Chapter 11
Introduction to
Calculus
11.4 Introduction to
Derivatives
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Objectives:
•
•
•
•
Find slopes and equations of tangent lines.
Find the derivative of a function.
Find average and instantaneous rates of change.
Find instantaneous velocity.
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Slopes and Equations of Tangent Lines
The average rate of change between any two points on the
graph of a function is the slope of the line containing the two
points. We call this line the secant line.
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Slopes and Equations of Tangent Lines (continued)
The figure shows how the secant line between points P and
Q changes as h approaches 0. The limiting position of the
secant line is called the tangent line to the graph of f at the
point P = (a, f(a)).
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Slope of the Tangent Line to the Curve at a Point
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Example: Finding the Slope of a Tangent Line
2
Find the slope of the tangent line to the graph of f ( x)  x  x
at (4, 12). At the point (4, 12), a = 4.
f (a  h)  f (a)
f (4  h)  f (4)
mtan  lim

lim
h 0
h 0
h
h
(4  h) 2  (4  h)    42  4 
 lim
The slope of the tangent
h
h 0
line to the graph of f
2
16  8h  h  4  h  12
 lim
at (4, 12) is 7.
h
h 0
h2  7h
h(h  7)
 lim
 lim(h  7)  0  7  7
 lim
h
h 0
h0
h
h 0
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Example: Finding the Slope-Intercept Equation of a
Tangent Line
Find the slope-intercept equation of the tangent line to the
graph of f ( x)  x at (1, 1).
We begin by finding the slope of the tangent line to the graph at (1, 1).
mtan  lim
h0
f ( a  h)  f ( a )
f (1  h)  f (1)
 lim
h
h
h0
 1  h  1 1  h  1
1 h 1
1 h 1
 lim 
 lim
 lim

h
h 0 
1  h  1 h0 h 1  h  1
h
h 0


1
1
1
 lim


h 0
1 h 1
1 0 1 2
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Example: Finding the Slope-Intercept Equation of a
Tangent Line
Find the slope-intercept equation of the tangent line to the
graph of f ( x)  x at (1, 1).
The tangent line is given to pass through (1, 1). We found the
slope of the tangent line to be 1 .
2
y  y1  m( x  x1 )
The slope-intercept equation of the
1
1
1
tangent line to the graph is y  x  .
y  1  ( x  1)
2
2
2
1
1
y 1  x 
2
2
1
1
y  x
2
2
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The Derivative of a Function
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Example: Finding the Derivative of a Function
2
Find the derivative of f ( x)  x  5 x at x. That is, find f '( x).
f ( x  h)  f ( x )
f '( x)  lim
h
h0
f '( x)  2 x  5
( x  h) 2  5( x  h)    x 2  5 x 
 lim
h
h 0
x 2  2 xh  h 2  5 x  5h  x 2  5 x
2 xh  h 2  5h
 lim
 lim
h
h
h 0
h 0
h(2 x  h  5)
 lim
 lim(2 x  h  5)  2 x  0  5  2 x  5
h
h0
h0
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Example: Finding the Derivative of a Function
(continued)
Find the slope of the tangent line to the graph of
f ( x)  x 2  5 x at x = – 1 and at x = 3.
f '( x)  2 x  5
f '(1)  2(1)  5  2  5  7
The slope of the tangent line at x = –1 is –7.
f '(3)  2(3)  5  6  5  1
The slope of the tangent line at x = 3 is 1.
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11
Average and Instantaneous Rates of Change
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Example: Finding Average and Instantaneous Rates of
Change
The function f ( x)  x3 describes the volume of a cube, f(x),
in cubic inches, whose length, width, and height each
measure x inches. If x is changing, find the average rate of
change of the volume with respect to x as x changes from 4
inches to 4.1 inches.
h  4.1  4  0.1
f (a  h)  f (a) f (4  0.1)  f (4) (4.1)3  43


 49.21
h
0.1
0.1
The average rate of change in the volume is 49.21 cubic
inches per inch as x changes from 4 to 4.1 inches.
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Example: Finding Average and Instantaneous Rates of
Change
(continued)
The function f ( x)  x3 describes the volume of a cube, f(x),
in cubic inches, whose length, width, and height each
measure x inches. If x is changing, find the average rate of
change of the volume with respect to x as x changes from 4
inches to 4.01 inches. h  4.01  4  0.01
f (a  h)  f (a) f (4  0.01)  f (4) (4.01)3  43


 48.1201
h
0.01
0.01
The average rate of change in the volume is 48.1201 cubic
inches per inch as x changes from 4 to 4.01 inches.
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Example: Finding Average and Instantaneous Rates of
Change (continued)
The function f ( x)  x3 describes the volume of a cube, f(x), in cubic
inches, whose length, width, and height each measure x inches. If x is
changing, find the instantaneous rate of change of the volume with
respect to x at the moment when x = 4 inches.
f ( x  h)  f ( x )
( x  h)3  x 3
f '( x)  lim
 lim
h
h
h 0
h 0
3
2
2
3
3
x  3x h  3xh  h   x

3 x 2 h  3 xh  h 2
 lim
 lim
h
h 0
h
h 0
h  3x 2  3xh  h 2 
 lim
 lim  3x 2  3xh  h2 
h
h 0
h0
 3x 2  3x 0  02  3 x 2
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Example: Finding Average and Instantaneous Rates of
Change (continued)
3
The function f ( x)  x describes the volume of a cube, f(x),
in cubic inches, whose length, width, and height each
measure x inches. If x is changing, find the instantaneous
rate of change of the volume with respect to x at the moment
when x = 4 inches.
The instantaneous rate of change of f is the derivative.
To find the instantaneous change of f at 4, find f '(4).
f '( x)  3x
2
f '(4)  3 4  3 16  48
2
The instantaneous rate of change of the volume at the
moment when x = 4 inches is 48 cubic inches per inch.
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Instantaneous Velocity
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Example: Finding Instantaneous Velocity
A ball is thrown straight up from ground level with an initial
velocity of 96 feet per second. The function
s(t )  16t 2  96t
describes the ball’s height above the ground, s(t), in feet, t
seconds after it is thrown. What is the instantaneous velocity
of the ball after 4 seconds?
Instantaneous velocity is given by the derivative of a function
that expresses an object’s position, s(t), in terms of time, t.
The instantaneous velocity of the ball at 4 seconds is s' (4).
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Example: Finding Instantaneous Velocity (continued)
The instantaneous velocity of the ball at 4 seconds is s' (4).
s(t )  16t 2  96t
s ( a  h)  s ( a )
s '(t )  lim
h
h 0
 lim
16(a  h) 2  96(a  h)   16a 2  96a 
h
h 0
h(32a  h  96)
 lim
h
h0
 32a  0  96  32a  96
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Example: Finding Instantaneous Velocity
A ball is thrown straight up from ground level with an initial
velocity of 96 feet per second. The function
s(t )  16t 2  96t
describes the ball’s height above the ground, s(t), in feet, t
seconds after it is thrown. What is the instantaneous velocity
of the ball after 4 seconds?
The instantaneous velocity of the ball at 4 seconds is s' (4).
s '(a)  32a  96
s '(4)  32 4  96  32
The instantaneous velocity
of the ball at 4 seconds
is –32 ft/sec.
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Example: Finding Instantaneous Velocity
A ball is thrown straight up from ground level with an initial
velocity of 96 feet per second. The function s(t )  16t 2  96t
describes the ball’s height above the ground, s(t), in feet, t
seconds after it is thrown. What is the instantaneous velocity
of the ball when it hits the ground?
The ball hits the ground when its s(t), its height above the ground, is 0.
s(t )  16t 2  96t
0  16t 2  96t
0  16t (t  6)
16t  0
t0
t 6  0
t6
The instantaneous velocity of the ball
when it hits the ground is s' (6).
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Example: Finding Instantaneous Velocity
(continued)
A ball is thrown straight up from ground level with an initial
velocity of 96 feet per second. The function s(t )  16t 2  96t
describes the ball’s height above the ground, s(t), in feet, t
seconds after it is thrown. What is the instantaneous velocity
of the ball when it hits the ground?
The instantaneous velocity of the ball
when it hits the ground is s' (6).
s '(a)  32a  96
s '(6)  32 6  96  96
The instantaneous velocity
of the ball when it hits the
ground is –96 ft/sec.
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