10.4: The Derivative
• The average rate of change is the ratio of the
change in y to the change in x
• The instantaneous rate of change of f at a is
the limit of the average rates of change of f over
shorter and shorter intervals around a
• Another name for the instantaneous rate of
change is THE DERIVATIVE
• Note: Sometimes the word “instantaneous” is
omitted
• Think of the odometer and the speedometer in
a car. The odometer measure the distance (in
miles) that the car travels and the
speedometer measure how fast (miles per
hour) the car travels at a particular time.
• The speedometer will give you the derivative at
a certain point.
• Ex: Before I headed to Virginia, I looked at the odometer:
50,155. The trip started at 2:00 PM and ended at 8:30 PM.
When I got to my parent-in-law’s house in VA, I checked my
odometer: 50,517.
• Find the average speed?
(50,517 - 50,155) / (8.5-2) = 362 / 6.5 = 55.7 mph
The average speed is 55.7 mph; however, during my trip, I didn’t
travel 55.7 mph all the time. Sometimes I went faster and
sometimes I went slower.
If I calculate the average speed at a smaller time interval (for
example, during 1 second interval or smaller than that), I will
have instantaneous average speed which is the derivative.
Given y = f (x), the average rate of change from x = a to x = a + h is
f ( a  h)  f ( a )
,h  0
h
The above expression is also called a difference quotient.
It can be interpreted as the slope of a secant.
f (a + h) – f (a)
h
Application:
The revenue generated by producing and selling calculators
is given by R(x) = x (75 – 3x) for 0  x  20.
What is the change in revenue if production changes from 9
to 12?
R(12) – R(9) = $468 – $432 = $36.
Increasing production from 9 to 12 will increase revenue by
$36.
The revenue is R(x) = x (75 – 3x) for 0  x  20.
What is the average rate of change in revenue (per unit change
in x) if production changes from 9 to 12?
To find the average rate of change we divide the change in
revenue by the change in production:
R (12)  R (9) 36

 12
12  9
3
Thus the average change in revenue is $12 when production is
increased from 9 to 12.
Example 1: given f(x)=x2
A)
Find the slope of the secant line for a =2 and h =1
Two ways to do this:
*** Use slope formula
We have (2, 4) and (3, 9) giving the above information
94 5
 5
3 2 1
*** Use different quotient
f ( a  h)  f ( a )
,h  0
h
f (2  1)  f (2) f (3)  f (2)


1
1
94 5

 5
1
1
TRY THE SAME PROBLEM WITH a=2, h=2. Do you get 6 for the answer?
Example 1- continue: given f(x)=x2
B)
Find and simplify the slope of the secant line for a = 2 and h is
any nonzero number.
*** Use different quotient
f (a  h)  f (a) f (2  h)  f (2)
, h  0
h
h
( 2  h) 2  2 2

h
4  4h  h 2  4

h
4h  h 2
h( 4  h)


h
h
 4h
Example 1- continue: given f(x)=x2
C)
Find the limit of the expression in part B.
f (2  h)  f (2)
lim
 lim (4  h)  4
h 0
h 0
h
D)
Find the slope of the graph and the slope of the tangent line at a=2
The slope obtained from the limit of slopes of secant lines in part C is
call the slope of the graph. Therefore, the slope of the graph and the
slope of the tangent line at a=2 is 4
E) Find the equation of the tangent line at 2
Slope m=4, (2,4)
y = mx + b
4 = 4(2) + b
-4 = b
Therefore the equation of the tangent line is y = 4x - 4
The Derivative
For y = f (x), we define the derivative of f at x, denoted
f ’ (x), to be
f ' ( x)  lim
h0
f ( x  h)  f ( x )
h
if the limit exists.
If f ’(a) exists, we call f differentiable at a.
If f ’(x) exist for each x in the open interval (a, b), then f is said
to be differentiable over (a, b).
Interpretations of the
Derivative
If f is a function, then f ’ is a new function with the following
interpretations:
■ For each x in the domain of f ’, f ’ (x) is the slope of the
line tangent to the graph of f at the point (x, f (x)).
■ For each x in the domain of f ’, f ’ (x) is the instantaneous
rate of change of y = f (x) with respect to x.
■ If f (x) is the position of a moving object at time x, then
v = f ’ (x) is the velocity of the object at that time.
Finding the Derivative
To find f ‘ (x), we use a four-step process:
Step 1. Find f (x + h)
Step 2. Find f (x + h) – f (x)
Step 3. Find f ( x  h)  f ( x)
h
Step 4. Find hlim
0
f ( x  h)  f ( x )
h
Example
Find the derivative of f (x) = x 2 – 3x
Step 1. f (x + h) = (x + h)2 – 3(x + h) = x2 + 2xh + h2 – 3x – 3h
Step 2. Find f (x + h) – f (x) =
x2 + 2xh + h2 – 3x – 3h – (x 2 – 3x) = 2xh + h2 – 3h
Step 3. Find
Step 4. Find
f ( x  h)  f ( x) 2 xh  h 2  3h

 2x  h  3
h
h
lim
h0
f ( x  h)  f ( x )
 lim (2 x  h  3)  2 x  3
h0
h
continue
Find the slope of the tangent to the graph of f (x) = x 2 – 3x
at x = 0, x = 2, and x = 3.
Solution: In example 2 we found the derivative of this
function at x to be
f ’ (x) = 2x – 3 Graphing Calculator:
Hence
Press: y=
f ’ (0) = -3
Type in the equation
f ’ (2) = 1, and
Graph
f ’ (3) = 3
Press: 2nd then Trace
Press 6 (dy/dx)
Type in the number
Enter
Practice: Find the derivatives
f ( x )  2 x  3
f ( x)  x
2
f ( x)  8 x  2 x
f ( x)  x  4
2
Example: application
The total sales of a company (in millions of dollars) t months from now
are given by S (t )  t  4
Find S(12) and S’(12), and interpret. Use these results to estimate the
total sales after 13 months and after 14 months.
Answer:
Use G.C:
S (12)  12  4  4
S’(12) = 0.125
In 12 months, the sales
will be 4 million dollars
In 12 months, the rate is
increasing at 0.125
million dollars ($125,000)
per month.
After 13 months, the sales will be 4.125 (4+0.125) million dollars
After 14 months, the sales will be 4.250 (4.125+0.125) million dollars
Nonexistence of the Derivative
Some of the reasons why the derivative of a function may
not exist at x = a are
■ The graph of f has a hole or break at x = a, or
■ The graph of f has a sharp corner at x = a, or
■ The graph of f has a vertical tangent at x = a.