Calculus I
Notes on Rates of Change and the Derivative.
Rates of Change:
We live in a constantly changing world, and we do have some quantities to talk about how fast something is changing, its rate of
change.
Rate of change in :
Respect to:
Is:
position
time
speed or velocity
speed
time
acceleration or deceleration
cost
number of items produced
marginal cost
revenue
number of items produced
marginal revenue
profit
number of items produced
marginal profit
radioactivity
time
decay rate
Since you are probably most familiar with velocity and acceleration from daily activities like driving, I will focus on them for
applications.
If we are going to make predictions of quantities related to a rate of change we have to be able to quantify it mathematically.
Example #1:
Let’s imagine a simple example of a spring without friction. Thus, it will continue to bounce up and down without stoping. The
bottom of the spring will move in simple harmonic motion.
Look at the figure to the right. As the spring bounces, the bottom of the spring will move
according to a sinusoidal wave. For simplicity, let’s assume the position of the bottom of
the spring, p, is given by
p  sin(t)
where t  time .
Looking at the figure to answer the following questions.
a) Where is the spring moving the slowest?
b) Where is it moving the fastest?
c) What are you using to answer a) and b)?
►
a) It would be moving the slowest to either the top or the bottom of the curve when it is
turning around.
b) It would be moving the fastest when it is halfway between the top and the bottom.
c) The more horizontal the curve, the slower it’s moving. While, the steeper the curve the
faster it’s moving.
□
So the steepness of the graph of a function determines the rate of change of that function. The steeper the curve, the faster it is
changing.
Tangent Lines:
So we need to quantify how fast a graph is increasing or decreasing. This leads to the following definition.
Tangent lines
An line is the tangent line to the graph of y  f (x) at x  c
if and only if,
1) the line passes though the point on the graph of y  f (x) at x  c
and
2) the line is increasing/decreasing at the same rate as the graph of y  f (x) at x  c .
Note, this definition of a tangent line might be different than ones you may have gotten in a geometry class for the tangent line of a
circle. Though, if you think about it, the definition above is a generalization of the one from geometry.
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #1 of 8
Now remember that the slope of a line is the measurement of the steepness of a line. Therefore, if we want to find the rate of change
of a function we need to find the slope of its tangent line at that point.
Example #2: Using the provided graph, find the rate of change of the graph at the given points. Assume the shown lines are tangent
lines.
►
Since the rate of change of a function is the slope of the tangent line at that point we just need
to find the slopes of the lines using the slope formula.
y  y1
m 2
x 2  x1
For the tangent line at (2,8):
The green line passes though (2,8) and (1,6). Thus,
m  1628  -2-1  2
and the rate of change at x  2 is 2.
For the tangent line at (5,5):
The blue line passes though (5,5) and (6,1). Thus,
m  1655  -14  - 4
and the rate of change at x  5 is -4.
□
Note, in the previous example we had a rate of change that was negative. What would that mean? Look back at the graph. The rate of
change was negative when the graph was decreasing, going down as you moved from left to right. While the graph was increasing,
going up as you moved from left to right, at the point when it had a positive rate of change.
Example #3: The provided graph is the graph of a position function, y = P(t), giving the position, y in cm, of an item attached to a
spring at a time, t in sec. A position of y=0 corresponds to the position that the spring would stay at rest if it wasn’t bouncing, its
equilibrium position. A positive position means the item is above the equilibrium position, and a negative position means it is below.
Estimate the velocity of the item based on the graph at the given points. Assume the shown
lines are tangent lines.
►
First, note that since specific points on each line aren’t given, we will have to make guesses
at what points are on each line. Thus, your answers may vary a little from my answers.
Also, remember that velocity is the rate of change of position with respect to time. So to find
the velocities, we will have to get the slopes of the lines.
a) The point of tangency looks like (1sec, 3cm) , and it also look like it’s going through
(3sec, 8.1cm) . Thus,
3cm
cm
velocity  8.1cm
 5.1cm
 2.55 sec
3sec 1sec
2sec
cm
Therefore, at this point it’s going up at about 2.55 sec
.
b) The point of tangency looks like (5sec, 0cm) , and it also look like it’s going through
( 7sec, -6.2cm) . Hence,
velocity 
-6.2cm  0cm
7sec 5sec

-6.2cm
2sec
cm
 -3.1 sec
cm
Therefore, at this point it’s going down at about 3.1 sec
.
c) The point of tangency looks like ( 7.5sec,-5cm) and it also look like it’s going through (8sec, -5cm) . Hence,
-5cm  5cm
0cm
cm
velocity  8sec
 0.5sec
 0 sec
 7.5sec
Therefore, at this point it’s at rest, but it’s only at rest for a moment before it begins to go back up.
□
So if we have the graph of the tangent line, we can at least estimate the rate of change of the function at a point, but how do we get
the exact slope of the tangent line or the exact rate of change if all we have is the function?
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #2 of 8
Average Rate of Change and Secant Lines:
The problem with finding the exact tangent line from just the function is that I need 2 points to get the slope and I only have 1 point
on the function that I know is also on the tangent line. Thus, before we can talk about how to get to the tangent line, we need to talk
about secant lines and what they tell us about rate of change.
Secant lines
An line is the secant line to the graph of y  f (x) at x  a and x  b
if and only if,
the line passes though the points on the graph of y  f (x) at both x  a and x  b .
So note the difference between the secant line and the tangent line. See figure to the right.
While the slope of the tangent line gives how fast the curve is increasing at that single point,
the slope of the secant line gives how fast the function would have to change on the interval,
[a,b], if it changed at a constant rate so that it would end up at the same point.
The slope of the secant line gives on average the rate of change of the function on an interval.
Average Rate of Change
The average rate of change, ARC, of f (x) on [a, b]
= the slope of the secant line passing through the points on y  f (x) at x  a and x  b
=
f (b)  f (a)
ba
Note that the formula for ARC is just a variation of the slope formula, m 
y2  y1
x 2  x1
. So when
you are finding ARC, you are finding the slope of the line through the points  a, f (a)  and  b, f (b)  .
Example #4: Find the exact average rate of change, ARC, of f (x)  6 tan(x) on 0, 3  .
►
f (0)  6 tan(0)  0
f  3   6 tan  3   6
ARC 
6 3 0
 0
3
 3  6
3
 18 3
□
Example #5: Find the exact average acceleration of v(t)  8t  3t 2 on 1hr,5hr  . v(t) is measured in mi
.
hr
►
v(1)  8(1)  3(1)2  5 mi
hr
v(5)  8(5)  3(5)2  -35 mi
hr
Averageacceleration 
-35 mi
 5 mi
hr
hr
5hr 1hr

-40 mi
hr
4hr
 -10 hrmi2
Note since v(1)>0 it was travelling forward at t = 1hr, v(5)<0, so it was travelling backwards at t = 5hr, and overall it’s accelerating
backwards on average at 10 hrmi2 .
□
Notice that the units for ARC are the units for the function divided by the units for the variable.
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #3 of 8
Difference Quotient:
Now we need to rewrite the formula for average rate of change.
Substitutions:
xa
Leads to:
b  x  x
and
ARC  f (b)b  fa (a )
changein x  x  b  a

f (x x)  f (x)
x
This last form is called the difference quotient.
Difference Quotient
The difference quotient of f (x) = DQ =
f (x  x)  f (x)
x
The main difference between average rate of change and the difference quotient is that we usually find ARC on a fixed interval, i.e.
[3,8], while with the difference quotient, we usually do it for an arbitrary x and x , but in the end they are both slopes of secant
lianes.
Example #6: Find the difference quotient for f (x)  3x  7 .
►
f (x  x)  3  x  x   7  3x  3x  7
DQ 
 3x  3x  7    3x  7 
x

3x  3x  7  3x  7 3x

x
x
3
So for f (x)  3x  7 , it’s DQ = 3 for any x and x .
□
Example #7: Find the difference quotient for f (x)  2x 2  3 .
►
f (x  x)  2  x  x   3  2  x 2  2x x  (x)2   3  2x 2  4x x  2(x) 2  3
2
 2x
DQ 
2
 4x x  2(x)2  3   2x 2  3
x

2x 2  4x x  2(x) 2  3  2x 2  3 4x x  2(x) 2

x
x
 4x  2x
So for f (x)  2x 2  3 , it’s DQ  4x  2x .
□
Since the average rate of change and the difference quotient are both slope of a secant line of the function, the difference quotient can
be used to find average rate of change.
Example #8: Use the difference quotient for f (x)  x 3 to find the average rate of change of f (x)  x 3 on  -2, 7  .
►
f (x  x)   x  x   x 3  3x 2 x  3x(x)2  (x)3
3
x
DQ 
3
 3x 2 x  3x(x) 2  (x)3    x 3 
x
 3x 2  3xx  (x) 2

x 3  3x 2 x  3x(x) 2  (x)3  x 3 3x 2 x  3x(x) 2  (x)3

x
x
So for f (x)  x 3 , it’s DQ  3x 2  3xx  (x)2 .
Using: x  -2 and x  7  -2=9
Then:
ARC  3(-2) 2  3(-2)(9)  (9) 2  12  54  81
 39
□
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #4 of 8
Approximating the Slope of the Tangent line:
For a well behaved functions, the difference quotient with a small x can be used to approximate the slope of the tangent line.
Example #9: Use the difference quotient for f (x) 
and x  0.3 to approximate the
1
x
slope of the tangent line at x  2 .
►
Slope of the tangent line
-1

2  2  0.3 
1
x  x
1
1
 x 

x    x   x  x  x 
f (x  x) 
DQ 

 x 1 x    x1 
x

x   x  x 
x x  x  x 
 x  x  x  x 


x  x  x
x x  x  x 
-1
2  2.3
-1
4.6
 -0.22

-x
-1

x x  x  x  x  x  x 
Thus, the slope of the tangent line is about -0.22.
□

As a rule you would expect a smaller x would give a better approximation for the slope of the tangent line. So let’s repeat example
#9 with a smaller x .
Example #10: Use the difference quotient for f (x) 
slope of the tangent line at x  2 .
►
From example #9.
-1
DQ 
x  x  x 
1
x
and x  0.1 to approximate the
Slope of the tangent line
-1
-1


2  2  0.1 2  2.1
-1
= -0.24
4.2
Thus, the slope of the tangent line is now about -0.24.
□

If we were to keep decreasing the size of x , we would expect to get better
approximations to the slope of the tangent line, but before we move on, let’s do an
example with a negative x .
Example #11: Use the difference quotient for f (x) 
slope of the tangent line at x  2 .
►
From example #9.
-1
DQ 
x  x  x 
1
x
and x  -0.1 to approximate the
Slope of the tangent line
-1
-1


2  2  -0.1 2 1.9 
-1
= -0.26
3.8
Thus, the slope of the tangent line is now about -0.26.
□

Using the answers from examples #10 and #11 with their graphs, we can deduce that the
tangent line of f (x)  1x at x  2 would be between -0.26 and -0.24. Thus, it would be
about -0.25, but we still can’t say what it is exactly.
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #5 of 8
The Derivative:
Since, we would expect the approximations would get better with smaller x 's , we need to use a limit to get the exact value. This
limit of the difference quotient is the derivative and will give the instantaneous rate of change or just rate of change.
The Derivative (Instantaneous Rate of Change)
The derivative of f (x) = the instantaneous rate of change of f (x)
 theslope of the tangent line at an arbitary x
 lim
x  0
f (x  x)  f (x)
x
There a number of notations for the derivative.
Notations for The Derivative
The derivative of f (x)  f '(x)  dxd f (x)   Dx f (x)
The derivative of y  y ' 
Be careful not to confuse
dy
dx
with
d
dx
.
dy
dx

dy
dx
is the derivative of y, and
d
dx
d
dx
 y  Dx  y 
is an operator that is used to denote “take the derivative of”
what is next.
Let’s start with getting a library of derivatives for some basic functions.
Example #12: Find the derivative of a) f (x)  x ; b) y  x 2 ;c) y  x 3 ;d) f (x)  1x ; e) f (x)  x ; and f) y  1 .
►
a) f (x)  x
b) y  x 2
2
f (x  x)  f (x)
x  x   x 2

f (x  x)  f (x)
f '(x)  dxd  x   lim
2
d
y '  dx  x   lim
 lim
x  0
x
x  0
x  0
x
x
x  x  x
x
2
2
2
 lim
 lim
x  2x x  (x)  x
2x x  (x) 2
x  0
x  0 x
 lim
 lim
 lim  2x  x 
x
x  0
x  0
x  0
x
x
 lim 1  1
x  0
 2x  0  2x
c) y  x 3
d) f (x) 
 x  x   x 3
f (x  x)  f (x)
 lim
x  0
x
x
3
2
2
3
2
x  3x x  3x(x)  (x)  x
 lim
x  0
x
2
3x x  3x(x) 2  (x) 3
 lim
x  0
x
2
 lim  3x  3x x  ( x) 2   3x 2  0  0
3
dy
dx

d
dx
 x 3   lim
x  0
d
dx
1
x
 lim
x  x  x
-x
 lim

x

0
x x  x  x 
x x  x  x 
 lim
-1
-1
-1

 2
x  x  x  x  x  0  x
x  0
x  0
 3x
 1    1x 
f (x  x)  f (x)
 lim x x
x  0
x
x
1
1
 x x    x   x  x  x 
x   x  x 
 lim
 lim
x  0

x

0

x
x
x


x
x
x  x  x 

  
f (x)  dxd  x1   lim
x 0
x  0
2
f) y  9
e) f (x)  x
f '(x) 
d
dx
f (x  x)  f (x)
x
1 1
0
 lim
 lim
x  0 x
x  0 x
 lim 0
 x   lim f (x  x)  f (x)  lim x  x  x
  x 0
x  0
x
x
y' 
 x  x  x   x  x  x 
x  x  x

  lim
 lim 
x  0

x

0
x  x  x  x 
 x   x  x  x 
 lim
x  0
x
x  x  x  x 
 lim
x  0
1
x  x  x

1
d
dx
1  lim
x 0
x  0
1
x


x  0  x 2 x 2x
0
□
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #6 of 8
Let’s summarize the answers from example #12 for some basic derivative that we can use in later problems.
A Few Basic Derivatives
d
x 1
dx  
 x 2   2x
d
 x 3   3x 2
dx 
d
dx
d
dx
d
dx
 1x  
-1
x2
 x   2xx
 
d
1 0
dx  
For the next example, let’s return to the problems we did in examples #9-#11.
Example #13: Find the exact slope of the tangent line to f (x) 
►
We know that
d
dx
 x1  
-1
x2
1
x
at x  2 .
. Thus,
slope of tangent line  (2)-12
 - 14
□
Hence, our estimate of -0.25 from example #10 ended up being the exact answer we
just didn’t know it at the time.
Example #14: Find the equation of the tangent line to f (x)  x 3 at x  -1 .
►
We know from previous classes that to find the equation of a line we usually need the slope and a point.
We know that dxd  x 3   3x 2 . Thus,
slope of the tangent line  3(-1)2
m3
Also, at x  -1; y  (-1)3  -1 , and we have the point (-1, -1) . Hence, using the point-slope formula,
y  y1  m  x  x1 
y  -1  3  x  -1
y  1  3  x  1
y  1  3x  3
y  3x  2
Therefore, the tangent line to f (x)  x 3 at x  -1 is y  3x  2 . You can graph both y  x 3 and y  3x  2 on your calculator to see
that y  3x  2 looks like the tangent line to f (x)  x 3 at x  -1 .
□
Let’s introduce a notation to help condense some of the work.
Evaluation Notation
f (x) evaluted at x  c
 f (x) x  c
 f (c)
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #7 of 8
Example #15: Find the rate of change of f (x)  x at x  25 .
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We know that dxd  x   2xx . Thus,
The rate of change  dxd  x 

x  25

x
2x
x  25

5
2(5)
5
10
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Example #16: If the height above ground, s(t), in meters of an object is given by s(t)  - 4.9t 2  20 where t is time in seconds. What it
the object’s velocity at t  2sec.
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- 4.9(t  t) 2  20   - 4.9t 2  20
- 4.9  t 2  2t t  (t) 2   20  4.9t 2  20
s(t  t)  s(t)
velocity  v(t)  s '(t)  lim
 lim
 lim
t  0
t  0
t  0
t
t
t
2
2
2
2
- 4.9t  9.8t t  4.9(t)  20  4.9t  20
-9.8t t  4.9( t)
 lim
 lim
 lim  -9.8t  4.9t   -9.8t  0
t  0
t  0
t  0
t
t
 -9.8t
velocity t  2sec  v(2)  -9.8(2)  -19.7 ms
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Note that just like average rate of change, the units for the derivative, f '(x) , are the units for f (x) divided by the units for the
variable, x.
Finally, since we don’t want to have to go back to the definition every time we need to take a derivative, next we will find shortcuts
to finding the derivative of certain cases. The shortcuts won’t handle all functions but they will work for the functions for now. Later
we will have to go back to the definition to develop more shortcuts.
To make the shortcuts more meaningful, I would recommend looking at the derivatives we found in example #12 before going to the
next set of notes, and see if you can see a pattern. All of these use the same shortcut.
Seminole State:Rickman
Notes on Rates of Change and the Derivative.
Page #8 of 8