Do I understand correctly that the growth "h" is at the end of reduction of limit omited because its value is so small it has practically no effect on value of slope?

Yes, because technically speaking we write "as the limit of h approaches zero" which means h keeps getting smaller and smaller until it is negligible.

How did we find the function at the end of example 2? f(x)=7 and f(x)=f(x-1)+5?

It seems to be an orphan that was not removed before the lesson was posted. It doesn't relate to the previous example and is not related to the next example.

In the example 1, I have not idea how do they get h^3+6h^2+12h+8 from f(2+h)−f(2), it is really confuse me. Can anyone explain?

They're simply evaluating f(x) for `x = 2 + h`. `(2 + h)^3` is equal to `h^3 + 6*h^2 + 12*h + 8`.

On example 3, when taking the derivative of 1/x, at the third step while using the formula - "h" is somehow transferred from the lower denominator to the denominator above it: x-(x+h)/x(x+h)/h --> x-(x+h)/xh(x+h) Which algebric rule relates to that? I couldn't figure this out.

This is a complex fraction simplified. ((x - (x-h)) / x(x+h)) / h is the same thing as (x - (x-h)) / x(x+h)) * 1/h, in other words a/b/c = a/b * 1/c = a/bc

In example 3, what if x is given but we'll have to find 'm' and 'c' with a given function? I'm confused

the slope can be found out between any 2 points on a straight line. now substitute a variable point on the line in slopes equation . you get your whole equation of line

In Example 4, after the first part is over, the answer found is 2. How is it equated to "2x" at the beginning of the solution of the second part?

The average rate of change is the slope of the secant line between two points on a graph. The secant line in the example is written in the form where the change in y is the numerator and the change in x is the denominator. We find that between the points (0, f(0)) and (2, f(2), we "rise" 2 units in the y direction for every unit we "run" in the x direction. 2 is simply the scalar multiple of x that the line must take to pass through both points on the graph of f(x).

What are the advantages and disadvantages of using the limit as *h* approaches *0* form over the limit as *x* approaches *a* form.

The limit as *h* approaches *0* form is known as the _formal_ definition of the derivative, and using it results in finding the derivative _function_, f'(x). The derivative function allows you to find the slope of the tangent line at any point of f(x). The limit as *x* approaches *a* form, or _alternate_ definition of the derivative, is used to find the derivative at a specific point *a*, or f'(a). This form is more useful when you only need to the derivative at one specific point because it is usually less complex to evaluate than formal definition.

Main content

Course: Calculus, all content (2017 edition) > Unit 2

Lesson 7: Using the formal definition of derivative

Tangent lines and rates of change

Google Classroom

How tangent lines are a limit of secant lines, and where the derivative and rate of change fit into all this.

Introduction

The position of a car driving down the street, the value of currency adjusted for inflation, the number of bacteria in a culture, and the AC voltage of an electric signal are all examples of quantities that change with time. In this section, we will study the rate of change of a quantity and how is it related geometrically to secant and tangent lines.

Secant and tangent lines

If two distinct points

P (x_{0}, y_{0})

and

Q (x_{1}, y_{1})

lie on the curve

y = f (x)

, the slope of the secant line connecting the two points is

m_{\sec} = \frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}}

If we let the point

x_{1}

approach

x_{0}

, then

Q

will approach

P

along the graph

f

. The slope of the secant line through points

P

and

Q

will gradually approach the slope of the tangent line through

P

x_{1}

approaches

x_{0}

. In the limit, the previous equation becomes

m_{\tan} = lim_{x_{1} \to x_{0}} \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}}

If we let

h = x_{1} - x_{0}

, then

x_{1} = x_{0} + h

and

h \to 0

x_{1} \to x_{0}

. We can rewrite the limit as

m_{\tan} = lim_{h \to 0} \frac{f (x_{0} + h) - f (x_{0})}{h}

When the limit exists, its value

m_{\tan}

is the slope of the tangent line to the graph of

f

at the point

P (x_{0}, y_{0})

Example 1

Find the slope of the tangent line to the graph of the function

f (x) = x^{3}

at the point

(2, 8)

Solution

Since

(x_{0}, y_{0}) = (2, 8)

, using the slope of the tangent line formula

m_{\tan} = lim_{h \to 0} \frac{f (x_{0} + h) - f (x_{0})}{h}

we get

\begin{aligned} m_{\tan} & = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} \\ = lim_{h \to 0} \frac{(h^{3} + 6 h^{2} + 12 h + 8) - 8}{h} \\ = lim_{h \to 0} \frac{h^{3} + 6 h^{2} + 12 h}{h} \\ = lim_{h \to 0} (h^{2} + 6 h + 12) \\ = 12. \end{aligned}

Thus, the slope of the tangent line is

12

. Recall from algebra that the point-slope form of the equation of the tangent line is

y - y_{0} = m_{\tan} \cdot (x - x_{0})

The point-slope formula gives us the equation

y - 8 = 12 \cdot (x - 2)

which we can rewrite as

y = 12 x - 16

Finding the slope at any point

Next we are interested in finding a formula for the slope of the tangent line at any point on the graph of

f

. Such a formula would be the same formula that we are using except we replace the constant

x_{0}

by the variable

x

. This yields

m_{\tan} = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

We denote this formula by

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

where

f^{'} (x)

is read "

f

prime of

x

." The next example illustrates its usefulness.

Example 2

f (x) = x^{2} - 3

, find

f^{'} (x)

and use the result to find the slopes of the tangent lines at

x = 2

and

x = - 1

Solution

Since

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

then

\begin{aligned} f^{'} (x) & = lim_{h \to 0} \frac{[(x + h)^{2} - 3] - [x^{2} - 3]}{h} \\ = lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} - 3 - x^{2} + 3}{h} \\ = lim_{h \to 0} \frac{2 x h + h^{2}}{h} \\ = lim_{h \to 0} (2 x + h) \\ = 2 x . \end{aligned}

To find the slope, we substitute

x = 2

and

x = - 1

into the result

f^{'} (x)

. We get

f^{'} (2) = 2 (2) = 4

and

f^{'} (- 1) = 2 (- 1) = - 2

Thus, slopes of the tangent lines at

x = 2

and

x = - 1

are

4

and

- 2

, respectively.

Example 3

Find the slope of the tangent line to the graph of

f (x) = 1 / x

at the point

(1, 1)

Solution

Using the slope of the tangent line formula

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

and substituting

f (x) = 1 / x

gives us

\begin{aligned} f^{'} (x) & = lim_{h \to 0} \frac{(\frac{1}{x + h}) - \frac{1}{x}}{h} \\ = lim_{h \to 0} \frac{\frac{x - (x + h)}{x (x + h)}}{h} \\ = lim_{h \to 0} \frac{x - x - h}{h x (x + h)} \\ = lim_{h \to 0} \frac{- h}{h x (x + h)} \\ = lim_{h \to 0} \frac{- 1}{x (x + h)} \\ = \frac{- 1}{x^{2}} . \end{aligned}

Substituting

x = 1

yields

f^{'} (1) = \frac{- 1}{(1)^{2}} = - 1

Thus, the slope of the tangent line at

x = 1

for the graph of

f (x) = 1 / x

m = - 1

. To find the equation of the tangent line, we use the point-slope formula,

y - y_{0} = m \cdot (x - x_{0})

where

(x_{0}, y_{0}) = (1, 1)

. The equation of the tangent line is

\begin{aligned} y - 1 & = - 1 \cdot (x - 1) \\ y & = - x + 1 + 1 \\ y & = - x + 2. \end{aligned}

Average speed

The primary concept of differential is calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of change of the distance traveled with respect to time. If a car travels

120

miles in

4

hours, his speed is

\frac{120 miles}{4 hours} = 30 mi/hr

This speed is called the average speed or the average rate of change of distance with respect to time. Of course, a car that travels

120

miles at an average rate of

30

miles per hour for

4

hours does not necessarily do so at constant speed. It may have slowed down or sped up during the

4

hour period.

However, if the car hits a tree, it would not be its average speed that determines the resulting damage but its speed at the instant of the collision. So here we have two distinct kinds of speeds, average speed and instantaneous speed.

The average speed of an object is defined as the object’s displacement

△ x

divided by the time interval

△ t

during which the displacement occurs:

v = \frac{△ x}{△ t} = \frac{x_{1} - x_{0}}{t_{1} - t_{0}}

The average speed is also the expression for the slope of a secant line connecting the two points. Figure 1 shows the

secant line

through the points

(t_{0}, x_{0})

and

(t_{1}, x_{1})

on the

position-versus-time curve

Thus we conclude that the average speed of an object between time

t_{0}

and

t_{1}

is represented geometrically by the slope of the secant line connecting the two points

(t_{0}, x_{0})

and

(t_{1}, x_{1})

. If we choose

t_{1}

close to

t_{0}

, then the average speed will closely approximate the instantaneous speed at time

t_{0}

Rates of change

The average rate of change of an arbitrary function

f

on an interval is represented geometrically by the slope of the secant line to the graph of

f

. The instantaneous rate of change of

f

at a particular point is represented by the slope of the tangent line to the graph of

f

at that point. Let's consider each case in more detail.

Average rate of change

The average rate of change of the function

f

over the interval

[x_{0}, x_{1}]

m_{\sec} = \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}}

Figure 2 shows the

secant line

through the points

(x_{0}, f (x_{0}))

and

(x_{1}, f (x_{1}))

on the

graph of f

. The slope of the secant line is the average rate of change

m_{\sec}

Instantaneous rate of change

The instantaneous rate of change of the function

f

at the point

x_{0}

m_{\tan} = f^{'} (x_{0}) = lim_{x_{1} \to x_{0}} \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}}

Figure 3 shows

tangent line

through the point

(x_{0}, f (x_{0}))

on the

graph of f

. The slope of the tangent line is the instantaneous rate of change

m_{\tan}

Example 4

Suppose that

y = x^{2} - 3

(a) Find the average rate of change of

y

with respect to

x

over the interval

[0, 2]

(b) Find the instantaneous rate of change of

y

with respect to

x

at the point

x = - 1

Solution

(a) Applying the formula for average rate of change with

f (x) = x^{2} - 3

and

x_{0} = 0

and

x_{1} = 2

yields

\begin{aligned} m_{\sec} & = \frac{f (x_{1}) - f (x_{0})}{x_{1} - x_{0}} \\ = \frac{f (2) - f (0)}{2 - 0} \\ = \frac{1 - (- 3)}{2} \\ = 2 \end{aligned}

This means the average rate of change over the interval

[0, 2]

is 2 units of increase in

y

for each unit of increase in

x

(b) From Example 2 above, we found that

f^{'} (x) = 2 x

, so

\begin{aligned} m_{\tan} & = f^{'} (x_{0}) \\ = f^{'} (- 1) \\ = 2 (- 1) \\ = - 2. \end{aligned}

This means that the instantaneous rate of change is negative. That is,

y

is decreasing at

x = - 1

. It is decreasing at a rate of

2

units in

y

for each unit of increase in

x

Source

CK-12: Tangent Lines and Rates of Change

License

Creative Commons Attribution-NonCommercial

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Miss H
Posted 8 years ago. Direct link to Miss H's post “What is the purpose of th...”
What is the purpose of the function that appears at the very end of example 2? f(x)= 7 | x=1, f(x-1)+5 | x>1
What is this related to?
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Not.Really.My.Mailbox
Posted 8 years ago. Direct link to Not.Really.My.Mailbox's post “Do I understand correctly...”
Do I understand correctly that the growth "h" is at the end of reduction of limit omited because its value is so small it has practically no effect on value of slope?
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(6 votes)
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- Abdelrahman Tarek
  Posted 8 years ago. Direct link to Abdelrahman Tarek's post “Yes, because technically ...”
  Yes, because technically speaking we write "as the limit of h approaches zero" which means h keeps getting smaller and smaller until it is negligible.
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anonymousg64
Posted 8 years ago. Direct link to anonymousg64's post “can someone explain to th...”
can someone explain to the math for substituting y=1/x in Example 3?
for some reason i cant see what happens in step one and two
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turdubaev.arstan
Posted 8 years ago. Direct link to turdubaev.arstan's post “How did we find the funct...”
How did we find the function at the end of example 2? f(x)=7 and f(x)=f(x-1)+5?
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(5 votes)
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- doctorfoxphd
  Posted 8 years ago. Direct link to doctorfoxphd's post “It seems to be an orphan ...”
  It seems to be an orphan that was not removed before the lesson was posted. It doesn't relate to the previous example and is not related to the next example.
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wwen
Posted 8 years ago. Direct link to wwen's post “In the example 1, I have ...”
In the example 1, I have not idea how do they get h^3+6h^2+12h+8 from f(2+h)−f(2), it is really confuse me. Can anyone explain?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Travis Bartholome
  Posted 8 years ago. Direct link to Travis Bartholome's post “They're simply evaluating...”
  They're simply evaluating f(x) for x = 2 + h. (2 + h)^3 is equal to h^3 + 6*h^2 + 12*h + 8.
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  (4 votes)
Juan Domene
Posted 8 years ago. Direct link to Juan Domene's post “In figure 1, it doesn't m...”
In figure 1, it doesn't make a lot of sense using a position vs time graph if you're talking about speed and that specific graph wasn't really a good choice since its slope can be negative and speed cannot be negative. You should use a displacement vs time graph instead or it's a bit confusing.
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- Jim Zack
  Posted 6 years ago. Direct link to Jim Zack's post “I agree with Juan. I thin...”
  I agree with Juan. I think that rather than using y to represent displacement (as it does in figure 1), a monotonically increasing function should be used to represent distance, since after all, we are looking at the rate of change in distance with respect to time.
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  (1 vote)
Evyatar Shamir
Posted 8 years ago. Direct link to Evyatar Shamir's post “On example 3, when taking...”
On example 3, when taking the derivative of 1/x,
at the third step while using the formula - "h" is somehow transferred from the lower denominator to the denominator above it:
x-(x+h)/x(x+h)/h --> x-(x+h)/xh(x+h)
Which algebric rule relates to that? I couldn't figure this out.
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(3 votes)
Answer
- Mateusz Jastrzębski
  Posted 8 years ago. Direct link to Mateusz Jastrzębski's post “This is a complex fractio...”
  This is a complex fraction simplified. ((x - (x-h)) / x(x+h)) / h is the same thing as (x - (x-h)) / x(x+h)) * 1/h, in other words a/b/c = a/b * 1/c = a/bc
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Ravai Tehillah
Posted 8 years ago. Direct link to Ravai Tehillah's post “In example 3, what if x i...”
In example 3, what if x is given but we'll have to find 'm' and 'c' with a given function? I'm confused
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(3 votes)
Answer
- Aditya
  Posted 8 years ago. Direct link to Aditya's post “the slope can be found ou...”
  the slope can be found out between any 2 points on a straight line. now substitute a variable point on the line in slopes equation . you get your whole equation of line
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  (3 votes)
visualiser
Posted 8 years ago. Direct link to visualiser's post “In Example 4, after the f...”
In Example 4, after the first part is over, the answer found is 2. How is it equated to "2x" at the beginning of the solution of the second part?
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- Mister Owl
  Posted 8 years ago. Direct link to Mister Owl's post “The average rate of chang...”
  The average rate of change is the slope of the secant line between two points on a graph. The secant line in the example is written in the form where the change in y is the numerator and the change in x is the denominator. We find that between the points (0, f(0)) and (2, f(2), we "rise" 2 units in the y direction for every unit we "run" in the x direction. 2 is simply the scalar multiple of x that the line must take to pass through both points on the graph of f(x).
  Comment on Mister Owl's post “The average rate of chang...”
  (3 votes)
Tanner Thygerson
Posted 8 years ago. Direct link to Tanner Thygerson's post “What are the advantages a...”
What are the advantages and disadvantages of using the limit as h approaches 0 form over the limit as x approaches a form.

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- Matthew Ponce de Leon
  Posted 8 years ago. Direct link to Matthew Ponce de Leon's post “The limit as *h* approach...”
  The limit as h approaches 0 form is known as the formal definition of the derivative, and using it results in finding the derivative function, f'(x). The derivative function allows you to find the slope of the tangent line at any point of f(x). The limit as x approaches a form, or alternate definition of the derivative, is used to find the derivative at a specific point a, or f'(a). This form is more useful when you only need to the derivative at one specific point because it is usually less complex to evaluate than formal definition.
  Button navigates to signup page
  (2 votes)