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Derivatives

It is all about slope!

Slope =Change in Y

Change in X

We can find an average slope between two points.

But how do we find the slope at a point?

There is nothing to measure!



But with derivatives we use a small difference ...

... then have it shrink towards zero.

Let us Find a Derivative!

We will use the slope formula:

Slope =Change in Y

=Δy

Change in X Δx

to find the derivative of a function y = f(x)

And follow these steps:

• Fill in this slope formula:Δy

=f(x+Δx) − f(x)

Δx Δx

• Simplify it as best we can,

• Then make Δx shrink towards zero.

Here we go:



Example: the function f(x) = x2

We know f(x) = x2, and can calculate f(x+Δx) :

Start with: f(x+Δx) = (x+Δx)2

Expand (x + Δx)2: f(x+Δx) = x2 + 2x Δx + (Δx)2

Start with the slope formula:f(x+Δx) − f(x)

Δx

Put in f(x+Δx) and f(x):x2 + 2x Δx + (Δx)2 − x2

Δx

Simplify (x2 and −x2 cancel): =2x Δx + (Δx)2

Δx

Simplify more (divide through by Δx): = 2x + Δx

And then as Δx heads towards 0 we get: = 2x

Result: the derivative of x2 is 2x

We write dx instead of "Δx heads towards 0", so "the derivative of" is

commonly written

x2 = 2x

"The derivative of x2 equals 2x"

or simply "d dx of x2 equals 2x"

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What does x2 = 2x mean?

It means that, for the function x2, the slope or "rate of change" at any point

is 2x.

So when x=2 the slope is 2x = 4, as shown here:

Or when x=5 the slope is 2x = 10, and so on.

Note: sometimes f’(x) is also used for "the derivative of":

f’(x) = 2x

"The derivative of f(x) equals 2x"

Let's try another example.

Example: What is x3 ?

We know f(x) = x3, and can calculate f(x+Δx) :

Start with: f(x+Δx) = (x+Δx)3

Expand (x + Δx)3: f(x+Δx) = x3 + 3x2 Δx + 3x (Δx)2 + (Δx)3






The slope formula:f(x+Δx) − f(x)

Δx

Put in f(x+Δx) and f(x):

x3 + 3x2 Δx + 3x (Δx)2 + (Δx)3 − x3

Δx

Simplify (x3 and −x3 cancel): =3x2 Δx + 3x (Δx)2 + (Δx)3

Δx

Simplify more (divide through by Δx): = 3x2 + 3x Δx + (Δx)2

And then as Δx heads towards 0 we get: x3 = 3x2

Have a play with it using the Derivative Plotter .

Derivatives of Other Functions

We can use the same method to work out derivatives of other functions (like

sine, cosine, logarithms, etc).

But in practice the usual way to find derivatives is to use:

Derivative Rules

Example: what is the derivative of sin(x) ?

On Derivative Rules it is listed as being cos(x)

Done.

Using the rules can be tricky!

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Example: what is the derivative of cos(x)sin(x) ?

You can't just find the derivative of cos(x) and multiply it by the derivative of

sin(x) ... you must use the "Product Rule" as explained on the Derivative

Rules page.

It actually works out to be cos2(x) - sin2(x)

So that is your next step: learn how to use the rules.

Notation

"Shrink towards zero" is actually written as a limit like this:

"The derivative of f equals the limit as Δx goes to zero of f(x+Δx) - f(x)

over Δx"

Or sometimes the derivative is written like this (explained on Derivatives as

dy/dx ):





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The Definition of the Derivative

In the first section of the last chapter we saw that the computation of the slope of a

tangent line, the instantaneous rate of change of a function, and the instantaneous

velocity of an object at all required us to compute the following limit.

We also saw that with a small change of notation this limit could also be written as,

(1)

This is such an important limit and it arises in so many places that we give it aname. We call it a derivative. Here is the official definition of the derivative.

Definition

The derivative of with respect to x is the function and is

defined as,

(2)

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Note that we replaced all the a’s in (1) with x ’s to acknowledge the fact that the

derivative is really a function as well. We often “read” as “ f prime

of x ”.

Let’s compute a couple of derivatives using the definition.

Example 1 Find the derivative of the following function using the definition of the derivative.

Solution

So, all we really need to do is to plug this function into the definition of the derivative, (1), and do some

algebra. While, admittedly, the algebra will get somewhat unpleasant at times, but it’s just algebra so

don’t get excited about the fact that we’re now computing derivatives.

First plug the function into the definition of the derivative.

Be careful and make sure that you properly deal with parenthesis when doing the subtracting.

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Now, we know from the previous chapter that we can’t just plug in since this will give

us a division by zero error. So we are going to have to do some work. In this case that means

multiplying everything out and distributing the minus sign through on the second term. Doing this gives,

Notice that every term in the numerator that didn’t have an h in it canceled out and we can now factor

an h out of the numerator which will cancel against the h in the denominator. After that we can

compute the limit.

So, the derivative is,




Solution

This one is going to be a little messier as far as the algebra goes. However, outside of that it will work in

exactly the same manner as the previous examples. First, we plug the function into the definition of the

derivative,

Note that we changed all the letters in the definition to match up with the given function. Also note

that we wrote the fraction a much more compact manner to help us with the work.

As with the first problem we can’t just plug in . So we will need to simplify things a

little. In this case we will need to combine the two terms in the numerator into a single rational

expression as follows.



Before finishing this let’s note a couple of things. First, we didn’t multiply out the

denominator. Multiplying out the denominator will just overly complicate things so let’s keep it

simple. Next, as with the first example, after the simplification we only have terms withh’s in them left

in the numerator and so we can now cancel an h out.

So, upon canceling the h we can evaluate the limit and get the derivative.



The derivative is then,


Solution

First plug into the definition of the derivative as we’ve done with the previous two examples.

In this problem we’re going to have to rationalize the numerator. You do

remember rationalization from an Algebra class right? In an Algebra class you probably only

rationalized the denominator, but you can also rationalize numerators. Remember that in rationalizing

the numerator (in this case) we multiply both the numerator and denominator by the numerator except

we change the sign between the two terms. Here’s the rationalizing work for this problem,

http://tutorial.math.lamar.edu/Classes/Alg/Radicals.aspx#Pre_Rad_Rationalize






Again, after the simplification we have only h’s left in the numerator. So, cancel the h and evaluate the

limit.



And so we get a derivative of,

Let’s work one more example. This one will be a little different, but it’s got a point

that needs to be made.

Example 4 Determine for

Solution

Since this problem is asking for the derivative at a specific point we’ll go ahead and use that in our

work. It will make our life easier and that’s always a good thing.



So, plug into the definition and simplify.

We saw a situation like this back when we were looking at limits at infinity. As in that section we can’t

just cancel the h’s. We will have to look at the two one sided limits and recall that

http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx#Limit_LimAtInf_Ex3






The two one-sided limits are different and so

doesn’t exist. However, this is the limit that gives us the derivative that we’re after.

If the limit doesn’t exist then the derivative doesn’t exist either.

In this example we have finally seen a function for which the derivative doesn’t ex ist

at a point. This is a fact of life that we’ve got to be aware of. Derivatives will notalways exist. Note as well that this doesn’t say anything about whether or not the

derivative exists anywhere else. In fact, the derivative of the absolute value function

exists at every point except the one we just looked at, .



The preceding discussion leads to the following definition.

Definition

A function is called differentiable at if

exists and is called differentiable on an interval if the derivative exists for each

point in that interval.

The next theorem shows us a very nice relationship between functions that are

continuous and those that are differentiable.

Theorem

If is differentiable at then is continuous

at .

See the Proof of Various Derivative Formulas section of the Extras chapter to see the proof

of this theorem.

Note that this theorem does not work in reverse. Consider

and take a look at,

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So, is continuous at but we’ve just

shown above in Example 4 that is not differentiable

at .

Alternate Notation

Next we need to discuss some alternate notation for the derivative. The typical

derivative notation is the “prime” notation. However, there is another notation that

is used on occasion so let’s cover that.

Given a function all of the following are equivalent and

represent the derivative of with respect to x .

Because we also need to evaluate derivatives on occasion we also need a notation

for evaluating derivatives when using the fractional notation. So if we want to

evaluate the derivative at x=a all of the following are equivalent.



Note as well that on occasion we will drop the (x) part on the function to simplify the

notation somewhat. In these cases the following are equivalent.

As a final note in this section we’ll acknowledge that computing most derivatives

directly from the definition is a fairly complex (and sometimes painful) process filled

with opportunities to make mistakes. In a couple of sections we’ll start developing

formulas and/or properties that will help us to take the derivative of many of the

common functions so we won’t need to resort to the definition of the derivative too

often.

This does not mean however that it isn’t important to know the definition of the

derivative! It is an important definition that we should always know and keep in the

back of our minds. It is just something that we’re not going to be working with all

that much.



THE DERIVATIVE

The rate of change of a functionat a specific value of x

The slope of a straight line

The slope of a tangent line to a curve

A secant to a curve

The difference quotient

The definition of the derivative

The derivative of f ( x )= x 2

Differentiable at x

Notations for the derivative A simple difference quotient

Section 2: Problems

The derivative of f ( x ) = 2 x − 5

The equation of a tangent to a curve

The derivative of f ( x ) = x 3

C ALCULUS IS CONCERNED WITH THINGS that do not change at a constant rate.

The values of the function called the derivative will be that varying rate of change.

Now, since we consider x to be the independent variable and y thedependent, then any change Δx in the value of x , will result in a change Δ y inthe value of y . In a straight line, the rate of change -- so many units of y foreach unit of x -- is constant, and is called the slope of the line.

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The slope of a straight line is this number:

Δ y

Δx = =

Change in y -coördinate

Change in x -coördinate.

( Topic 8 of Precalculus.)

A straight line has one and only one slope.

If x represents time, for example, and y represents distance, then a

straight line graph that relates them indicates constant speed. 45 miles perhour, say -- at every moment of time.

The slope of a tangent line to a curveCalculus however is concerned with rates of change that are not constant.

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If this curve represents distance Y versus time X , then the rate ofchange -- the speed -- at each moment of time is not constant. The questionthat calculus asks is: "What is the rate of change at exactly the point P ?" If

we can name the slope of the tangent line to the curve at that point, then that will be the answer. And the method for finding that slope -- that number -- was the remarkable discovery by both Isaac Newton (1642-1727) andGottfried Leibniz (1646-1716). That method is the one for finding what iscalled the derivative.

A secant to a curve

A tangent is a straight line that just touches a curve. A secant is a straight line

that cuts a curve. Hence, consider the secant line that cuts the curve atpoints P and Q. Then the slope of the secant PQ is theaverage rate of changebetween those two points. For example, if

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then on changing from x 1 to x 2, the function has changed an average of4 units of y for every 5 units of x . But once again, the question calculus asksis: How is the function changing exactly at x 1? What is the slope

of the tangent to the curve at P ?

We cannot however evaluateΔ y

Δx

exactly at P -- because

Δ y and Δx

would then both be 0, and the value

of

Δ y

Δx

would be

completely

ambiguous.

Therefore we will consider shorter and shorter intervals of Δx , which will result in a sequence of secants --

-- a sequence of slopes. And we will define the tangent at P to be thelimit ofthat sequence of slopes.

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That slope, that limit, will be the value of what we will call the derivative.

The difference quotient

Let y = f ( x ) be a continuous function, and let the coördinates of a fixedpoint P on the graph be ( x , f ( x )). ( Topic 4 of Precalculus.) Let x now changeby an amount Δx . Then the new x -coördinate is x + Δx .It is the x -coördinate of Q on the graph.

But when the value of x changes, there is a resulting change Δ y in the value of y , that is, in the value of f ( x ). Its new value is f ( x + Δx ). The

coördinates of Q are ( x + Δx , f ( x + Δx )). Then

We now have the definition of the slope of the tangent line at P :

The slope of the tangent line at P is the limit of the change in the functiondivided by the change in the independent variable

as that change approaches 0.

Since Δx -- not x -- is the variable that approaches 0, x remains constant,and that limit will be a function of x . Since it will be derived from f ( x ), we call

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it the derived function or the derivative of f ( x ). To remind us that it was derivedfrom f ( x ), we denote it by f ' ( x ) -- " f-prime of x ."

This quotient --

-- is called the Newton quotient, or the difference quotient. Calculating andsimplifying it is a fundamental task in differential calculus.

Again, the difference quotient is a function of Δx . But to simplify the written calculations, then instead of writing Δx , we will write h .

Δx =h

Δ y = ( x + h ) − f ( x )

The difference quotient then becomes:

We now express the definition of the derivative as follows.

DEFINITION 5. By the derivative of a function f ( x ), we mean the followinglimit, if it exists:

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We call that limit the function f ' ( x ) -- " f -prime of x " -- and when thatlimit exists, we say that f itself is differentiable at x , and that f has a derivative.

And so we are taking the limit of the difference quotient as h approaches

0. When that limit exists, that means that the difference quotient can bemade as close to that limit -- " f ' ( x )" -- as we please. ( Lesson 2. )

As for x , we are to regard it as fixed. It is the specific value at which weare evaluating f ' ( x ).

In practice, we have to simplify the difference quotient beforeletting h approach 0. We have to express the numerator --

f ( x + h ) − f ( x )

-- in such a way that we can divide it by h .

Thus, the derivative is a rule that assigns to each value of x the slope ofthe tangent line at the point ( x , f ( x )) on the graph of f ( x ).

As an example, we will apply the definition to prove that the slope ofthe tangent to the function f ( x ) = x 2, at the point ( x , x 2 ), is 2x .

THEOREM. f ( x ) = x 2

implies

' ( x ) = 2x .

Proof. Here is the difference quotient, which we will proceed to simplify:

1)( x + h )2 − x 2

h

2) =x 2 + 2xh + h 2 − x 2

h

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3) =2xh + h 2

h

4) = 2x + h .

In going from line 1) to line 2), we squared the binomial x + h . (Lesson18 of Algebra.)

In going to line 3), we subtracted the x 2s. That is, we subtracted f ( x ).

In going to line 4), we divided the numerator by h . ( Lesson 20 of Algebra. )

We can do that because h is never equal to 0, even when we take thelimit (Lesson 2).

We now complete the definition of the derivative and take the limit:

f ' ( x ) = (2x + h )

= 2x .

This is what we wanted to prove.

Whenever we apply the definition, we have to algebraically manipulatethe difference quotient so that we can simply replace h with 0. In fact, theentire theory of limits, with all its complexities and subtleties, was invented tojustify just that. (Poor Newton and Leibniz were criticized for offeringjustifications that the 19th century inventors of limits didn't like.) We mayput h = 0 here because the difference quotient reduces to 2x + h , and istherefore a polynomial in h .

Problem. Let f ( x ) = x 2, and calculate the slope of the tangent to the graph --

a) at x = 5.

Since f ' ( x ) = 2x , then at x = 5 the slope of the tangent line is 10.

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b) at x = −3. −6.

c) at x = 0. 0.

Differentiable at x

According to the definition, a function will be differentiable at x if acertain limit exists there. Graphically, this means that the graph at that valueof x will have a tangent line. At which values, then, would a function not bedifferentiable?

Where it does not have a tangent line

Above are two examples. The function on the left does not have aderivative at x = 0, because the function is discontinuous there. At x = 0there is obviously no tangent.

As for the graph on the right, it is the absolute value function, y = |x |.(Topic 5 of Precalculus.) And it is not possible to define the tangent lineat x = 0, because the graph makes an acute angle there. In fact, the slope ofthe tangent line as x approaches 0 from the left, is −1. The slope approachingfrom the right, however, is +1. The slope of the tangent line at 0 -- which

would be the derivative at x = 0 -- therefore does not exist . ( Definition 2.2. )

The absolute value function nevertheless is continuous at x = 0. For,the left-hand limit of the function itself as x approaches 0 is equal to the right-hand limit, namely 0. This illustrates that continuity at a point is no guarantee

of differentiability -- the existence of a tangent -- at that point.(Conversely, though, if a function is differentiable at a point -- if there is

a tangent -- it will also be continuous there. The graph will be smooth andhave no break.)







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Since differential calculus is the study of derivatives, it is fundamentallyconcerned with functions that are differentiable at all values of their domains.Such functions are called differentiable functions.

Can you name an elementary class of differentiable functions?

To see the answer, pass your mouse over the colored area.To cover the answer again, click "Refresh" ("Reload").Think about this yourself first!

Polynomials.

Notations for the derivative

Since the derivative is this limit: then the symbol for taking

that limit is (Read: "dee- y , dee-x .")

For example, if

y =x 2,

then, as we have seen,

= 2x .

"Dee- y , dee-x -- the derivative of y with respect to x -- is 2x ."

We also write y ' ( x ) = 2x .

" y -prime of x is equal to 2x ."

This symbol by itself: d ("dee, dee-x ") , is called the differentiating







dx

operator . We are to take the derivative of what follows it. For example,

d

dx ( x ) signifies the derivative with respect to x of f ( x ).

d

dt (4t 3 − 5) signifies the derivative with respect to t of (4t 3 − 5).

And so on.

A simple difference quotient

The difference quotient is a version ofΔ y

Δx . And at times we

will use the latter. That is, the change in the value of afunction y = f ( x ) is y + Δ y . Hence the difference quotient is

At times it will be convenient to express the difference quotient as

Note: As Δx approaches 0 -- as the point Q moves closer to P along thecurve -- then Δ y , or equivalently, Δ f also approaches 0. That is,




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The derivative. Lesson 5, Section 2: Problems

Back to Section 1

The derivative of f ( x ) = 2 x − 5

The equation of a tangent to a curve

The derivative of f ( x ) = x 3

Problem 1. Let f ( x ) = 2x − 5.

a) Write the difference quotient and simplify it.

To see the answer, pass your mouse over the colored area.To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first!

f (x + h ) − f (x )

h =

2(x + h ) − 5 − (2x − 5)

h

=2x + 2h − 5 − 2x + 5

h

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=2h

h

= 2.

b) Evaluate f ' ( x ) at x = 9 and at x = −9.

' ( x ) = 2

= 2,

according to Theorem 4 of Lesson 2.

The rate of change of f ( x ) is 2 for all values of x . f ' ( x ) is constant. Butthat should be obvious. y = 2x − 5 is the equation of a straight line

whose slope is 2. ( Topic 9 of Precalculus.) And the value of the slope of astraight line is the rate of change of y with respect to x . So many units of y for

each unit of x .

There is no tangent to a straight line, because a tangent, by

definition, touches a curve at one point only.

Example. The equation of a tangent to a curve.

a) Calculate the slope of the line that is tangent to y = x 2 at the pointa) on the curve where x = 4.

b) What is the equation of that line?

Solution. a) The slope of the tangent to the curve at x = 4 is the value of thea) derivative at x = 4. The derivative of y = x 2 is 2x . Therefore ata) x = 4, the slope of the tangent is 8.

b) The equation of a straight line has this form:

y = ax + b ,

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where a is the slope of the line. Therefore, since a = 8, the equation is

y = 8x + b .

To find the value of b , we can now proceed as in Solution 1 to Problem1 in Lesson 34 of Algebra. Since x = 4 in the function y = x 2, then y =

16. The coördinate pair (4, 16) will solve that equation:

16 = 8· 4 + b

= 32 + b .

Therefore,

b =−16.

The equation of the tangent line is

y = 8x − 16.

See Problem 2f) below.

Problem 2 .

a) Calculate the derivative of f ( x ) = x 3. Follow the sequence ofa) Problem 1.

a) [Hint: ( a + b )3 = a 3+ 3a 2b + 3ab 2 + b 3. Topic 25 of Precalculus.]

( x + h) − f ( x )

h =

(x + h )3 − x 3

h

= x 3 + 3x 2h + 3xh 2 + h 3 − x 3 h

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=3x 2h + 3xh 2 + h 3

h

= 3x 2 + 3xh + h 2.

f ' ( x ) = (3x 2 + 3xh + h 2)

= 3x 2.

b) Evaluate the slope of the tangent to y = x 3 at x = 4.

The slope at x is 3x 2. Therefore, at x = 4, the slope is 3· 16= 48.

c) Evaluate the slope of the tangent to y = x 3 at x = −2.

3· (−2)2 = 3· 4 = 12.

d) What is the rate of change of f ( x ) = x 3 at x = −1.

3· (−1)2 = 3· 1 = 3.

At x = −1, the function is increasing at the rate of 3 unitsof y per unit of x .

e) What is the rate of change of that function at x = 5.

3· 52 = 3· 25 = 75.

At x = 5, the function is increasing at the rate of 75 units of y perunit of x .

f) What is the equation of the tangent to y = x 3 at x = 5.



At x = 5, the slope of the tangent is 75. Therefore the equation ofthe tangent will be

y = 75x + b .

To find b , proceed as in the Example above.

When x = 5, then y = x 3 = 125, so that the pair (5, 125) solve thatequation.

125 = 75· 5 + b .

Therefore, b = −250. The equation of the tangent is

y = 75x − 250.

Problem 3. Prove: The straight line that is tangent to y = x 2 at the point( a , a 2 ), bisects the distance of a from the origin.

Let x be the x -intercept of the tangent line. Then we are to provethat x = a /2.

The vertical leg of that right triangle is a 2. The horizontal leg is a − x . Therefore the slope of that line is

a2

a − x .

But the slope of that line is 2a because the derivative of x 2is 2x . Therefore,

a2

a − x = 2a







a2 = 2a(a − x )

a = 2(a −

x ) = 2a −

2 x

2 x = 2a − a = a

x =a

2

Problem 4.

a) Show:d

dx

1

x = −

1

x 2 .

d

dx

1

x =



=

To see how the difference quotient was simplified, see Lesson 3 ofPrecalculus, Problem 11c.

b) What is the rate of change of the function at x = 4?

At x = 4, − 1

x 2 = −

1

16 . The function is decreasing at the rate

of 116

of a unit of y per unit of x .

c) What is the rate of change of the function at x = ¼?

At x = ¼, − 1

x 2 = −16. The function is decreasing at the

rate of 16 units of y per unit of x .

Look at the graph. The closer to 0, the greater the rate of change. Thefurther away from 0, the smaller the rate of change.

At every point of that graph, the tangent has a negative slope --thederivative is always negative. As we move from left to right, the values ofthat function always decrease.

We have the following result, then:

d

dx

1

x

= − 1

x 2

.

That is,

d

dx x −1 = −x −2.

http://www.themathpage.com/aPreCalc/functions.htm#calc



http://www.themathpage.com/acalc/continuous-function.htm#over






http://www.themathpage.com/acalc/derivative-2.htm#1over-x








This has the form

d

dx x n = nx n −1

In what follows we will be exploiting that form.

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