## Summary of Topics in Chapter 3 ofApplied Calculusand Chapter 10 ofFinite Mathematics & Applied CalculusTopic: Introduction to the Derivative

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Limits | Estimating Limits Numerically | Estimating Limits Geometrically | Computing Limits Algebraically | Continuous Functions
| Average Rate of Change | Derivative | Numerical Approach | Geometric Approach | Algebraic Approach | Velocity | Derivatives of Powers, Sums, and Constant Multiples | Marginal Analysis

Limits

We write

 limx→a f(x) = L
or  f(x) → L as x → a
to mean that f(x) approaches the number L as x approaches (but is not equal to) a from both sides.

A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a. We write

 limx→a+ f(x) = L
or  f(x) → L as x → a+
and
 limx→a- f(x) = L
or  f(x) → L as x → a-
to mean that f(x) → L as x approaches a from the right or left, respectively. For limxaf(x) to exist, the left and right limits must both exist and must be equal.

We write

 limx→+∞ f(x) = L
and
 limx→-∞ f(x) = L
to mean that f(x) → L as x gets arbitrarily large or becomes a negative number with arbitrarily large magnitude, respectively.
Examples

1. As x → 3, the quantity 3x2-4x+2 approaches 17, and hence

 limx→3 (3x2-4x+2) = 27
Notice this is just the value of the function at x = 3 (see "Algebraic Approach" below).

2. Notice that the function

 x2 - 9x - 3
is not defined at x = 3. However, for other values of x, it simplifies to
 x2 - 9x - 3 = (x - 3)(x + 3)x - 3 = x + 3,
and, as x → 3, this quantity approaches 6. Therefore
 x2 - 9x - 3 → 6 as x → 3,

or  limx→3 x2 - 9x - 3 = 6

There are many more examples in the on-line tutorial on limits computed algebraically.

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Estimating Limits Numerically
 To analyze a limit of the form limx→a f(x) or limx→±∞ f(x) numerically:
• Make a table of values for f(x) using values of x that approach a closely from either side.
• If the limit exists, then the values of f(x) will approach the limit as x approaches a from both sides.
• The more accurately you wish to evaluate this limit, the closer to a you will need to choose the values of x.
• For a limit as x → +∞, use positive values of x getting larger and larger.
• For a limit as x-∞, use negative values of x getting larger and larger in magnitude.
Examples
 1. To estimate limx→3 x2 - 9x - 3 , constuct a table with values of x close to 3 on either side:

x approaching 3 from the left

x approaching 3 from the right
x
2.9
2.99
2.999
2.9999
 f(x) = x2 - 9x - 3
5.9
5.99
5.999
5.9999
 3
 3.0001 3.001 3.01 3.1 6.0001 6.001 6.01 6.1

Since the values of f(x) appear to be approaching 6 as x approaches 3 from either side, we estimate that the limit is 6.

 2. To estimate limx→ +∞ x2 - x + 12x2 - 3 , constuct a table with values of x approaching +∞:

x approaching +∞   →
x
10
100
1000
10,000
 f(x) = x2 - x + 12x2 - 3
0.461929
0.495124
0.499501
0.49995
 +∞

Since the values of f(x) appear to be approaching 0.5 as x approaches 3 from either side, we estimate that the limit is 0.5.

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Estimating Limits Geometrically
 To analyze a limit of the form limx→a f(x) or limx→±∞ f(x) geometrically:
• Draw the graph of f(x) either by hand or using technology, such as a graphing calculator.
• If you want to calculate a limit as xa for a real number a, position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the left of x = a.
• Move the point along the graph toward x = a from the left and read the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit
 limx→a- f(x)
• Repeat the above two steps, but this time starting from a point on the graph to the right of x = a, and approach x = a along the graph from the right. The value the y-coordinate approaches (if any) is then
 limx→a+ f(x)
• If the left and right limits both exist and have the same value L, then
 limx→a+ f(x)
exists and equals L.
• If you want to calculate a limit as x → +∞, position your pencil point (or the graphing calculator "trace" cursor) on a point near the right edge of the graph, and move the pencil along the graph to the right, estimating the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit
 limx→+∞+ f(x)
For x-∞, start toward the left edge, and move your pencil toward the left.
Example The following picture shows the graph of f(x). First, we calculate the limit of f(x) as x → 0 from the left:
 limx→0- f(x)

Thus,
 limx→0- f(x)
= -1

Next, we calculate the limit of f(x) as x → 0 from the right:
 limx→0+ f(x)

Thus,
 limx→0+ f(x)
= 3
Since the left- and right limits disgaree, we conclude that
 limx→0 f(x)
does not exist.

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Computing Limits Algebraically:
Limits as xa
 To compute a limit of the form limx→a f(x) algebraically:
1. Check to see whether f is a closed form function. These are functions specified by a single formula involving constants, powers of x, radicals, exponentials and logarithms, combined using arithmetic operations and composition of functions.
1. If a is in the domain of f, then limxa  f(x) = f(a).
2. If a is not in the domain of f, but f(x) can be reduced by simplification to a function with a in its domain, then (a) applies to the reduced form of the function.
3. If a is not in the domain of f, and you cannot simplify the function as in (b), then simplify as much as possible, and evaluate the limit by the numerical approach, or, if you know its graph, by the graphical approach.
2. If f is not closed-form, and a is a point at which the definition of f changes, compute the left limit and right limit seperately, and check whether they agree.

Limits as x±∞
 To compute a limit of the form limx→±∞ f(x) algebraically:

If x is approaching ±∞, then check to see whether f(x) is a ratio of polynomials. If it is, then you can ignore all but the highest powers of x in the numerator and denominator. This simpler function will have the same limit as f.

Examples
 1 Consider the limit limx→1 x2 - 9x - 3 .
 Notice that the function f(x) = x2 - 9x - 3 is closed form, and a = 1 is in its domain.
Therefore, the limit is obained by substituting x = 1 (point(a) opposite):
 limx→1 x2 - 9x - 3 = 1 - 91 - 3 = 4
 2 Now consider limx→3 x3 - 9x - 3 .
This time, a = 3, which is not in the domain of f, and so we need to first simplify f(x) to reduce it to a function that does have 3 in its domain:
 limx→3 x2 - 9x - 3 = limx→3 (x - 3)(x + 3)x - 3 = limx→3 x + 3.
Now, x = 3 is in the domain of f, and so we find the limit by putting x = 3:
 limx→3 x2 - 9x - 3 = limx→3 x + 3 = 3 + 3 = 6.
 3 Consider limx→+∞ x3 + x2 - 92x3 - x - 3 .
Here f(x) is a ratio of polynomials, so we ignore everything except the highest power of x in the numerator and denominator:
 limx→+∞ x3 + x2 - 92x3 - x - 3
=  limx→+∞ x3 + x2 - 92x3 - x - 3
=  limx→+∞ x32x3
=  limx→+∞ 12 = 12
(Cancel the x2)

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Continuous Functions

A function f is continuous at a if limxaf(x) exists, and is equal to f(a).

The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.

Examples

The function f(x) = 3x2-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers).

The function

g(x)= 4x2+1x - 3
is also a closed form function, and hence continuous on its domain (all real numbers excluding 3).

On the other hand, the function

h(x)=  -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if 1 < x ≤ 2
is not a closed-form function, and is in fact discontinuous at x = 1. (See the tutorial on continuity.)

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Average Rate of Change of f(x) over [a, b]: Difference Quotient

The average rate of change of f(x) over the interval [a, b] is

 Average rate of change = ΔfΔx = f(b) - f(a)b - a .
We also call this average rate of change the difference quotient of f(x) over the interval [a, b]. Its units of measurement are units of f(x) per unit of x.
Alternative Formulation: Average Rate of Change of f(x) over [a, a+h]
(Replace b above by a+h.)

The average rate of change of f(x) over the interval [a, a+h] is

 Average rate of change = f(a+h) - f(a)h .

Units: The units of the average rate of change are units of f per unit of x.

Example

Let f(x) = 2x2 - 4x + 1. Then the average change of f(x) over the interval [2, 4] is

Average rate of change = f(4) - f(2)4 - 2
= 17 - 12 = 8

Interpretation: If, say f(x) represents the annual profit of your company (in millions of dollars) and x represents the year since January 1990, then the units of measurement of the average rate of change are millions of dollars per year. Thus, your company made an average annual profit of \$8 million per year over the period January 1992 to January 1994.

Use the handy little utility below to compute the average change of the above function f(x) over other intervals. Enter the x-coordinates (a and b in the formula), leave everything else blank, and press "Compute." (You can also change the function to anything you like, using standard technology formatting.)

 f(x) = a =     b = Ave. Rate of Change:

You can also use the function evaluator to compute average rates of change.

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Instantaneous Rate of Change of f(x) at x = a: Derivative

The instantaneous rate of change of f(x) at x = a is given by taking the limit of the average rates of change (computed by the difference quotient) as h approaches 0.

 Instantaneous rate of change = limh→0 f(a+h) - f(a)h .
We also call this instantaneous rate of change the derivative of f(x) evaluated at x = a, and write it as f'(a) (read "f prime of a"). Its units of measurement are units of f(x) per unit of x. Thus,
 f'(a) = limh→0 f(a+h) - f(a)h .

Note:

• f'(a) = Instantaneous rate of change of f at the point a.
• f'(x) = Instantaneous rate of change of f at the point x.
Hence, the derivative f'(x) is a function of x.

Since f'(x) is a limit, it may or may not exist. That is, the quantities [f(x+h) - f(x)]/h may or may not approach a fixed number as h approaches zero. If everything works out fine and the limit exists, then we say that f is differentiable at x. Otherwise, we say that f is not differentiable at x.

On this page, we summarize three ways of obtaining the derivative of a function at a point: numerical, graphical, and algebraic.

Examples

Let f(x) = 2x2 - 4x + 1, as above. Then the instantaneous change of f(x) at x = 2 is

f'(2) = 4.
(We shall see where this answer came from below.)

Interpretation
If, say f(x) represents the annual profit of your company (in millions of dollars) and x represents the year since January 1990, then the units of measurement of the instantaneous rate of change are millions of dollars per year. Thus, your company's annual profit was increasing at a rate of \$4 million per year at the start of 1992.

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Derivative as Rate of Change: Numerical Approach

To compute an approximate value of f'(a) (for a given value of a) numerically, one can use either:

1. A table of values
2. A quick approximation
The first approach shows better and better approximations, sometimes allowing you to guess the exact value, while the second method gives a quick estimate.

Using a Table
In a table, compute a succession of values of difference quotients

 f(a+h) - f(a)h
for smaller and smaller values of h, and decide what number these values are approaching. (See the example opposite.)

A Quick Approximation
Use a single small value of h and compute the difference quotient:

 f'(a) ≈ f(a + 0.0001) - f(a)0.0001 Forward Difference Quotient
Here, we chose h = 0.0001. The smaller h, the better the approximation. (See the example opposite.)

Another Quick Approximation: Balanced Difference Quotient
The following formula often gives a better estimate of the derivative

 f'(a) ≈ f(a+0.0001) - f(a-0.0001)0.0002 Balanced Difference Quotient

Derivative Calculator (Balanced Difference Quotient Approximation)
Enter a function, enter the point a, adjust h as you want, and press "Compute".

 f(x) = a =     h = f'(a)

Examples

Continuing with the example f(x) = 2x2 - 4x + 1, let us compute an approximate value of f'(2).

Using a Table: The difference quotient (with a = 2) is

 f(2+h) - f(2)h
=
2(2+h)2-4(2+h)+1 - (2(2)2-4(2)+1)

h
The following table shows the value of this difference quotient for several values of h.

 h 1 0.1 0.01 0.001 Difference Quotient 6 4.2 4.02 4.002

As h gets smaller, we see that the value gets closer and closer to 4, so we conclude

f'(2) ≈ 4.

Using A Quick Approximation (Forward Difference Quotient): We use the formula (with a = 2)

 f'(2) ≈ f(2+0.0001) - f(2)0.0001 = f(2.0001) - f(2)0.0001 = 1.00040002 - 10.0001 = 4.0002.
(You could have used the little utility at the top of the page to do this calculation.)

Notice that the "quick approximation" method does not give the exact answer, but the balanced difference quotient will in this case (see opposite).

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Derivative as Slope: Geometric Approach

Secant and Tangent Lines
The slope of the secant line through the points on the graph of f where x = a and x = a+h is given by the slope of the line PQ in the following diagram:

 msec = Slope of secant line through P and Q = f(a+h) - f(a)h

This is also the formula for the average rate of change of f over [a,a+h]. So,

Slope of Secant = Average rate of change

The slope of the tangent line through the point on the graph of f where x = a is obtained by moving the point Q closer to P; in other words, by letting h approach 0:

 mtan = slope of tangent = limh→0 f(a+h) - f(a)h = f'(a)

This is also the formula for the instantaneous rate of change of f at the point a. So,

Slope of Tangent = Instantaneous rate of change = Derivative

We can approximate the slope of the tangent through the point where x = a by using the balanced difference quotient,

 mtan ≈ f(a+0.0001) - f(a0.0001)0.0002 .

Zooming In
We can also interpret the derivative, or slope of the tangent, at a given point on the graph as the slope of the (almost) straight line obtained by "zooming-in" to that point on the curve.(See opposite.)

Extra Topic: Graph of the Derivative
Press here for on-line on how to obtain the graph of f' from the graph of f.

Examples

Continuing with the example f(x) = 2x2 - 4x + 1,

 Slope of secant line through points where x = 2 and x = 3 = Average rate of change of f(x) over [2, 3] = 6   (see calculation above)
 Slope of tangent line through point where x = 2 = Instantaneous rate of change of f(x) at x = 2 = 4   (see calculation above)

Here is the graph with these two lines shown.

Zooming In

Here is an illustration of zooming in to a point on a graph where x = 0.75.

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Computing the Derivative Algebraically

To compute the derivative of a function algebraically, proceed as follows.

1. Write down the definition of the derivative,
 f'(x) = limh→0 f(x+h) - f(x)h .
2. Substitute for f(x+h) and f(x)
You can use an actual value for x if you are asked, say, to compute f'(3), or just leave it as x if you are asked for the derivative function f'(x) .
3. Simplify the numerator in order to factor out an "h." Then cancel the "h"s and take the limit to obtain the answer.
Sometimes, you need to do more than just simplify the numerator...
Example

Going back to our first example, f(x) = 2x2 - 4x + 1, let us now calculate f'(x) algebraically by following the steps in the adjacent window.

f'(x) =  limh→0 f(x+h) - f(x)h
=  limh→0 (2(x+h)2-4(x+h)+1) - (2x2-4x+1) h
=  limh→0 2x2+4xh+2h2-4x-4h+1-2x2+4x-1h
=  limh→0 4xh+2h2-4hh
=  limh→0 h(4x+2h-4)h
=  limh→0 (4x+2h-4)
=4x-4

Thus,  f'(x) = 4x-4.

Go to the tutorial on average rates of change for practice in computing average rates of change algebraically (whatg we did above up to the last step), or to tutorial on computing the derivative algebraically and scroll down to the box called "Computing the Derivative Algebraically." The "Help" button brings up the complete solution.

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Velocity

For an object moving in a straight line with position s(t) at time t, the average velocity from time t to time t+h is given by the difference quotient

 vaverage = s(t+h) - s(t)h .

The instantaneous velocity at time t is given by

 v(t) = s'(t) = limh→0 s(t+h) - s(t)h .
Examples

Suppose the position of a moving object is given by

s(t) = t2 -2t+4 miles
at time t hours. Then its velocity at time t is given by
s'(t) = 2t-2 miles per hour.
Thus, for example, its velocity at time t = 3 hours is
s'(3) = 4 miles per hour.

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Derivatives of Powers, Sums, and Constant Multiples

Power Rule
If f(x) = xn, then f'(x) = nxn-1. This is true for any real n. In differential notation, this reads

 ddx
xn
=nxn - 1

Sum and Constant Multiple Rules
If f'(x) and g'(x) exist, and c is a constant, then

(a) [f(x)   ±   g(x)]' = f'(x)   ±   g'(x)
(B) [cf(x)]' = cf'(x) .
In differential notation, these rules are
 (a) ddx [f(x) ± g(x)] = ddx [f(x)] ± ddx [g(x)]
 (B) ddx [cf(x)] = c ddx [f(x)]

In words:

The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

The derivative of c times a function is c times the derivative of the function.

Examples

 ddx
x3.2
=3.2x2.2
 ddx
4x - 1
= - 4x - 2
 ddx
3x1.1 + 4x - 2
= 3.3x0.1 - 8x - 3
 ddx
 1x
=
 ddx
x - 1
= - x - 2
=
- 1x2

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Marginal Analysis

If Q(x) represents any quantity such as cost, revenue, profit or loss on the sale of x items, then Q'(x) is called the marginal quantity. Thus, for instance, the marginal cost measures the increase in total cost per item. This is effectively the cost of each additional item.

The marginal cost is distinct from the average cost, which measures the average of the total cost of the first x items. Average cost is given by

C(x) =
 C(x)x
=
Total Cost

Number of Items
Examples

Suppose the cost of (the first) x items is given by

C(x) = 4x0.2 - 0.1x pounds Sterling.

Then the marginal cost is
C'(x) = 0.8x-0.8 - 0.1 pounds Sterling per unit.
In particular, C'(3) ≈ 0.23 (pounds Sterling per unit) is the approximate cost of the third unit (or the fourth unit).

The average cost of the first three units is

C(3) = C(3)3
 4.68293
1.56 pound Sterling/item

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Last Updated: May 2007