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Functions, Graphs, LimitsDifferential Calculus

One-Variable CalculusPOLI 270 - Mathematical and Statistical Foundations

Sebastian M. Saiegh

Department of Political ScienceUniversity California, San Diego

September 30, 2010

Sebastian M. Saiegh One-Variable Calculus


One-Variable Calculus

1 Functions, Graphs, LimitsFunctionGraphsFunctional FormsLimits

2 Differential CalculusRate of ChangeDerivatives



FunctionGraphsFunctional FormsLimits

Functional Relationships

Political Science, economics, sociology, as well as biology,chemistry, and physics are frequently concerned with binaryrelationship between elements of different sets.

For example, we may be interested in the dependence of onequantity upon others. (i.e. a manufacturer might want toknow how profit varies with production level).

Among these relationships, we are interested in those wherethere is a correspondence between every element in one of thesets and one and only one element in the other set.

A relationship of this kind is called a functional relationships orsimply a function.




Function: Definition

Definition

Function. A function defines a rule which assigns to each elementin a set A a unique element in a set B.

In traditional calculus a function is defined as a relationshipbetween two variables.

Denote the variables by x and y .f : A→ B, assigns to each x ∈ A a unique element y ∈ B.

If every value of x is associated with exactly one value of y , then yis said to be a function of x .




Domain, Image, and Range

The most common way to denote a function is to replace the termy , called the image of the element x and to write y = f(x) (thesymbol f(x) is the value of f at x and it is usually read as “f of x”.).

If f is a function from A to B and S ⊂ A, we say that f isdefined on the set S .

The largest set on which f is defined is, of course, the set A.

We call A the domain of f.

The set f(A) is called the range of f.




Domain, Image, and Range (cont.)

For example, suppose A = {1, 2, 3}, and B = {√

2, 4}, we candefine f such that,




Domain, Image, and Range (cont.)

Because a number on the left leads to exactly one number on theright, we can say that the numbers on the right (i.e. the images)are a function of those on the left.

The arrows give us a “mapping” of domain to range.

As you can see, more than one number on the left can lead tothe same number on the right, but not vice versa




Vocabulary of Functions

It is customary to use x for what is called the independent variable,and y for what is called the dependent variable because its valuedepends on the value of x .

Values that can be taken by the independent variable are,thus, called the variable’s domain. Values that can be takenby the dependent variable are called the range.

Note also that in our example, we defined the relationship betweenthe elements of the sets A and B using arrows, which represent our“correspondence rules”.

We may define a function, thus, by simply establishing a setof rules.




Vocabulary of Functions (cont.)

Alternatively, we may specify our rule using a mathematicalequation, such as y = x2.

This equation defines a function from R to itself.

For each x ∈ R there exists a unique y ∈ R which satisfies therule y = x2.

The domain of this function is R.The range of this function is [0,∞) (recall that the product oftwo negative numbers is positive).

We can also write this function as f(x) = x2: the dependentvariable is the square of x .




Explicit and Implicit Functions

Many functions are also given by an algebraic formula.

For example, y = f(x) = x2 − x + 1.In this form, the expression is called an explicit function of x .

The equation may also be expressed as x2 − x − y − 1 = 0, and inthis case we refer to it as an implicit function of x because theexplicit form is implied by the equation.




Explicit and Implicit Functions (cont.)

So, if f(x) = x2 − x + 1, what is the value of this function whenx = −1, x = 0, x = 12 ?

We find the value by inserting the designated x value into theformula:

f(−1) = (−1)2 − (−1) + 1 = 1 + 1 + 1 = 3

f(0) = (0)2 − (0) + 1 = 1

f

(1

2

)=

(1

2

)2−(

1

2

)+ 1 =

1

4− 1

2+ 1 =

3

4.




Evaluating Functions

Our chief interest will be in rules for evaluating functions definedby formulas.

If the domain is not specified, it will be understood that thedomain is the set of all real numbers for which the formulaproduces a real value, and for which it makes sense.

For example, given the equation y = 1x ,1x is defined for all values

of x except zero; so the range is all real numbers except zero.




Evaluating Functions (cont.)

Consider now the equation y 2 = x .

This equation does not define a function from R to itself.

Take any value for which x < 0, and you will find that there isno value of y associated with such value of x .

Does this function define a function from [0,∞) to R?

Again, the answer is no.

This time is certainly true that there is a correspondence betweenevery element in the set [0,∞) and an element in the range.However, the equation does not assigns to each element in thedomain a unique element in the range.




Evaluating Functions (cont.)

The equation y 2 = x , though, does define a function from [0,∞)to [0,∞).

Given any x ∈ [0,∞), there is a unique y ∈ [0,∞) whichsatisfies y 2 = x .

Recall the discussion about roots from last week: we observe that,for each x ≥ 0, f(x) =

√x .




Cartesian Representations

It may be an overstatement to say that “a picture is worth athousand words,” yet a picture of a function is very useful.

This picture is called the graph of the function.

Equations with two variables can be represented by curves on theCartesian plane.

We start by constructing coordinate axes: we construct a pairof mutually perpendicular intersecting lines, one horizontal,the other vertical.




Cartesian Representations (cont.)

The horizontal line is often called the x-axis.

Values of the independent variable are usually represented bypoints along this axis.

We call the vertical line the y-axis.

Values of the dependent variable are represented by pointsalong this axis.

The point where these lines meet, called the origin, represents zero.

The scale of the y-axis does not need to be the same as thatfor the x-axis. In fact, y and x can have different units, suchas distance and time.





We can represent one specific pair of values associated by thefunction in the following way:

Let a represent some particular value for the independentvariable x , and let b indicate the corresponding value ofy = f(x). Thus, b = f(a)

We can now draw a line parallel to the y-axis at distance a fromthe axis, and another line parallel to the x-axis at distance b.

The point P at which these two lines intersect is designatedby the pair of values (a, b) for x and y , respectively.





The number a is called the x-coordinate of P, and the number b iscalled the y-coordinate of P.

In the designation of a typical point by the notation (a, b) we willalways designate the x-coordinate first and the y-coordinatesecond.




Graphs: Examples

For example, given f(x) = x − 1, its graph is the set of points(a, b) in the plane such that y = x − 1.

Notice that the function is defined: each vertical line cuts thegraph in one and only one place.




Graphs: Examples

Similarly, the equation y = x2 defines a function from R to itself.

Again, each vertical line cuts the graph in one and only one place.




Graphs: Examples

Consider now the correspondence rule

g(x) =

{2x − 3 if x ≤ 42 if x ≥ 4.




Graphs: Examples

This correspondence rule does not define a function: the verticalline of the equation x = 4 meets the graph in two points: (4, 2),and (4, 5).In words, the number 4 has two images. But, if we eliminate thesecond image of 4, for example:

h(x) =

{2x − 3 if x ≤ 42 if x > 4.

Then, h is a defined function.




Plotting Functions

The most direct way to plot the graph of a function y = f(x) is tomake a table of reasonably spaced values of x and of thecorresponding values of y = f(x).

Then each pair of values can be represented by a point. Agraph of the function is obtained by connecting the pointswith a smooth curve.

A function whose graph is unbroken (i.e. it can be drawn withoutlifting the pencil from the paper) is said to be continuous, whileone whose graph has a gap or a hole is discontinuous.




Discontinuous Functions

For the functions we will encounter in this class, discontinuitiesmay arise in one of the following two ways:

A function defined in several pieces will have discontinuities ifthe graphs of the individual pieces are not connected to eachother.

A function defined as a quotient will have a discontinuitywhenever the denominator is zero.




Discontinuous Functions (cont.)

For example, the graph of the function f(x) = x2+x−2x−2 , looks like

this:




Plotting Functions: Symmetry

The graphical representation of certain function can be simplified ifwe take into account its symmetry; the graph for |x | is:




Common Functions

Constant Function: it assigns a single fixed number k to everyvalue of the independent variable, x . Hence, f(x) = k .

The graph is just a horizontal line.




Common Functions (cont.)

Identity Function: A function f is identical over R if and only if theimage of every real number is that same real number.

Formally, if ∀x : f(x) = x .




Common Functions (cont.)

Absolute Value Function.

This is no other than the graph for |x |.




Linear Functions

A function defined by an equation in the form y = mx + b, wherem and b are constants, is called a linear function because its graphis a straight line.

This is a graph of a typical linear function.Sebastian M. Saiegh One-Variable Calculus



Linear Functions (cont.)

If a line is not vertical, its direction with respect to the coordinateaxes in the plane is described by a number called the slope of theline.

The slope is measured by marking two points P1 = (x1, y1)and P2 = (x2, y2) on the line and computing the ratio

Slope =change in y coordinate

change in x coordinate

or,

Slope =y2 − y1x2 − x1




Linear Functions (cont.)

Here is another way to find the slope of a straight line if itsequation is given.

If the linear function is in the form y = mx + b, then theslope is given by the expression in the previous slide.

Substituting for y , we have

Slope =(mx2 + b)− (mx1 + b)

x2 − x1=

mx2 −mx1x2 − x1

=m(x2 − x1)

x2 − x1= m.

So, for example the slope of y = 7x − 5 is simply m = 7.




Polynomials

If a0, a1, a2, ..., an are all real numbers, then the equationy = a0 + a1x + a2x

2 + ...+ anxn defines a function from R to itself.

Any value of x which is substituted on the right hand sidegenerates a unique corresponding value of y .

If an 6= 0, we call this function a polynomial of degree n.

A polynomial of degree 0 is called a constant.




Rational Functions

The quotient of two polynomials is called a rational function.

Suppose that g and h are polynomial functions. Let S denotethe set R with all the values of x for which h(x) = 0 removed.

Then, the equation

y =g(x)

h(x)

defines a function from S to R.

Such a function is a rational function.




Cubic Polynomial

Example

The function from R to itself defined by the equationy = x3 − 3x2 + 2x is called a polynomial function of degree 3 (or,more loosely, a ‘cubic polynomial’).




Rational Function: Example

Example

Let S be the set R with 2 and −2 removed. Then, the equationy = x

2+4x2−4 , (x 6= ±2) defines a function from S to R.




Algebra of Functions

If S ⊂ R and f and g are two functions from S to R, then wedefine the function f + g to be that function from S to R whichsatisfies (f + g)(x) = f(x) + g(x) (x ∈ S).

If λ is any real number, we define λf to be the function fromS to R which satisfies (λf)(x) = λf(x) (x ∈ S).Again, we define the functions fg and f/g by(fg)(x) = f(x) � g(x) (x ∈ S) and(f/g)(x) = f(x)/g(x) (x ∈ S).

For the latter definition to make sense, of course, it is essentialthat g(x) 6= 0 for all x ∈ S .




Algebra of Functions: Composition

A somewhat less trivial way of combining functions is to employthe operation of composition.

Let S and T be subsets of R and suppose that g : S → T andf : T → R.Then we define the composite function f ◦ g : S → R byf ◦ g(x) = f(g(x)) (x ∈ S).

Sometimes f ◦ g is called a ‘function of a function’.




Algebra of Functions: Composition (cont.)

Example

Let f : R→ R be defined by f(x)= x2−1x2+1

(x ∈ R) and let g : R→ Rbe defined by g(x)=x3.Then, f ◦ g(x) : R→ R is given by the formula

f ◦ g(x) = f(g(x)) = {g(x)}2 − 1

{g(x)}2 + 1=

x6 − 1x6 + 1

.




Inverse Functions

Suppose that A and B are sets and that f is a function from A toB.

This means that each element a ∈ A has a unique imageb = f(a) ∈ B.

We say that f−1 is the inverse function to f if f−1 is a functionfrom B to A which has the property that x = f−1(y) if and only ify = f(x).




Inverse Functions (cont.)

Not all functions have inverse functions.

In fact, it is clear that a function f : A→ B has an inversefunction f−1 : B → A if and only if each b ∈ B is the image ofa unique a ∈ A. (Otherwise f−1 could not be a function).

A function which has this property is said to be a 1 : 1correspondence between A and B.




Bounded Functions

Let f be defined on S . We say that f is bounded above on S by theupper bound h if and only if, for any x ∈ S , f(x) ≤ h.

This is the same as saying that the set f(S) = {f(x) : x ∈ S}is bounded above by h.

If f is bounded above on S , then it follows from thecontinuum property that it has a smallest upper bound (orsupremum) on S .




Bounded Functions (cont.)

Suppose that

k = supx∈S

f(x) = sup f(S).





It may or may not be true that, for some η ∈ S , f(η) = k .

If such value of η does exist, we say that k is the maximum of f onthe set S and that this maximum is attained at the point η.





Similar remarks apply to lower bounds and minima.

If a function f is both bounded above and below on the set S ,then we simply say that f is bounded on the set S .

From last week’s class, it follows that a function f is bounded on aset S if and only if, for some b, it is true that, for any x ∈ S ,|f(x)| ≤ b.





Example

Let f : (0,∞)→ R be defined by f(x) = 1x (x > 0).

This function is unbounded above on (0, 1]. It is, however,bounded below on (0, 1] and attains a minimum of 1 at the pointx = 1.




Limits of Functions

The idea of a limit is at the heart of calculus.

A derivative, the fundamental concept of differential calculus,is a limit.

An integral, the fundamental concept of integral calculus, is alimit.

We will first approach the study of the concept of limit in anintuitive rather than a formal way.

Next, I will give a precise mathematical definition.




Limits of Functions (cont.)

Limits describe what happens to a function f(x) as its variable xapproaches a particular number c.

To illustrate this concept, suppose you want to know whathappens to the function f(x) = x

2+x−2x−1 as x approaches 1.

Note that f(x) is not defined at x = 1.

Yet, we can get a feel for the situation by evaluating f(x) usingvalues of x that get closer and closer to 1 from both the left andthe right.





The following table summarizes the behavior of f(x) for x near 1:

The function values in this table suggest that f(x) approaches thenumber 3 as x gets closer and closer to 1 from either side.





This behavior may be described by saying “the limit of f(x) as xapproaches 1 equals 3” and expressed as

limx→1

f(x) = 3.

In more general terms, if a function f(x) is defined for values of xabout some fixed number c , and if, as x is confined to smaller andsmaller intervals about c , the values of f(x) cluster more and moreclosely about some specific number L, the number L is called thelimit of f(x) as x approaches c . It is customary to express this as,

limx→c

f(x) = L.





Note that the intervals we use lie around the point of interest c ,but that the point itself is not included. In fact, f(c) may beentirely different from limx→c f(x).

The graph of f(x) = x2+x−2x−1 is a line with a “hole” at (1,3), and

the points (x , y) approach this hole as x approaches 1 from eitherside.




Limit: Definition

So far we have discussed limits in an intuitive way. Now we areready for a precise definition of a limit.

Definition

Limit. Let f(x) be defined in an interval about x = c , but notnecessarily at x = c . If there is a number L such that to eachpositive number � there corresponds a positive number δ such that|f(x)− L| < � provided 0 < |x − c | < δ, we say that L is the limitof f(x) as x approaches c and write

limx→c

f(x) = L.




Properties of Limits

If limx→c f(x) and limx→c g(x) exist, then

limx→c

[f(x) + g(x)] = limx→c

f(x) + limx→c

g(x)

limx→c

[f(x)− g(x)] = limx→c

f(x)− limx→c

g(x)

limx→c

[k f(x)] = k limx→c

f(x) for any constant k

limx→c

[f(x)g(x)] = [ limx→c

f(x)][ limx→c

g(x)]




Limit of a constant function

For any number c and a constant k

limx→c

k = k

and

limx→c

x = c

That is, the limit of a constant is the constant itself, and the limitof f(x) = x as x approaches c is c.




Limit of a constant function (cont.)




Limits of other characteristic functions

Limit of the identity function: given the function f(x) = x ,

limx→c

x = c

Limit of a linear function: given f(x) = α + βx , (β 6= 0).limx→c

(α + βx) = α + βc

Limits of Polynomials and Rational Functions: if p(x) andq(x) are polynomials, then

limx→c

p(x) = p(c)

and

limx→c

p(x)

q(x)=

p(c)

q(c)if q(c) 6= 0.




Continuity

Earlier on, we defined a continuous function as one whose graph isan unbroken curve with no holes or gaps.

Alternatively, continuity may be thought of as the quality ofhaving parts that are in immediate connection with oneanother.

Not all the functions have this property, but those which do havespecial features that make them extremely important in thedevelopment of calculus.




Continuity (cont.)

We are ready now to define continuity in terms of limits.

Definition

Continuity. A function f is continuous at c if

(a) f(c) is defined

(b) limx→c f(x) exists

(c) limx→c f(x) = f(c).

If f(x) is not continuous at c , it is said to have a discontinuitythere.



Rate of ChangeDerivatives

Differentiation

Differentiation is a mathematical technique of exceptional powerand versatility.

It is one of the two central concepts in calculus and has avariety of applications in the social sciences, including theoptimization of functions, and the analysis of rates of change.

One of the main ideas of differential calculus is the concept ofthe derivative.

The derivative of a function is simply another function thatdescribes the rate at which a dependent variable changes withrespect to the rate at which the independent variable changes.




Differentiation

To introduce such concept it is customary to look at two problems,one from physics and the other one pertaining geometry.

The former is the calculation of the instantaneous velocity ofa moving object.

The latter is to find the exact slope of the tangent to afunction’s curve at any specified point along the curve.

Both problems lead to the same calculation: the limit of asequence of ratios when the denominator tends to zero.




Average Rate of Change of a Function

Earlier today, we defined a function as a relationship between twovariables.

It is interesting to see now what happens to one of thevariables in response to changes in the other one.

Denote, as before, the variables by x and y .

If x increases from an initial value x1 to a new value x2, thenet change in x is x2 − x1.If the corresponding y values are y1 and y2, the net change iny is y2 − y1.




Average Rate of Change of a Function (cont.)

We define the average rate of change of y to be:

average rate of change =change in y

change in x=

y2 − y1x2 − x1

.

This average provides a useful summary of what happens to y as aresult of the change in x .




Average Rate of Change of a Function: Example

Say you are in the business of producing widgets, and your weeklycosts c depend on the production level x .

If costs rise from $48,500 to $57,800 when production isincreased from 500 to 650 units, the average rate of change ofcosts is,

change in c

change in x=

57, 800− 48, 500650− 500

=9300

150= 62 dollars per unit

On the average, costs increase by $62 for each additional widgetproduced when production is raised from 500 to 650 units.




Rate of Change: Notation

It is customary to use the following notation to describe changes ina variable.

If x goes from x1 to x2, this change is called an increment inx and is indicated by the symbol

∆x = (final value)− (initial value) = x2 − x1.

The symbol ∆x is read as “delta x”. The corresponding incrementin the dependent variable y is ∆y = y2 − y1, and the formula forthe average rate of change can be rewritten as ∆y∆x .




Rate of Change: Notation (cont.)

Notice that we always take (final value) minus (initial value) indiscussing an increment in some variable and in forming averages.

Example

If y = f(x) = 2 + x + x2, wan can find the average rate of changein y as x increases from 0 to 1 in the following way. Clearly∆x = x2 − x1 = 1− 0 = 1. And, y1 = f(x1) = 2 + (0) + (0)2 = 2;y2 = f(x2) = 2 + (1) + (1)

2 = 4. Therefore, the correspondingincrement in y is ∆y = y2 − y1 = 4− 2 = 2, and the average rateof change is

∆y

∆x=

2

1= 2.




Rate of Change: Example

Example

A demographic study shows that the number y of residents in aparticular city is described by the function f(x) = 100, 000 � 2

x10 ,

where x represents years elapsed since 1950.

Suppose that we want to find the average rate of change in y (theaverage population growth rate) over the period from 1965 to1975:The years 1965 and 1975 corresponds to x values x1 = 15 andx2 = 25, so the increment in x is ∆x = x2 − x1 = 25− 15 = 10years.




Rate of Change: Example (cont.)

The corresponding y values and increment ∆y are:

y1 = f(x1) = 100, 000(21510 ) = 100, 000 � 2

32 = 100, 000(2

√2) = 282, 800,

y2 = f(x2) = 100, 000(22510 ) = 100, 000 � 2

52 = 100, 000(4

√2) = 565, 600,

∆y = y2 − y1 = 565, 600− 282, 800 = 282, 800,

So, the average growth rate over this particular 10-year period is

∆y

∆x=

282, 800

10= 28, 280 residents per year .





We can use this last example to illustrate the graphicalinterpretation of the average rate of change.





The initial and final values of x are marked on the horizontal axis.

Over them sit the points Q1 = (x1, y1) and Q2 = (x2, y2)corresponding to the initial and final situations in 1965 and1975.

The fact that ∆x = 10 means that x moves 10 units to the righton the horizontal axis.

Because y values represent the height of the graph above thex axis, the increment ∆y = 282, 800 is the amount the graphrises as a result of this change in x .





If we draw a straight line L between Q1 and Q2, the average rateof change ∆y∆x is just the slope of this straight line.

This geometric interpretation is true for any function y = f(x).

The average rate of change ∆y∆x is always equal to the slope of thestraight line joining the corresponding initial and final points onthe graph:

average rate =∆y

∆x=

amount graph rises (or falls)

horizontal distance covered.




Rate of Change at a single moment

Average rates of change are important, but in many situations theyare not quite appropriate.

Suppose you drive 150 miles in 3 hours. Your average speedfor the trip is 50 mph. But your speed can be 70 mph some ofthe time, and 40 mph at other moments.

If you try to argue your way out of a speeding ticket, the fact thatyour average speed for the trip was 50 mph, even if provable, doesnot interest the police officer.

He is concerned with your speed at the moment he spottedyour car.




Rate of Change at a single moment (cont.)

This example indicates the need for a concept of rate of change ata single moment: the time of apprehension in this case.

We will proceed now to describe this concept -theinstantaneous rate of change of one variable with respect toanother.




Instantaneous Rate of Change

Suppose the relationship between two variables is given by somefunction y = f(x).

If x1 is a fixed “base value” of the independent variable x ,consider what happens as x changes by an amount ∆x fromx1 to x2 = x1 + ∆x .

For most functions, the average rates of change ∆y∆x will beabout the same as long as ∆x is small.

In fact, as the increment ∆x is made smaller and smaller, theaverages tend toward a “limit value” interpreted as theinstantaneous rate of of change of y when x = x1.

Namely, the average rates hover about the instantaneous rate andget closer to it as ∆x gets close to zero.




Instantaneous Rate of Change (cont.)

Example

Let f(x) = x2, and consider the base value x1 = 1. We are lookingfor the instantaneous rate of change.

The change in x is x2 − x1 = ∆x . The corresponding y values andincrement ∆y are

y1 = f(x1) = (1)2 = 1

y2 = f(x2) = (1 + ∆x)2 = 1 + 2(∆x) + (∆x)2

and

∆y = y2 − y1 = 2(∆x) + (∆x)2.





The formula for the average rate of change

∆y

∆x=

2(∆x) + (∆x)2

∆x

simplifies if we cancel ∆x in the numerator and denominator forany non-zero increment:

∆y

∆x= 2 + ∆x





The following table lists different values for ∆y∆x as ∆x becomessmaller:

Note that when ∆x is small, the averages are all close to 2.





In fact, from the equation we can see that the averages ∆y∆xapproach the value 2 as the increment in ∆x approaches zero; theconstant term 2 stays fixed while the second term becomes smallerand smaller.

Therefore, we are led to assign the value 2 as the limit valueof the averages ∆y∆x . This limit value tells us how fast y ischanging when x = 1.

For any continuous function defined by, say, y = f(x), if we ask“At what rate does y change as x changes?”, we can find theanswer by taking the following limit:

lim∆x→0

∆y

∆x.





In the preceding example, the averages ∆y∆x = 2 + ∆x tend toward2 as ∆x gets smaller and smaller; we can express this as

lim∆x→0

∆y

∆x= 2.

or

lim∆x→0

(2 + ∆x) = 2.




Instantaneous Rate of Change (The “Delta Process”)

Definition

Given a function y = f(x) and some base point x , consider thevalues of f at nearby points x + ∆x for small nonzero increments∆x . Then compute:

(1.) The value of f at the base point, namely y1 = f(x)

(2.) The value of f at the nearby point, namely y2 = f(x + ∆x)

(3.) The change in y : ∆y = y2 − y1 = f(x + ∆x)− f(x)(4.) The average rate of change going from x to x + ∆x :

∆y∆x =

y2−y1∆x =

f(x+∆x)−f(x)∆x

(5.) The limit value: lim∆x→0∆y∆x as the increment ∆x → 0.

We call this limit value the instantaneous rate of change of y atthe base point x .




Instantaneous Rate of Change: Examples

This limit value can often be computed for an arbitrary base point.

Example

Instantaneous rate of change of f(x) = x2 at an arbitrary basepoint x .

At base point x and nearby point x + ∆x , the y values are

y1 = x2

y2 = (x + ∆x)2 = x2 + 2x(∆x) + (∆x)2

and the increment in y is

∆y = y2 − y1 = x2 + 2x(∆x) + (∆x)2 − x2

= 2x(∆x) + (∆x)2




Instantaneous Rate of Change: Examples (cont.)

Therefore the average rate of change is

∆y

∆x=

2x(∆x) + (∆x)2

∆x= 2x + (∆x)

for any nonzero increment ∆x . As ∆x gets smaller and smaller,the base point x does not vary, so the first term 2x stays fixed.

The second term becomes very small as ∆x approaches zero,so the instantaneous rate of change is given by the limit value

lim∆x→0

∆y

∆x= lim

∆x→0(2x + ∆x) = 2x

for any base point x . If x = 1 the rate of change takes thevalue 2(1)=2, as in the previous example.





Example

Lets calculate the instantaneous rate of change of f(x) = 1x at anarbitrary base point x using the delta process.

First, notice that the function is undefined at x = 0, so it ismeaningless to discuss the instantaneous rate of change there.

For base points x 6= 0, we must examine the behavior of ∆y∆xfor small nonzero increments ∆x .





We know that at the base point x ,

y =1

x

and that at the nearby point x + ∆x ,

y =1

x + ∆x

so that,

∆y

∆x=

[ 1x+∆x −1x ]

∆x

for all small increments ∆x .





The limit value of ∆y∆x is not at all apparent from this formula.

But if we simplify the expression algebraically, some of the ∆xcancel, and it is easier to see what happens as ∆x gets small.

∆y

∆x=

1

∆x�[ 1

x + ∆x− 1

x

]=

1

∆x�[x − (x + ∆x)

x(x + ∆x)

]=

1

∆x�[ −(∆x)

x2 + x(∆x)

]=

∆x

∆x�[ −1

x2 + x(∆x)

]=

−1x2 + x(∆x)





As ∆x approaches zero, all terms involving ∆x become very small,and the remaining terms stay fixed.

With this in mind, notice how the denominator in the aboveequation behaves:

lim∆x→0

(x2 + x(∆x)) = x2.

Because the numerator stays equal to −1 and thedenominator approaches the value x2, the quotientapproaches −1

x2as ∆x approaches zero.

Thus, the instantaneous rate of change of this function is

lim∆x→0

∆y

∆x=−1x2.




Derivative of a Function

The instantaneous rate of change of a function y = f(x) is soimportant that it is given a special name.

It is called the derivative of f(x) and is denoted by the symbolf ′(x).

The derivative measures how fast the value of f(x) is changingat the base point being considered.

At some base points, f is changing rapidly; at some others, itchanges slowly.

Thus, the derivative f ′(x) depends on the base point x , and soshould be thought of as a new function “derived from” the originalfunction f(x).




Derivative of a Function (cont.)

Just like Balzac’s character in “The Deputy of Arcis”, who wastalking prose without knowing it, we have already computed thederivatives of some simple functions in today’s class using the deltaprocess:

If f(x) = x2, its derivative (instantaneous rate of change) isf ′(x) = 2x

If f(x) = 1x , its derivative (instantaneous rate of change) isf ′(x) = −1

x2




Derivative: Definition

Definition

Given a function f(x), we define the derivative f ′(x) at base pointx to be the limit value of averages

f ′(x) = lim∆x→0

( f(x + ∆x)− f(x)∆x

)= lim

∆x→0

∆y

∆x

provided this limit value exists. In that case we say f isdifferentiable at x .




Derivative: Notation

Various symbols other than f ′(x) are usually used for thederivative of y = f(x).

In this class we will use the most popular symbols,

f ′(x), dfdx

, dydx

, ddx

(f(x)), Df

interchangeably.




Differentiation Rules

The basic method for calculating derivatives is the delta processexplained before, which can be time consuming.

I shall apply the delta process once and for all to obtaindifferentiation rules that allow us to write down at a glancethe derivatives of simple functions such as polynomials.

Thereafter, I will appeal to these rules rather than to the deltaprocess whenever possible.




Differentiation Rules (cont.)

The first rule gives the derivative of a simple power of x ,f(x) = x r , when r is a fixed constant.

Two special cases should be mentioned first:

If r = 0, then f(x) = x0 = 1 (constant function everywhereequal to 1).

If r = 1, then f(x) = x1 = x (identity function).

The derivatives of these functions are very easy to find by the deltaprocess (thus, I will not do it here – you can try it at home if youare bored):

If f(x) = 1, then f ′(x) = 0

If f(x) = x , then f ′(x) = 1





We also already know that:

If f(x) = x2, its derivative is f ′(x) = 2x

Lets work out the derivative of y = x3 at an arbitrary base point:At base point x and nearby point x + ∆x , the y values are

y1 = x3

y2 = (x + ∆x)3 = x3 + 3x2(∆x) + 3x(∆x)2 + (∆x)3


∆y = y2 − y1 = x3 + 3x2(∆x) + 3x(∆x)2 + (∆x)3 − x3

= 3x2(∆x) + 3x(∆x)2 + (∆x)3






∆y

∆x=

3x2(∆x) + 3x(∆x)2 + (∆x)3

∆x= 3x2 + 3x(∆x) + (∆x)2

for small increments ∆x .

Now let ∆x . get smaller and smaller. The base point x doesnot vary, so the first term 3x2 stays fixed.

The last two terms involve ∆x and obviously become smaller andsmaller as ∆x approaches zero, so that the instantaneous rate ofchange is given by

lim∆x→0

∆y

∆x= 3x2

for any base point x .





Now that we have worked out the derivatives of a few powers x r ,we can put this information together in the following table:

Notice that a regular pattern seems to be emerging. If f(x) = x4,we might guess (correctly) that f ′(x) = 4x3.





More generally, if f(x) = x r (and r is a positive integer), it seemsreasonable to conjecture that f ′(x) = rx r−1, because this formulayields the right result for r = 0, 1, 2, 3, 4.

This formula is actually valid for any value of r , whether ornot it is a positive integer.




Differentiation: Power Rule

Power Rule. If f(x) = x r , its derivative is

ddx

(x r ) = rx r−1

for any fixed exponent r




Differentiation Rules: Constants

So far we have looked at equations in which as x changes its value,so does y .

Our next step is to find out what effect on the process ofdifferentiating is caused by the presence of constants, that is,of numbers which don’t change when x or y changes its value.

Example

Lets calculate the instantaneous rate of change of y = x3 + 5 at anarbitrary base point x .




Differentiation Rules: Constants (cont.)

Just as before, at base point x and nearby point x + ∆x , the yvalues are

y1 = x3 + 5

y2 = (x + ∆x)3 + 5 = x3 + 3x2(∆x) + 3x(∆x)2 + (∆x)3 + 5


∆y = y2 − y1 = x3 + 5 + 3x2(∆x) + 3x(∆x)2 + (∆x)3 − (x3 + 5)= 3x2(∆x) + 3x(∆x)2 + (∆x)3


3x2 + 3x(∆x) + (∆x)2




Differentiation Rules: Constants (cont.)

The last two terms involve ∆x and obviously become smaller andsmaller as ∆x approaches zero, so that the instantaneous rate ofchange is given by

lim∆x→0

∆y

∆x= 3x2

for any base point x .

So the 5 has quite disappeared. It added nothing to thechange in x , and does not enter into the derivative.

If we had put 7, or any other number, it would have disappeared.

So if we take the letter a, or b, or c to represent any constant,it will simply disappear when we differentiate.




Derivative of a Constant

Consider the graph of a constant function, y = c : it is a horizontalline, and the slope is zero.

Thus, for example if f(x) = 5, then f ′(x) = 0.

Derivative of a Constant. For any constant c , its derivativeis

ddx

(c) = 0

That is, the derivative of a constant is zero.




Derivative of a Constant (cont.)

Consider now what happens if we have a function y = c � x , wherec is a constant.

We can find the derivative of this function using the deltaprocess: if x is a fixed base point and ∆x a nonzeroincrement, the average rate of change for y = c � x is

y(x + ∆x)− y(∆x) = c(x + ∆x)− cx = (cx + c∆x)− cx = c∆x

Therefore,

dy

dx= lim

∆x→0

∆y

∆x= lim

∆x→0

c∆x

∆x= c .





Similarly, say you multiply a function f(x) by a constant c . Youwould obtain a new function c � f(c).

For example, if f(x) = x3 and c = 5, c � f(c) = 5x3. We canfind the derivative of this function using the delta process: ifx is a fixed base point and ∆x a nonzero increment, theaverage rate of change for y = c � f(x) is

∆y

∆x=

c � f(x + ∆x)− c � f(x)∆x

= c �f(x + ∆x)− f(x)

∆x= c �

∆f

∆x





By definition of the derivative f ′(x), the averages ∆f∆x approachf ′(x) as ∆x gets small.

Clearly, c � ( ∆f∆x ) must then approach c � f′(x), so that

dy

dx= lim

∆x→0

∆y

∆x= lim

∆x→0c �

∆f

∆x= c � f ′(x) = c �

df

dx.




The Constant Multiple Rule

The Constant Multiple Rule. For any constant c , itsderivative is

ddx

(cf) = c dfdx

This rule expresses the fact that the curve of the functiony = cf(x) is c times as steep as the curve y = f(x).




The Constant Multiple Rule (cont.)

Example

Let f(x) = −7x5. The derivative of this function is f ′(x) = −35x4.

d

dx(−7x5) = (−7) d

dx(x5) = −7(5x4) = −35x4.




Derivative of Sums

Suppose we decide now to combine two functions f(x) and g(x)and form their sum f + g by adding their values for each x :(f + g)(x) = f(x) + g(x)

Thus, for example, if f(x) = x2 and g(x) = x , we obtainf + g = x2 + x .

The next rule states that a sum can be differentiated term byterm.




The Sum Rule

The Sum Rule. The derivative of a sum is the sum of thederivatives

ddx

(f + g) = dfdx

+ dgdx




The Sum Rule (cont.)

Example

Let y = x2 + 3x5.

We know that ddx (x2) = 2x , and that ddx (3x

5) = 15x4.

According to the sum rule, we can simply add thesederivatives to get the derivative of the sum x2 + 3x5.

Namely,

d

dx(x2 + 3x5) =

d

dx(x2) +

d

dx(3x5) = 2x + 15x4.




Differentiation Rules: Polynomials

By combining the sum rule with the power and constant multiplerules, we can differentiate any polynomial.

Example

Let f(x) = 3x3 + 5x + 7.

d

dx=

d

dx(3x3) +

d

dx(5x) +

d

dx(7)

= 9x2 + 5x0 + 0

= 9x2 + 5




Differentiation Rules: Multiplication

Suppose that we want now to differentiate the producty = x2(3x + 1).

We may be tempted to differentiate the factors x2 and 3x + 1separately and then multiply our answers.

That is, since f ′(x2) = 2x and f ′(3x + 1) = 3, we mayconjecture that f ′(x) = 6x .

However this answer is wrong.

To see this, we can rewrite the function as y = 3x3 + x2 andobserve that the derivative is 9x2 + 2x and not 6x .




Product Rule

The derivative of a product, thus, is not the product of theindividual derivatives.

Here is the correct formula for the derivative of a product.

The Product Rule. If f(x) and g(x) are differentiable, theirproduct (f � g)(x) has derivative

ddx

(fg) = f(x) dgdx

+ g(x) dfdx




Derivative of a Product

The derivative of a product is the first factor times the derivative ofthe second plus the second factor times the derivative of the first.

According to the product rule, thus, the derivative of theproduct in the previous example is:

d

dx[x2(3x + 1)] = x2

d

dx(3x + 1) + (3x + 1)

d

dx(x2)

= x2(3) + (3x + 1)(2x)

= 9x2 + 2x

which is precisely the result that we obtained before when theproduct was multiplied out and differentiated as a sum.




Derivative of a Quotient

Recall that the quotient of two functions f(x) and g(x) is definedby taking the quotient of their values for each x :

f

g(x) =

f(x)

g(x).

This process yields such functions as 1x where f(x) = 1 andg(x) = x , and y = x

(x2+1), where f(x) = x and g(x) = x2 + 1.

We also know that the quotient function f(x)g(x) is defined

wherever g(x) 6= 0.




Quotient Rule

The derivative of a quotient is not the quotient of the individualderivatives. As with products, there is a differentiation formula forquotients.

The Quotient Rule. If f(x) and g(x) are differentiable, their

quotient f(x)gx has derivative

ddx

(f(x)g(x)

)=

g(x) dfdx−f(x)dgdx

(g(x))2

for all x where g(x) 6= 0.




Quotient Rule (cont.)

The quotient rule is probably the most complicated formula wehave seen so far.

Here is one way to remember it: “low dee high, high dee low,low low”.

Or, you may start by squaring the denominator, which is thedenominator in the original quotient, and then take that as thefirst term in the numerator.

Once you do that, you are half way through.

The final step is to think of the product rule, except that itcontains a minus sign. So, you just switch the terms. Bingo!





Example

Let f(x) = 1 and g(x) = x2. Find ddx (1x2

).

Here f(x) = 1 and g(x) = x2, so that dfdx = 0,dgdx = 2x . Thus

d

dx

( 1x2

)=

g(x) dfdx − f(x)dgdx

(g(x))2

=(x2)(0)− (1)(2x)

(x2)2

=−2xx4

= − 2x3

for all x 6= 0.





I just picked a very simple example, but the quotient rule issomewhat cumbersome. So, it is better not to use it unnecessarily.

Example

Let y = 23x2− x3 +

45 +

x+1x .

We can rewrite the function as y = 23 x−2 − 13 x +

45 + 1 + x

−1, andthen apply the power rule term by term to obtain

dy

dx=

2

3(−2x−3)− 1

3+ 0 + 0 + (−1)x−2

= −43

x−3 − 13− x−2

= − 43x3− 1

3− 1

x2




Derivatives of Composite Functions

In many situations, a quantity is given as a function of one variablewhich, in turn, can be though of a function of a second variable.

For example, if f(x) = x3 and g(x) = 5x + 1, then thecomposite function y = f(g(x)) = (5x + 1)3.

This function y = (5x + 1)3 may look too complicated at firsthand to tackle directly.




Derivatives of Composite Functions (cont.)

However, there is a very simple way to handle these functions:write some symbol such as u for the expression 5x + 1; then theequation becomes

y = u3.

Then,

dy

du= 3u2,

and for the expression u = 5x + 1,

du

dx= 5.





Now, we just have to calculate

dy

dx=

dy

du

du

dx,

that is

dy

dx= 3u2(5)

= (3)(5x + 1)2(5)

= 15(5x + 1)





This “trick” is known as the “chain rule”.

Let the composite function g(h(x)) be the function formedfrom functions g(u) and h(x) by substituting h(x) for u in theformula for g(u).




Chain Rule

The Chain Rule. If g(u) and h(x) are differentiable func-tions,

ddx

g(h(x)) = g ′(h(x))h ′(x)




Chain Rule

To see that this in nothing more than a restatement of the aboveexample, suppose that y = g(h(x)). Then, y = g(u), whereu = h(x)and by the chain rule,

dy

dx=

dy

du

du

dx= g ′(u)h ′(x) = g ′(h(x))h ′(x).




Chain Rule (cont.)

Example

Let f(x) =√

x2 + 3x + 2. We can find the derivative of thisfunction using the chain rule.

Think of f(x) as the composite function g(h(x)), where

g(u) =√

u = u12 , and h(x) = x2 + 3x + 2.




Chain Rule (cont.)

Then,g ′(u) =

√u = 12 u

− 12 , and h ′(x) = 2x + 3,

and, by the chain rule,

f ′(x) = g ′(h(x))h ′(x)

=1

2(x2 + 3x + 2)−

12 (2x + 3)

=2x + 3

2√

x2 + 3x + 2


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Differential CalculusRate of ChangeDerivatives

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