1.7 derivative

(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
run of the points.
Slopes and Derivatives

(x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
Δx = x2 – x1
run of the points.
Δy
Δx=The slope m is the ratio of the “rise” to the “run”.

(x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
Δx = x2 – x1
run of the points.
Δy
The slope measures the tilt of a line in relation to the
horizon, that is, the steepness in relation to the x
axis.

(x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
Δx = x2 – x1
run of the points.
Δy
axis. Therefore horizontal lines have its steepness or
slope = 0 .

(x1, y1)
(x2, y2)
Δy=y2–y1=rise
Δx=x2–x1=run
Geometry of Slope
Δy = y2 – y1
Δx = x2 – x1
run of the points.
Δy
* http://www.mathwarehouse.com/algebra/linear_equation/interactive-
slope.php
axis. Therefore horizontal lines have its steepness or
slope = 0 . Steeper lines have “larger” slopes*.

Algebra of Slope

Algebra of Slope
Δy = y2 – y1
= the difference in the outputs y

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
= the difference in the inputs x

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
Algebraically the slope
m = Δy/Δx = Δy : Δx is the ratio
the difference in the outputs the difference in the inputs:

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
The units of this ratio are (units of y) / (units of x).

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
This is also the amount of change in y for each unit
change in x.

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
change in x. The ratio “2 eggs : 3 cakes” is the same
as “2/3 egg per cake”.

Algebra of Slope
Δy = y2 – y1
Δx = x2 – x1
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.

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),

versus the distance y traveled, or (x, y). We have
(6, 75000), (3, 75300).

The slope m = (75,300 – 75,000) / (3 – 6)
= –100 mpg.
(6, 75000), (3, 75300).

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.
(6, 75000), (3, 75300).

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

the distance y traveled, or (t, y). we have
(0, 75000), (2, 75300).

The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
(0, 75000), (2, 75300).

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.
(0, 75000), (2, 75300).

The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
(0, 75000), (2, 75300).
c. Compare the measurements of the time t versus
the amount of fuel x, or (t, x),

The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
(0, 75000), (2, 75300).
the slope n = (6 – 3) / (0 – 2)
= –1.5 gph.
the amount of fuel x, or (t, x). We have
(0, 6), (2, 3). The rate of change is

The slope n = (75,300 – 70,000) / (2 – 0)
= 150 mph.
(0, 75000), (2, 75300).
So the fuel–to–time rate of the fuel consumption per
hour is 1.50 gallon per hour.
the amount of fuel x, or (t, x). We have
(0, 6), (2, 3). The rate of change is
the slope n = (6 – 3) / (0 – 2)
= –1.5 gph.

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.

x
P
y= f(x)
Q
This may be seen in the figure
shown at points P and Q.

x
P
y= f(x)
Q
Obviously the “slope” at the point P should be positive,
and the “slope” at the point Q should be negative.

x
P
y= f(x)
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.

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).

x
P
y= f(x)
Derivatives

x
P
y= f(x)
Derivatives
The slope at a point P is also
called the derivative at P.

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.

x
P
y= f(x)
Derivatives
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.

x
P
y= f(x)
Derivatives
Furthermore, it is well defined because we are able to
compute the slopes of tangents at different locations
algebraically.

x
P
y= f(x)
Derivatives
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.

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.

Example B.
Given f(x) = x2 – 2x + 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

Example B.
Given f(x) = x2 – 2x + 2
f(x+h) – f(x)
h
(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))

Example B.
Given f(x) = x2 – 2x + 2
f(x+h) – f(x)
h
(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))

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+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)

(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)

(2+h, f(2+h)).
f(2+h) – f(2)
h
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)

(2+h, f(2+h)).
f(2+h) – f(2)
h
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)

(2+h, f(2+h)).
f(2+h) – f(2)
h
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)

(2+h, f(2+h)).
f(2+h) – f(2)
h
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)

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

(2, 2)
2
y = x2–2x+2

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

(2+h, f(2+h)
(2, 2)
2 2 + h
h
f(2+h)–f(2)
the value h.
y = x2–2x+2

(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
the value h. As h varies, we obtained
different cords with different slopes.
h
y = x2–2x+2

(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
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

(2+h, f(2+h)
(2, 2)
2 2 + h
f(2+h)–f(2)
toward the tangent line. These cords
have slopes 2 + h.
slope = 2 + h
h
y = x2–2x+2

(2+h, f(2+h)
(2, 2)
2 2 + h
slope = 2 + h
f(2+h)–f(2)
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

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

Connect the cord at the base–point
(x, f(x)) to the other end point at
(x+h, f(x+h)).

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
(x+h, f(x+h)).
h
y = x2–2x+2

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
(x+h, f(x+h)). The difference quotient is
h
f(x+h) – f(x)
h
y = x2–2x+2

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
h
f(x+h) – f(x)
h
=
(x+h)2 – 2(x+h) + 2 – [x2 – 2x + 2]
h
y = x2–2x+2

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
h
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

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
h
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

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
h
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

(x+h, f(x+h)
(x, f(x))
x x + h
f(x+h)–f(x)
h
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

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

(x, f(x))
x
y = x2–2x+2

(x, f(x))
x
y = x2–2x+2
slope at x
= 2x – 2

(x, f(x))
x
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

(x, f(x))
x
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

(x, f(x))
x
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

(x, f(x))
x
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

(x, f(x))
x
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = x2–2x+2
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

(x, f(x))
x
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = 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

(x, f(x))
x
f(x) = x2 – 2x + 2 is f ’(x) = 2x – 2.
y = 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

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

We will examine derivative geometrically

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)

* 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.

1.7 derivative

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