The Derivative Function
Objective: To define and use the
derivative function
Definition 2.2.1
/
f
• The function
defined by the formula
f ( x  h)  f ( x )
f ( x)  lim
h 0
h
/
• is called the derivative of f with respect to x. The
domain of f / consists of all x in the domain of f for
which the limit exists.
• Remember, this is called the difference quotient.
Example 1
• Find the derivative with respect to x of f ( x)  x  1
and use it to find the equation of the tangent line to
f ( x)  x 2  1 at x  2.
2
• Note: The independent variable is x. This is very
important to state. Later, we will be taking derivatives
with respect to other independent variables.
Example 1
f ( x)  x 2  1
• Find the derivative with respect to x of
and use it to find the equation of the tangent line to
f ( x)  x 2  1 at x  2.
2
2
f
(
x

h
)

f
(
x
)
[(
x

h
)

1
]

[
x
 1]
/
f ( x)  lim

h 0
h
h
Example 1
• Find the derivative with respect to x of f ( x)  x  1
and use it to find the equation of the tangent line to
f ( x)  x 2  1 at x  2.
2
2
2
f
(
x

h
)

f
(
x
)
[(
x

h
)

1
]

[
x
 1]
/
f ( x)  lim

h 0
h
h
2
2
2
x

2
xh

h

1

x
 1 h( 2 x  h )
/
f ( x)  lim

 2x  h  2x
h 0
h
h
Example 1
• The slope of the tangent line to y  x  1 at x  2
/
f
(2)  4. When x  2, y  5 , so the equation of
is
the tangent line at x  2 is
2
y  5  4( x  2)
or
y  4x  3
Example 1
• We can also use the other formula to find the
derivative of f ( x)  x 2  1 .
f ( x1 )  f ( x0 )
lim
x1  x0
x1  x0
[ x1  1]  [ x0  1] x1  x0
lim

x1  x0
x1  x0
x1  x0
2
2
2
2
( x1  x0 )( x1  x0 )
 x1  x0  2 x0
x1  x0
Example 2
a) Find the derivative with respect to x of
f ( x)  x 3  x
3
3
f
(
x

h
)

f
(
x
)
[(
x

h
)

(
x

h
)]

[
x
 x]
/
f ( x)  lim

h 0
h
h
Example 2
a) Find the derivative with respect to x of
f ( x)  x 3  x
3
3
f
(
x

h
)

f
(
x
)
[(
x

h
)

(
x

h
)]

[
x
 x]
/
f ( x)  lim

h 0
h
h
[ x 3  3x 2 h  3xh2  h3  x  h]  [ x 3  x]
lim
h 0
h
Example 2
a) Find the derivative with respect to x of
f ( x)  x 3  x
3
3
f
(
x

h
)

f
(
x
)
[(
x

h
)

(
x

h
)]

[
x
 x]
/
f ( x)  lim

h 0
h
h
[ x 3  3x 2 h  3xh2  h3  x  h]  [ x 3  x]
lim
h 0
h
3x 2 h  3xh2  h3  h h(3x 2  3xh  h 2  1)
lim

 3x 2  1
h 0
h
h
Example 2
• We can use the other formula to find the derivative
of f ( x)  x3  x .
f ( x1 )  f ( x0 )
lim
x1  x0
x1  x0
[ x  x1 ]  [ x0  x0 ] [ x  x0 ]  [ x1  x0 ]
lim

x1  x0
x1  x0
x1  x0
3
1
3
3
1
3
( x  x )( x  x1 x0  x0 )  ( x1  x0 ) ( x1  x0 )[( x1  x1 x0  x0 )  1]
lim 1 0 1

x1  x0
x1  x0
x1  x0
2
2
2
lim ( x1  x0 x1  x0  1)  3x0  1
2
x1 x0
2
2
2
Example 2
• Lets look at the two graphs together and discuss the
relationship between them.
Example 2
/
• Since f ( x) can be interpreted as the slope of the
tangent line to the graph y  f (x) at x it follows that
f / ( x) is positive where the tangent line has
positive slope, is negative where the tangent line has
negative slope, and zero where the tangent line is
horizontal.
Example 3
• At each value of x, the tangent line to a line is the line
itself, and hence all tangent lines have slope m. This
is confirmed by:
f ( x  h)  f ( x) m( x  h)  b  [mx  b]
lim

h 0
h
h
mx  mh  b  mx  b mh
lim

m
h 0
h
h
Example 4
• Find the derivative with respect to x of f ( x)  x
• Recall from example 4, section 2.1 we found the slope
of the tangent line of y  x was 1 , thus,
2 x
lim
x  x0
x  x0
x  x0

x  x0
( x  x0 )( x  x0 )
• Memorize this!!!!
f ( x) 
/
1
2 x

1
x0  x0
Example 4
• Find the derivative with respect to x of f ( x)  x
• Find the slope of the tangent line to f ( x)  x
at x = 9.
• The slope of the tangent line at x = 9 is
1
f (9) 

2 9 6
/
1
Example 4
• Find the derivative with respect to x of f ( x)  x
• Find the slope of the tangent line to f ( x)  x
at x = 9.
• Find the limits of f / ( x) as x  0 and as x  
and explain what those limits say about the graph of
f.
Example 4
• Find the limits of f / ( x) as x  0 and as f ( x)  x
and explain what the limits say about the graph of f .
• The graphs of f(x) and f /(x) are shown. Observe that
f / ( x)  0 if x  0 , which means that all tangent lines
to the graph of y  x have positive slopes, meaning
that the graph becomes more and more vertical as
x  0 and more and more horizontal as x  .
Instantaneous Velocity
• We saw in section 2.1 that instantaneous velocity
was defined as
f (t  h)  f (t )
vinst  lim
h 0
h
• Since the right side of this equation is also the
definition of the derivative, we can say
f (t  h)  f (t )
v(t )  f (t )  lim 
h 0
h
/
• This is called the instantaneous velocity function, or
just the velocity function of the particle.
Example 5
• Recall the particle from Ex 5 of section 2.1 with
position function s  f (t )  1  5t  2t 2 . Here f(t) is
measured in meters and t is measured in seconds.
Find the velocity function of the particle.
Example 5
• Recall the particle from Ex 5 of section 2.1 with
position function s  f (t )  1  5t  2t 2 . Here f(t) is
measured in meters and t is measured in seconds.
Find the velocity function of the particle.
f (t  h)  f (t )
[1  5(t  h)  2(t  h) 2 ]  [1  5t  2t 2 ]
v(t )  lim 
 lim
h 0
h 0
h
h
 2[t 2  2th  h 2  t 2 ]  5h  4th  2h 2  5h
 lim 

h 0
h
h
 lim (4t  2h  5)  5  4t
h 0
Differentiability
• Definition 2.2.2 A function is said to be differentiable
at x0 if the limit
f ( x0  h)  f ( x0 )
f ( x0 )  lim 
h 0
h
/
exists. If f is differentiable at each point in the open
interval (a, b) , then we say that is differentiable on
(a, b), and similarly for open intervals of the form
. (a,), (, b)and (,) .In the last case, we say that
it is differentiable everywhere.
Differentiability
• Definition 2.2.2 A function is said to be differentiable
at x0 if the limit
f ( x0  h)  f ( x0 )
f ( x0 )  lim 
h 0
h
/
exists. When they ask you if a function is
differentiable on the AP Exam, this is what they want
you to reference.
Differentiability
• Geometrically, a function f is differentiable at x if the
graph of f has a tangent line at x. There are two
cases we will look at where a function is nondifferentiable.
1. Corner points
2. Points of vertical tangency
Corner points
• At a corner point, the slopes of the secant lines have
different limits from the left and from the right, and
hence the two-sided limit that defines the derivative
does not exist.
Vertical tangents
• We know that the slope of a vertical line is
undefined, so the derivative makes no sense at a
place with a vertical tangent, since it is defined as the
slope of the line.
Differentiability and Continuity
• Theorem 2.2.3 If a function f is differentiable at x,
then f is continuous at x.
• The inverse of this is not true. If it is continuous, that
does not mean it is differentiable (corner points,
vertical tangents).
Differentiability and Continuity
• Theorem 2.2.3 If a function f is differentiable at x,
then f is continuous at x.
• Since the conditional statement is true, so is the
contrapositive:
• If a function is not continuous at x, then it is not
differentiable at x.
Other Derivative notations
• We can express the derivative in many different
ways.
d
dy
/
f ( x)  [ f ( x)] 
y
dx
dx
/
• Please note that these expressions all mean the
derivative of y with respect to x.
Other formulas to use
• There are several different formulas you can use to
find the derivative of a function. The only ones we
will use are:
f ( x  h)  f ( x )
f ( x)  lim 
h 0
h
/
f ( x)  f ( x0 )
f ( x)  lim 
x  x0
x  x0
/
f ( w)  f ( x)
f ( x)  lim 
w x
w x
/
Homework
•
•
•
•
Section 2.2
Page 152-153
1-25 odd, 31
For numbers 15,17,19, use formula 13, not formula
12.
Example
• Evaluate:
4( x  h)3  3( x  h) 2  6  4 x 3  3x 2  6
lim
h 0
h
Example
• Evaluate:
4( x  h)3  3( x  h) 2  6  4 x 3  3x 2  6
lim
h 0
h
• This is the definition of the derivative of 4 x 3  3x 2  6 .
• The answer is 12 x 2  6 x . You are not supposed to do
any work, just recognize this!