We learned in the previous section that to find a function (the
derivative of a function) that can be used find the instantaneous
rate of change at a point on the original function we could use the
limit process. In this Chapter we will find the derivative of a
function using some short cuts. This however does not imply that
the limit process cannot be used instead of these short cuts.
Differentiation Rule #1Theorem 2.2 – The Constant Rule
The derivative of a constant function is 0. (Yippee)
Since the derivative function is a function representing a slope formula, it makes sense
that the slope formula would equal zero since a constant function is horizontal over the
set of real numbers.
Examples:
f ( x)  3
f ( x)  0
y  5
dy
0
dx
s t   4
s(t )  0
p (t )  3
p(t )  0
Differentiation Rule #2
Theorem 2.3 - The Power Rule:
If "n" equals a rational number then:
If f ( x)  x n 
d n
( x )  n  x n 1
dx
f ( x)  n  x n 1
For f to be differentiable at x=0, n must be a number such that x n 1
is defined on an interval containing 0.
Examples:
If f ( x)  x1  f ( x)  1 x 0  1
If y  x 2 
y
 2  x1  2 x
x
1
4
1

1  4 1 1  34
1

If s  t   t  s  t   t  s (t )  t
 t  3
4
4
4t 4
4
1
dy
2
 y  x 2 
 2  x 21  2 x 3   3
2
x
dx
x
Showing Why the power Rule Works using the difference quotient and the limit process:
If y 
Suppose you were asked to find the derivative of the function y  x3 . Using the limit approach to
differentiation we would get the following:
f ( x  x)  f ( x)
( x  x)3  x 3
x 3  3x 2 x  3xx 2  x 3  x 3
 lim
 lim
x 0
x 0
x 0
x 0
x
x
x
2
2
3
2
2
3x x  3xx  x
x(3x  3xx  x )
lim 
 lim
 3x 2
x 0

x

0
x
x
lim x3  lim
n
In general, when you differentiate a power function x this factoring out of delta x from
the numerator will always leave you with one term without a delta x. Then when you
find the limit as delta x approaches zero all those delta x terms go to zero so you will
always be left with the term without delta x which will always have the form n  x n 1 .
Differentiation Rule #3
Theorem 2.4 – The Constant Multiple Rule
If f is a differentiable function and "c" is a real number, then c  f is also differentiable
d
and
c  f ( x)  c  f ( x)
dx
Examples:
d
( x)  2 1x 0  2
dx
4t 2 4 2
4
4
8
f (t ) 
 t  f (t )  f (t )   2t1  t
5
5
5
5
5
1
3

3
3
 1 
y
 3 x 2  3    x 2   3
x
 2
2x 2
y  2 x  y  2
1
4
1 1
4 5
4  x5
x
4
4x
5
5

f ( x)  5 
 f ( x) 


3
3
3
3 x
1
5
4
15 x
4
5
Differentiation Rule #4
Theorem 2.5 – The Sum and Difference Rule
The sum or difference of 2 differentiable functions is also differentiable and the
derivative of the sum of difference is the sum or difference of the derivatives of the
two functions.
d
 f ( x)  g ( x)  f ( x)  g ( x)
dx
Examples of Rule 4
f ( x)  2 x 2  3 x  f ( x)  4 x  3
g ( x) 
2 2
4
x  4 x  2  g ( x)  x  4
3
3
Derivative Rule # 5
Theorem 2.6 – Derivatives of the Sine and Cosine Functions
d
sin x  cos x
dx
d
cos x   sin x
dx
Recall the derivative of a function is a function you can use to calculate the slope of a tangent line at
any point on the function.
Why is the derivative of the 𝐬𝐢𝐧 𝒙 = 𝒄𝒐𝒔𝒙 ?
(From 0
𝜋
2
) The sine function graph would have tangent lines with positive slopes because the
𝜋
function increases from (From 0 2 ) and the cosine function (From 0
which correspond to the slope values of the sine function.
𝜋
2
) produces positive outputs
Why is the derivative of the 𝐜𝐨𝐬 𝒙 = −𝒔𝒊𝒏𝒙?
(From 0
𝜋
2
) The cosine function graph would have tangent lines with negative slopes because the
𝜋
function decreases from (0 2 ) but the sine function produces positive outputs from (From 0
you need to take the opposite of those positive outputs to get the appropriate negative slopes.
𝜋
2
). So
Examples:
y  2sin x  y  2 cos x
f ( x) 
cos x 1
1
 cos x  f ( x)   sin x
3
3
3
y  4 cos x  3sin x 
dy
 4sin x  3cos x
dx
Rates of Change
If you recall, the first question of calculus was how to find the instantaneous rate of
change. We know that distance changes over time but many other things besides
distance can change over time. We can use derivatives of any position function to find
the instantaneous rate of change of anything that changes.
Any function whose position changes relative to time relative to the origin at a
particular time can be represented by the general position function:
s change in distance(position)

 average velocity
t
change in time
Suppose the position of a rock dropped from a height of 40 feet measured in
feet “s”after time “t” measured in seconds is given by the function
s(t )  16t 2  40 .
What if you wanted to find the instantaneous rate of change of the position
function given at t = 1 second.
This function is known as the instantaneous velocity at time 1 and the
velocity function is the derivative of the position function. To calculate
instantaneous velocity at a time “t” you need to evaluate the velocity
function for “t”. We will use the previous example to find the velocity of the
rock at time = 1 second.
Match the given y  f ( x) graphs with the corresponding y  f ( x) graphs. Keep in mind
the graph of y  f ( x) represents the slopes of the tangent line of the original function.
s(t )  16t 2  40
s(t )  v(t )
s(t  t )  s(t )
 32t  s(t )
t 0
t
The instantaneous velocity at t = 1  v(1)  32(1)  32 ft / sec
Example:
velocity  v(t )  lim
At t=0 a diver jumps from a platform diving board that is 32 feet
above the water. The position of the diver at time “t” seconds is
given by the position function s(t )  16t  16t  32 .
2
a. When does the diver hit the water?
b. What is the divers velocity at impact?
Solution:
To find the time when the diver hits the water you need to set the position function
equal to 0 and solve for t.
0  16t 2  16t  32
0  16(t 2  t  2)  0  16(t  2)(t  1)
t  2(reasonable) or t  1 (not reasonable)
To find the velocity at impact t=2 seconds, you must find the formula for velocity
which is the derivative of the position function s(t )  16t 2  16t  32 .
s(t )  16t 2  16t  32
d
s(t )  32t  16
dt
v(2)  32(2)  16  64  16  48 ft / sec
v(t ) 
Homework Section 2.2
Page 15-117 (10-18, 31-36, 41-46, 93, 94, 95)