1. Slopes and Derivatives

2. (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
= the difference in the
heights of the points.
Δx = x2 – x1
= the difference in the
run of the points.
Slopes and Derivatives

3. (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
= the difference in the
heights of the points.
Δx = x2 – x1
= the difference in the
run of the points.
Δy
Δx=The slope m is the ratio of the “rise” to the “run”.
Slopes and Derivatives

4. (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
= the difference in the
heights of the points.
Δx = x2 – x1
= the difference in the
run of the points.
Δy
Δx=The slope m is the ratio of the “rise” to the “run”.
The slope measures the tilt of a line in relation to the
horizon, that is, the steepness in relation to the x
axis.
Slopes and Derivatives

5. (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
= the difference in the
heights of the points.
Δx = x2 – x1
= the difference in the
run of the points.
Δy
Δx=The slope m is the ratio of the “rise” to the “run”.
The slope measures the tilt of a line in relation to the
horizon, that is, the steepness in relation to the x
axis. Therefore horizontal lines have its steepness or
slope = 0 .
Slopes and Derivatives

6. (x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
= the difference in the
heights of the points.
Δx = x2 – x1
= the difference in the
run of the points.
Δy
Δx=The slope m is the ratio of the “rise” to the “run”.
* http://www.mathwarehouse.com/algebra/linear_equation/interactive-
slope.php
The slope measures the tilt of a line in relation to the
horizon, that is, the steepness in relation to the x
axis. Therefore horizontal lines have its steepness or
slope = 0 . Steeper lines have “larger” slopes*.
Slopes and Derivatives

7. Algebra of Slope
Slopes and Derivatives

8. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Slopes and Derivatives

9. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Slopes and Derivatives

10. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:
Slopes and Derivatives

11. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:
The units of this ratio are (units of y) / (units of x).
Slopes and Derivatives

12. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:
The units of this ratio are (units of y) / (units of x).
This is also the amount of change in y for each unit
change in x.
Slopes and Derivatives

13. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:
The units of this ratio are (units of y) / (units of x).
This is also the amount of change in y for each unit
change in x. The ratio “2 eggs : 3 cakes” is the same
as “2/3 egg per cake”.
Slopes and Derivatives

14. Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y
Δx = x2 – x1
= the difference in the inputs x
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:
The units of this ratio are (units of y) / (units of x).
This is also the amount of change in y for each unit
change in x. The ratio “2 eggs : 3 cakes” is the same
as “2/3 egg per cake”. The slope m is “the amount of
change of y if x changes by one unit.
Slopes and Derivatives

15. Example A. We had 6 gallons of gas at the start of a
trip and the odometer was registered at 75,000 miles.
Two hours later, the gas gauge indicated there were
3 gallons left and the odometer was at 75,300 miles.
Let x = the gas–tank indicator reading
y = the odometer reading
t = the time in hours. Find the following slopes.
Slopes and Derivatives

16. Example A. We had 6 gallons of gas at the start of a
trip and the odometer was registered at 75,000 miles.
Two hours later, the gas gauge indicated there were
3 gallons left and the odometer was at 75,300 miles.
Let x = the gas–tank indicator reading
y = the odometer reading
t = the time in hours. Find the following slopes.
a. Compare the measurements of the fuel amount x
versus the distance y traveled, or (x, y),
Slopes and Derivatives

17. Example A. We had 6 gallons of gas at the start of a
trip and the odometer was registered at 75,000 miles.
Two hours later, the gas gauge indicated there were
3 gallons left and the odometer was at 75,300 miles.
Let x = the gas–tank indicator reading
y = the odometer reading
t = the time in hours. Find the following slopes.
a. Compare the measurements of the fuel amount x
versus the distance y traveled, or (x, y). We have
(6, 75000), (3, 75300).
Slopes and Derivatives

18. The slope m = (75,300 – 75,000) / (3 – 6)
= –100 mpg.
Example A. We had 6 gallons of gas at the start of a
trip and the odometer was registered at 75,000 miles.
Two hours later, the gas gauge indicated there were
3 gallons left and the odometer was at 75,300 miles.
Let x = the gas–tank indicator reading
y = the odometer reading
t = the time in hours. Find the following slopes.
a. Compare the measurements of the fuel amount x
versus the distance y traveled, or (x, y). We have
(6, 75000), (3, 75300).
Slopes and Derivatives

19. The slope m = (75,300 – 75,000) / (3 – 6)
= –100 mpg.
So the distance–to–fuel rate or the fuel efficiency
is 100 miles per gallon.
Example A. We had 6 gallons of gas at the start of a
trip and the odometer was registered at 75,000 miles.
Two hours later, the gas gauge indicated there were
3 gallons left and the odometer was at 75,300 miles.
Let x = the gas–tank indicator reading
y = the odometer reading
t = the time in hours. Find the following slopes.
a. Compare the measurements of the fuel amount x
versus the distance y traveled, or (x, y). We have
(6, 75000), (3, 75300).
Slopes and Derivatives

20. b. Compare the measurements of the time t versus
the distance y traveled, or (t, y).
Slopes and Derivatives

21. b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
Slopes and Derivatives

22. The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
Slopes and Derivatives

23. So the distance–to–time rate (or the velocity) per hour
is 150 miles per hour.
The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
Slopes and Derivatives

24. So the distance–to–time rate (or the velocity) per hour
is 150 miles per hour.
The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
c. Compare the measurements of the time t versus
the amount of fuel x, or (t, x),
Slopes and Derivatives

25. So the distance–to–time rate (or the velocity) per hour
is 150 miles per hour.
The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
the slope n = (6 – 3) / (0 – 2)
= –1.5 gph.
c. Compare the measurements of the time t versus
the amount of fuel x, or (t, x). We have
(0, 6), (2, 3). The rate of change is
Slopes and Derivatives

26. So the distance–to–time rate (or the velocity) per hour
is 150 miles per hour.
The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
b. Compare the measurements of the time t versus
the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).
So the fuel–to–time rate of the fuel consumption per
hour is 1.50 gallon per hour.
c. Compare the measurements of the time t versus
the amount of fuel x, or (t, x). We have
(0, 6), (2, 3). The rate of change is
Slopes and Derivatives
the slope n = (6 – 3) / (0 – 2)
= –1.5 gph.

27. Question: What are the reciprocals of the above
rates and what do they measure?
Slopes and Derivatives

28. Question: What are the reciprocals of the above
rates and what do they measure?
Slopes measure steepness of straight lines.
Slopes and Derivatives

29. Question: What are the reciprocals of the above
rates and what do they measure?
Slopes measure steepness of straight lines.
We want to extend this system of geometric
measurement to measuring “steepness” of curves.
Slopes and Derivatives

30. Question: What are the reciprocals of the above
rates and what do they measure?
Slopes measure steepness of straight lines.
We want to extend this system of geometric
measurement to measuring “steepness” of curves.
A curve has “ups” and “downs”
so a curve has different
“slopes” at different points.
Slopes and Derivatives

31. Question: What are the reciprocals of the above
rates and what do they measure?
x
P
y= f(x)
Slopes measure steepness of straight lines.
We want to extend this system of geometric
measurement to measuring “steepness” of curves.
Q
A curve has “ups” and “downs”
so a curve has different
“slopes” at different points.
This may be seen in the figure
shown at points P and Q.
Slopes and Derivatives

32. Question: What are the reciprocals of the above
rates and what do they measure?
x
P
y= f(x)
Slopes measure steepness of straight lines.
We want to extend this system of geometric
measurement to measuring “steepness” of curves.
Q
A curve has “ups” and “downs”
so a curve has different
“slopes” at different points.
This may be seen in the figure
shown at points P and Q.
Obviously the “slope” at the point P should be positive,
and the “slope” at the point Q should be negative.
Slopes and Derivatives

33. Question: What are the reciprocals of the above
rates and what do they measure?
x
P
y= f(x)
Slopes measure steepness of straight lines.
We want to extend this system of geometric
measurement to measuring “steepness” of curves.
Q
A curve has “ups” and “downs”
so a curve has different
“slopes” at different points.
This may be seen in the figure
shown at points P and Q.
We define the “slope at a point P on the curve y = f(x)”
to be “the slope of the tangent line to y = f(x) at P” .
Obviously the “slope” at the point P should be positive,
and the “slope” at the point Q should be negative.
Slopes and Derivatives

34. Slopes and Derivatives
Derivatives

35. Slopes and Derivatives
Derivatives
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).

36. Slopes and Derivatives
x
P
y= f(x)
Derivatives
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).

37. Slopes and Derivatives
x
P
y= f(x)
Derivatives
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).
The slope at a point P is also
called the derivative at P.

38. Slopes and Derivatives
x
P
y= f(x)
Derivatives
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).
The slope at a point P is also
called the derivative at P.
By “well defined” we mean that the geometric notion of
“the tangent line at P” is intuitive and unambiguous so
its slope is unambiguous.

39. Slopes and Derivatives
x
P
y= f(x)
Derivatives
By “well defined” we mean that the geometric notion of
“the tangent line at P” is intuitive and unambiguous so
its slope is unambiguous.
We will examine the notion of the “tangent
line” in the next section. For now, we
accept the tangent line intuitively as a
straight line that leans against y = f(x) at P
as shown.
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).
The slope at a point P is also
called the derivative at P.

40. Slopes and Derivatives
x
P
y= f(x)
Derivatives
By “well defined” we mean that the geometric notion of
“the tangent line at P” is intuitive and unambiguous so
its slope is unambiguous.
Furthermore, it is well defined because we are able to
compute the slopes of tangents at different locations
algebraically.
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).
The slope at a point P is also
called the derivative at P.

41. Slopes and Derivatives
x
P
y= f(x)
Derivatives
By “well defined” we mean that the geometric notion of
“the tangent line at P” is intuitive and unambiguous so
its slope is unambiguous.
Furthermore, it is well defined because we are able to
compute the slopes of tangents at different locations
algebraically. We will carry out these computations for
the 2nd degree example from the last section.
If f(x) is an elementary function,
then the “slope” of the tangent is
well defined for “most” of the
points on its graph y = f(x).
The slope at a point P is also
called the derivative at P.

42. Example B.
Given f(x) = x2 – 2x + 2
a. Find the slope of the cord connecting the points
(x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2.
Slopes and Derivatives

43. Example B.
Given f(x) = x2 – 2x + 2
a. Find the slope of the cord connecting the points
(x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2.
Substitute the values of x and h,
we are find the slopes of the cord
connecting (2, f(2)=2) and
(2.2, f(2.2)). (2.2, f(2.2))
(2, 2)
2 2.2
0.2
Slopes and Derivatives

44. Example B.
Given f(x) = x2 – 2x + 2
a. Find the slope of the cord connecting the points
(x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2.
f(x+h) – f(x)
h
Substitute the values of x and h,
we are find the slopes of the cord
connecting (2, f(2)=2) and
(2.2, f(2.2)). Its slope is
f(2.2) – f(2)
0.2
= (2, 2)
2 2.2
0.2
(2.2, f(2.2))
Slopes and Derivatives

45. Example B.
Given f(x) = x2 – 2x + 2
a. Find the slope of the cord connecting the points
(x, f(x)) and (x+h,f(x+h)) with x = 2 and h = 0.2.
f(x+h) – f(x)
h
Substitute the values of x and h,
we are find the slopes of the cord
connecting (2, f(2)=2) and
(2.2, f(2.2)). Its slope is
f(2.2) – f(2)
0.2
=
=
2.44 – 2
0.2
= 2.2
(2, 2)
2 2.2=
0.44
0.2
0.44
0.2
slope m = 2.2
(2.2, f(2.2))
Slopes and Derivatives

46. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
Slopes and Derivatives

47. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
Slopes and Derivatives

48. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
We are to simplify the difference quotient formula
with f(x) = x2 – 2x + 2 at x = 2.
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
Slopes and Derivatives

49. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
f(2+h) – f(2)
h
We are to simplify the difference quotient formula
with f(x) = x2 – 2x + 2 at x = 2.
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
Slopes and Derivatives

50. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
f(2+h) – f(2)
h
We are to simplify the difference quotient formula
with f(x) = x2 – 2x + 2 at x = 2.
=
[(2+h)2 – 2(2+h) + 2] – (2)
h
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
Slopes and Derivatives

51. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
f(2+h) – f(2)
h
We are to simplify the difference quotient formula
with f(x) = x2 – 2x + 2 at x = 2.
=
[(2+h)2 – 2(2+h) + 2] – (2)
h
h2 + 2h
h
=
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
Slopes and Derivatives

52. b. Given f(x) = x2 – 2x + 2, simplify the formula for the
slope of the cord connecting the points (2, 2) and
(2+h, f(2+h)).
f(2+h) – f(2)
h
We are to simplify the difference quotient formula
with f(x) = x2 – 2x + 2 at x = 2.
=
[(2+h)2 – 2(2+h) + 2] – (2)
h
h2 + 2h
h
= h + 2
=
(2+h, f(2+h)
(2, 2)
2 2 + h
h
slope = h + 2
f(2+h)–f(2)
Slopes and Derivatives

53. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
Slopes and Derivatives

54. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2, 2)
2
y = x2–2x+2
Slopes and Derivatives

55. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h.
(2, 2)
2
y = x2–2x+2
Slopes and Derivatives

56. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h.
y = x2–2x+2
Slopes and Derivatives

57. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h.
y = x2–2x+2
Slopes and Derivatives

58. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h. As h varies, we obtained
different cords with different slopes.
h
y = x2–2x+2
Slopes and Derivatives

59. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h. As h varies, we obtained
different cords with different slopes.
As h gets larger, the cords deviates
away from the tangent line at (2, 2)
and as h gets smaller the
corresponding cords swing and settle
toward the tangent line. h
y = x2–2x+2
Slopes and Derivatives

60. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h. As h varies, we obtained
different cords with different slopes.
As h gets larger, the cords deviates
away from the tangent line at (2, 2)
and as h gets smaller the
corresponding cords swing and settle
toward the tangent line. These cords
have slopes 2 + h.
slope = 2 + h
h
y = x2–2x+2
Slopes and Derivatives

61. c. Now we deduce the slope at the point (2, 2) in the
following geometric argument.
(2+h, f(2+h)
(2, 2)
2 2 + h
slope = 2 + h
f(2+h)–f(2)
The cord is fixed at the base–point
(2, 2) and the other end depends on
the value h. As h varies, we obtained
different cords with different slopes.
As h gets larger, the cords deviates
away from the tangent line at (2, 2)
and as h gets smaller the
corresponding cords swing and settle
toward the tangent line. These cords
have slopes 2 + h. Hence as
h “shrinks” to 0, the slope or the
derivative at (2, 2) must be 2.
h
y = x2–2x+2
Slopes and Derivatives

62. d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
Slopes and Derivatives

63. Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)).
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
Slopes and Derivatives

64. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)).
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
y = x2–2x+2
Slopes and Derivatives

65. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
y = x2–2x+2
Slopes and Derivatives

66. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
y = x2–2x+2
Slopes and Derivatives

67. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
2xh – 2h + h2
h
=
y = x2–2x+2
Slopes and Derivatives

68. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
2xh – 2h + h2
h
= 2x – 2 + h
=
y = x2–2x+2
slope = 2x–2+h
Slopes and Derivatives

69. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
2xh – 2h + h2
h
= 2x – 2 + h
=
As h shrinks to 0, the slopes of the
cords approach the value 2x – 2
y = x2–2x+2
slope = 2x–2+h
Slopes and Derivatives

70. (x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)). The difference quotient is
h
d. Simplify the difference quotient of f(x) then find the
slope at the point (x, f(x)) – as a formula in x.
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
2xh – 2h + h2
h
= 2x – 2 + h.
=
As h shrinks to 0, the slopes of the
cords approach the value 2x – 2
which must be the slope at (x, f(x)).
y = x2–2x+2
slope = 2x–2+h
Slopes and Derivatives

71. Let’s summarize the result.
Slopes and Derivatives

72. Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
Slopes and Derivatives

73. Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
y = x2–2x+2
Slopes and Derivatives

74. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
y = x2–2x+2
Slopes and Derivatives

75. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
y = x2–2x+2
Slopes and Derivatives

76. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
y = x2–2x+2
slope at x
= 2x – 2
Slopes and Derivatives

77. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x)
y = x2–2x+2
slope at x
= 2x – 2
Slopes and Derivatives

78. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”.
y = x2–2x+2
slope at x
= 2x – 2
Slopes and Derivatives

79. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”. So the derivative of
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
slope at x
= 2x – 2
Slopes and Derivatives

80. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”. So the derivative of
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
The name derivative came from the
fact that f’(x) = 2x – 2 is derived from
f(x) = x2 – 2x + 2.
slope at x
= 2x – 2
Slopes and Derivatives

81. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”. So the derivative of
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
The name derivative came from the
fact that f’(x) = 2x – 2 is derived from
f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells
us the slopes of f(x) = x2 – 2x + 2.
slope at x
= 2x – 2
Slopes and Derivatives

82. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”. So the derivative of
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
The name derivative came from the
fact that f’(x) = 2x – 2 is derived from
f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells
us the slopes of f(x) = x2 – 2x + 2.
For example the slope at x = 2 is f ’(2) = 2,
at x = 3 is f ’(3) = 4, etc…
slope at x
= 2x – 2
Slopes and Derivatives

83. (x, f(x))
x
Let’s summarize the result.
If f(x) = x2 – 2x + 2, then the slope at
the point (x, f(x)) is 2x – 2.
This slope–formula is the derivative of
f(x) and it is written as f ’(x) – it’s read
as “f prime of x”. So the derivative of
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
The name derivative came from the
fact that f’(x) = 2x – 2 is derived from
f(x) = x2 – 2x + 2. The derivative f ’(x) = 2x – 2 tells
us the slopes of f(x) = x2 – 2x + 2.
For example the slope at x = 2 is f ’(2) = 2,
at x = 3 is f ’(3) = 4, etc… The derivative f ’(x) is the
extension of the concept of slopes of lines to curves.
slope at x
= 2x – 2
Slopes and Derivatives

84. The first topic of the calculus course is derivatives.
Slopes and Derivatives

85. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
Slopes and Derivatives

86. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
* developing the computation techniques for finding
derivatives of all elementary functions
Slopes and Derivatives

87. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
* developing the computation techniques for finding
derivatives of all elementary functions
Slopes and Derivatives
We will examine derivative geometrically

88. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
* developing the computation techniques for finding
derivatives of all elementary functions
Slopes and Derivatives
We will examine derivative geometrically by
* developing the relations between derivatives f ’(x)
and the shape of the graph of y = f(x)

89. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
* developing the computation techniques for finding
derivatives of all elementary functions
Slopes and Derivatives
We will examine derivative geometrically by
* developing the relations between derivatives f ’(x)
and the shape of the graph of y = f(x)
* developing the computation procedures using f ’(x)
to locate special positions on y = f(x)

90. The first topic of the calculus course is derivatives.
We will examine derivative algebraically by
* developing the language of derivative which
makes the concept more rigorous
* developing the computation techniques for finding
derivatives of all elementary functions
Slopes and Derivatives
We will examine derivative geometrically by
* developing the relations between derivatives f ’(x)
and the shape of the graph of y = f(x)
* developing the computation procedures using f ’(x)
to locate special positions on y = f(x)
We will also investigate the applications of the
derivatives which include problems of optimization,
rates of change and numerical methods.