (less precise but still instructive) proof of theProduct Rule:

We know that (near a)

f(a + x) f(a) + f(a)x

g(a + x) g(a) + g(a)xTherefore,

(f g)(a + x) = f(a + x)g(a + x) (f(a) + f(a)x)(g(a) + g(a)x)

= f(a)g(a) + [f(a)g(a) + g(a)f(a)]x + (f(a))(g(a))(x)2

f(a)g(a) + [f(a)g(a) + g(a)f(a)]x

Notice that the coefficient of the linear term in this expression is f(a)g(a) +g(a)f(a). That is,

(f g)(a) = f(a)g(a) + g(a)f(a)


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After adding, subtracting and multiplying functions, we should divide them

next. As is the case with numbers, when you start dividing you have to be abit more careful. Whereas you can add, subtract or multiply any two num-bers, you cannot divide them if the second one is zero. Something just likeit happens when we try to divide one differentiable function, f by another,g. The new function that results when we do this division is denoted by the

symbolf

g, and its value at an input x is

f

g

(x) =

f(x)

g(x)

The domain of this new function does not include those input values at whichthe denominator g gives the value 0.

Quotient Rule: If f and g are differentiable functions, then the quo-tient function f /g is differentiable at the points x where g(x) = 0; andat those points,

d

dx

f(x)

g(x)

=

g(x) ddx

[f(x)] f(x) ddx

[g(x)]

[g(x)]2

or, equivalently, f

g

=fg gf

g2

Again, when we prove this theorem we will see why the Quotient Rule hasthis strange form.

Example 11. Prove the Quotient Rule.

Click here for help


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26 CONTENTS


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Example 12. Calculate the derivative ofex

x

2.

Click here for help

Example 13. Find an equation of the tangent line to the graph y =

x

x + 1at the point (1, 0.5).

Click here for help
http://quickneasycalculus.blogspot.com/2007/07/example-12.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-13.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-13.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-12.html


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28 CONTENTS

Example 14. Determine, from the expression we found for its derivative,

whether the graph has a low point, high point, or inflection point at x = 1.

Click here for help

Here is a picture of this graph, generated using Maple by entering the com-mand

> plot(sqrt(x)/(x+1), 0..2);

0.5

0.4

0.3

0.2

0.1

0

x

21.510.50


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Example 15. (From Section 3.7 in Stewart) If a tank holds 5000 liters of

water, which drains from the bottom of the tank in 40 minutes, then Torri-cellis law give the volume V of the water remaining after t minutes as

V = 5000

1 t

40

20 t 40

Find the rate at which water is draining from the tank after 5 minutes, andafter 30 minutes.

Click here for help


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30 CONTENTS

The Chain Rule

(Section 3.4 in Stewart)


If you go to the course web site and click computer-assisted learning in thecolumn on the left, it will take you to MathQs a set of interactive learningdevices. One of these is on the topic of the chain rule. Try it out!

We have studied derivatives of constant multiples, sums, differences, products

and quotients of functions. Another important way to combine functions isthrough composition. If f and g are functions, the composite f g is thefunction that for the input x gives the output f(g(x)). In other words,

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

Concept Question 7. Suppose f(x) =

x and g(x) = x2 + 1. Then thecomposition f g is

A. x + 1B.

x(x2 + 1)

C.

x2 + 1

D. x

x2 + 1

Click here for help


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THE CHAIN RULE 31

Concept Question 8. We want to express the function k(x) = ex3+5 as a

composition of two functions f andg; that is we want k = fg. What shouldf and g be?

A. f(x) = x3 + 5 and g(x) = ex

B. f(x) = x3 and g(x) = 5 + ex

C. f(x) = ex and g(x) = x3 + 5

D. f(x) = ex3+5 and g(x) = x3 + 5

Click here for help

Though composition is, conceptually, probably the most complicated way to

combine functions, the theorem for calculating the derivative of a compositionis one of the simplest:

The Chain Rule: If f and g are both differentiable and F = f gis the composite function defined by F(x) = f(g(x)), then F is alsodifferentiable, and F is given by the product

F(x) = (f g)(x) = f(g(x))g(x)

Another way to put this: If y = f(x) and u = g(x) are both differen-tiable, then

dydx

= dydu

dudx

Notice, however, that the two factors on the right are not fractions! Eachrepresents a derivative.


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32 CONTENTS

The idea behind the chain rule is very simple. The following diagram illus-

trate this. Imagine starting at some initial input x, as illustrated on the leftof the diagram. The function g turns x into u, which f then turns into y.Composing the two functions f and g is a little like building a box aroundthe two components f and g to make a new input-output machine calledF (dashed line in the diagram). Suppose we change x by a small amount x.This produces a small change u in the output of g, where the relationshipbetween the two small changes is given (approximately) by the derivative:

u

x g(x)

Since u is also the input to f, the small change u in its input in turnproduces a small change y in the output of f, again related to u by thederivative of f:

y

u f(u)

The derivative ofF at x is (approximately) the ratio of the two small changes:

F(x) yx

=y

u

u

x

g fx yu = g(x)

ux yproduces produces

u

x g(x) y

u f(u)

so,


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THE CHAIN RULE 33

y

x f(g(x)) g(x) . y

u u

x

While the equations in this discussion are all in the form of approximate

equalities, in that the ratiosy

u,

u

xand

y

xall represent average rates

of change, and therefore only approximate the instantaneous rates, theseapproximations get better and better as x (and therefore also u) tend to

zero. In the limit, therefore the equality becomes an exact equality combiningderivatives.

Notice that limu0

y

urepresents the derivative of f at u = g(x); that is, it

is equal to f(u) = f(g(x)). This accounts for the form of the first factor inthe formula for the chain rule:

F(x) = f(g(x))g(x)

Example 16. Calculate the derivative of

x2 + 1

Click here for help


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34 CONTENTS

Example 17. Find the derivative of 4ex3+5.

Click here for help

Example 18. Find the derivative of F(x) = e(x2+3x+8)7

Click here for help


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INVERSE FUNCTIONS 35

Inverse Functions


Our next goal and final goal in this short introduction to derivatives is topresent the formulas for the derivatives of logarithmic functions. In orderto do this we have included a brief review of the theory of inverse functions(this section) followed by the definition of logarithmic functions (the nextsection), before we discuss their derivatives in the last section.

Note the following indication that there is computer help for this topic on thecourse web site. This computer help is interactive and designed not only to

test whether you remember the facts about inverse functions, but also to helpyou understand them.


A function f is one-to-one if no two inputs (values in its domain) have thesame output (values in its range). The following graphs illustrate this.

y

f(x)

x1 x2

Not one-to-one

The graph shows that the points x1 and x2 produce the same output(i.e. f(x1) = f(x2)), therefore the function f(x) is not one-to-one. We cangeometrically check to see if a function is one-to-one if no horizontal lineintersects the graph of the function at more than one point.


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36 CONTENTS

g(x)

satisfies horizontal line test

x

y

One-to-one

The graph of g(x) illustrates that there are no two distinct points in itsdomain that produce the same output (i.e. g(x1) = g(x2), whenever x1 = x2).Suppose we have a function f with domain D and range R. We want to lookfor a function g that undoes what f does. If such a function g exists forf, then it is called the inverse of f. In symbols this can be written as

g(y) = x f(x) = y.This can be illustrated with the following diagram:

g

fx

x

y

y

D R

Theng(f(x)) = x,

andf(g(x)) = x.


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Concept Question 9. Suppose f(x) = x + 5. What should the inverse

function g be?

A. g(y) = y + 5

B. g(x) = 1/(x + 5)

C. g(x) = x 5

Click here for help


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38 CONTENTS

Example 19. Suppose f(x) = 1/x, then what is its inverse g(x)?

Click here for help

Concept Question 10. Suppose f(x) = x2, then what is the inverse of f?

A. g(x) =

x

B. It has no inverse

C. g(x) = x2

Click here for help
http://quickneasycalculus.blogspot.com/2007/07/example-19_7832.htmlhttp://quickneasycalculus.blogspot.com/2007/07/concept-question-10.htmlhttp://quickneasycalculus.blogspot.com/2007/07/concept-question-10.htmlhttp://quickneasycalculus.blogspot.com/2007/07/example-19_7832.html


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40 CONTENTS

(b) What does the inverse function tell you?


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Graphs of Inverse Functions

Suppose g is the inverse of f and that D and R are the domain and range off, respectively. Then R and D are the domain and range of g, respectively.This means that the domain of f equals the range ofg and that the range off is the same as the domain of g.

f

y

x

y1

x1

Example 21. Sketch the graph of the inverse of f on the axes given above.

Click here for help


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42 CONTENTS

Usually we write f1, rather than g, for the inverse. Note that this means

that f1

(x) is not the same as (f(x))1

! The former is the inverse functionapplied to x, while the latter is the reciprocal of f(x).

To understand the effect of the inverse function g of a function f whosegraph is before you, you do not have to reflect the graph of f to produce thegraph of g! The effect of the function f is shown by going from a point x1on the horizontal axis via a vertical line to the graph of f, and then movinghorizontally towards the vertical axis, reaching it at y1 = f(x1). The effect ofg is seen on the same picture (without drawing the graph of g) by reversingthis procedure: start at a point y1, then travel horizontally to the graph off and then vertically to reach the horizontal axis at x1 = g(y1).


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LOGARITHMIC FUNCTIONS 43

Logarithmic Functions


One of the most important applications of the theory of inverse functions isthe introduction of Logarithmic Functions.

From its graph we know that ax (a > 1) is increasing, and therefore has aninverse. The inverse of the function ax is the logarithm to the base a,

loga(x).

Recall the relationship between a function f and its inverse g. We expressedit in three different but equivalent ways:

g(y) = x f(x) = y ,

g(f(x)) = x ,

f(g(x)) = x .

Concept Question 11. Letf(x) = ax and g(x) = loga(x). Which of the

following statements is true? ( means if and only if)[i. ] loga(y) = x ax = y [A. ] i,ii, and iii[ii. ] a(loga)

x = x [B. ] ii, iii, and v

[iii. ] loga(ax) = x [C. ] i, iii, and iv

[iv. ] aloga(x) = x [D. ] i, ii, and iv

[v. ] loga(y) = ax x = y [E. ] iii, iv, and v

Click here for help


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44 CONTENTS

The graph of loga x:

loga x

ax

1

1

y

x

What is the domain of logax ?

What is the range of logax ?

Counterpart of Laws of Exponents

1. loga(xy) = loga(x) + loga(y);

2. loga(x/y) = loga(x) loga(y);3. loga(x

y) = y loga(x).

You might say that logarithms turn products into sums, quotients into dif-ferences and powers into products. These laws are a direct consequence ofthe laws of exponents on page 57 in the textbook. MAKE SURE YOUKNOW THESE LAWS VERY WELL.


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Example 22. Evaluate log5 200 3log5 2.

Click here for help

Example 23. Solve for x in the expression log6 x + log6(x + 1) = 1.

Click here for help


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46 CONTENTS

The natural logarithm is the logarithm loge(x).

Special notation: ln x (n for natural). Note the following:

ln x = y ey = x; ln ex = x;

eln x = x (if x > 0);

ln(e) = 1.

Example 24. Solve for x in the expression ln x2x1 = 1 + ln x3x1.Click here for help


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Example 25. Since a = eln(a), therefore ax = eln(a)x. Use this to find a

formula for the derivative of ax

.

Click here for help


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48 CONTENTS

Derivatives of Log Functions


Example 26. We know that x = aloga(x). Using the Chain Rule for theright hand side, differentiate both sides and then use the result to calculatethe derivative of loga x.

Click here for help

Notice that in this proof, as in several other proofs in this introduction todifferentiation, we glossed over some fine points. In this case we used theChain Rule, which, if you go back to check, assumes that both functions inthe composition are differentiable in the first place. Here we quietly assumedthat we knew this to be true, when in fact we did not prove that loga(x) isa differentiable function. In a more rigorous treatment of Calculus this issuewould be addressed.

A special case for this formula is

d

dx(ln(x)) =

1

x.

You should MEMORIZE these formulas, as they will appear many times.


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DERIVATIVES OF LOGARITHMIC FUNCTIONS 49

Example 27. Graph the function ln |x|.

Click here for help

1 2 31234

Example 28. Find a formula for ddx ln |x|.

Click here for help


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50 CONTENTS

Example 29. Calculate the derivative of ln(| x3 + 2x 7 |).

Click here for help

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