Section 2.1The Derivative & the Tangent Line Problem

Unit 2 – Differentiation

Objectives:

1. Find the slope of the tangent line to a curve at a point.2. Use the limit definition to find the derivative of a function.3. Understand the relationship between differentiability and continuity.

Roots of Calculus

» Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century.

1. The tangent line problem (Section 1.1 & 2.1)2. The velocity and acceleration problem

(Section 2.2 & 2.3)3. The minimum and maximum problem

(Section 3.1)4. The area problem (Section 1.1 and 4.2)

The Tangent Line

» Recall the tangent line problem from 1.1:

Given a function f and a point P on its graph, find the equation of the tangent line at point P.

In order to solve this problem, it is necessary to find the slope of the tangent line at P.

𝑃𝑥

𝑦𝑓 (𝑥)

point of tangencytangent line

The Tangent Line

» We discussed using the slope formula and limits, but we didn’t go into detail on how to use the slope formula.

The slope can be approximated by finding the slope of a secant line that passes through the point of tangency and the slope formula:

As Q moves closer to P, the slope of the secant line better approximates that of the tangent line. When Q and P coincide, we have the exact slope of the tangent line.

𝑃𝑥

𝑓 (𝑥)

𝑄

The Tangent Line

» To use the slope formula, we must modify it slightly.

x

y{ ¿∆ 𝑥

(𝑥1, 𝑦 1 )(𝑥2 , 𝑦2 )

f

We will start by letting .

Here the Greek letter is used to represent change. So means the change in .

On a graph, this amounts to the change between the x-values of the two points.

𝒎=𝒚𝟐−𝒚𝟏

∆ 𝒙

The Tangent Line

» Now we will put the points into function notation.

x

y{ ¿∆ 𝑥

(𝑐 , 𝑓 (𝑐))(𝑐+∆𝑥 , 𝑓 (𝑐+∆ 𝑥))

f

Since & are points on the function , we can put them into function notation. Let’s say . The y-value will then be .

So, becomes .

The second point is a little harder. Recall we defined . Now that equation reads So, . That means

So, becomes .

The Tangent Line

» Now we will redefine the change in y.

x

y{ ¿∆ 𝑥

(𝑐 , 𝑓 (𝑐))(𝑐+∆𝑥 , 𝑓 (𝑐+∆ 𝑥))

f

The change in y, or would originally be defined as

Now that we are in function notation though, this becomes:

So, the entire slope formula is now:

This formula is called the difference quotient.

¿ }∆ 𝑦

The Tangent Line

» Now that we have an equation for the slope of a secant line, how do we apply it to finding the slope of a tangent?

𝑃𝑥

𝑓 (𝑥)

𝑄¿∆ 𝑥

¿∆ 𝑥

{ ¿∆ 𝑥

We can see that as the points get closer together, shrinks in distance.

What we can do then, is find the limit as .

Definition of Tangent Line with Slope m

If is defined on an open interval containing , and if the limit

exists, then the line passing through with slope is the tangent line to the graph of at the point

This is also called the slope of the graph of at .

Vertical Tangent Lines

The definition of a tangent line to a curve does not cover the possibility of vertical tangent lines. Instead we can use this definition:

If is continuous at and

the vertical line passing through is a vertical tangent line to the graph of .

Example 1

» Find the slope of the graph of at the point

Now obviously since this is a linear equation, we know the answer is 1/3, but that isn’t the point. What we would like to do is use the limit that we just found to find the slope. So, we let and apply the limit.

Example 2

» Find the slopes of the tangent lines to the graph of at the points & You may be thinking this will require doing the same process twice in order to find the slope at two different points, and that would work, but if we do the process arbitrarily, then it will save us time. So, we will instead use the point

Example 2 continued…

» Find the slopes of the tangent lines to the graph of at the points &

So, for any point on the function , the slope is given by .

Example 2 continued…

» Find the slopes of the tangent lines to the graph of at the points &

For any point on the function , the slope is given by .

Now, to find the slope at we let , so .

For the point we let , so .

What we have just found is a formula for the slope of the function. This formula has a special name called the derivative of the function.

Definition of the Derivative of a Function

The derivative of at is given by

provided the limit exists. For all for which this limit exists, is a function of .

Definition of the Derivative of a Function

The derivative simply creates a slope function that finds the slope at any point. You input a point, the output is the slope.

The process of finding the derivative of a function is called differentiation. A function is differentiable at if its derivative exists at and is differentiable on an open interval if it is differentiable at every point in the interval.

Definition of the Derivative of a Function

Here we see the derivative of a function being shown in motion with tangent lines drawn at each point along the curve. Green denotes positive slopes, red shows negative slopes, and black shows a slope of 0.

Notations

» In addition to which is read as “f prime of x,” other notations are used to denote the derivative of .

» The most common are:

𝑓 ′ (𝑥 ) , 𝑑𝑦𝑑𝑥 , 𝑦′ , 𝑑𝑑𝑥 [ 𝑓 (𝑥 )] ,𝐷𝑥 [ 𝑦 ] , �̇�

Of these the most popular are Lagrange’s notation, and Leibniz’s notation, which is read as “the derivative of y with respect to x” or simply “dy, dx”.

Example 3

» Find the derivative of

Example 4

» Find for . Then find the slope of the graph of at the points and Discuss the behavior of at

In order to continue finding the limit, we must rationalize the numerator.

Example 4 continued…

» Find for . Then find the slope of the graph of at the points and Discuss the behavior of at

Example 4 continued…

» Find for . Then find the slope of the graph of at the points and Discuss the behavior of at

At at

At which is undefined. By looking at the graph we see a vertical tangent line at

x

y

Example 5

» Find the derivative with respect to for the function .

.

Alternative Limit Form of a Derivative

» The following is an alternative form of derivative.

𝑓 ′ (𝑐 )=lim𝑥→𝑐

𝑓 (𝑥 )− 𝑓 (𝑐)𝑥−𝑐

x

y

{ ¿𝑥−𝑐(𝑐 , 𝑓 (𝑐))

(𝑥 , 𝑓 (𝑥 ))

f

𝑓 (𝑥 )− 𝑓 (𝑐 )¿ }

This form requires that the left and right hand limits be the same, thus requiring that the function is continuous.

Theorem 2.1Differentiability Implies Continuity

» If is differentiable at , then is continuous at .

Because of the alternative (and equivalent) form of the derivative on the previous slide, the requirement is made that in order for a function to be differentiable at a particular point, it must be continuous at that point.

The converse is NOT true. A function that is continuous at a particular point is NOT necessarily differentiable at that point. The following examples will illustrate this further.

Example 6

» Show that the function is NOT differentiable at

lim𝑥→−3+¿ 𝑓 (𝑥 )− 𝑓 (−3 )

𝑥− (−3 )= lim

𝑥→− 3+¿ |𝑥+3|−|− 3+3|𝑥+3

= lim𝑥→−3+ ¿ |𝑥+3|

𝑥+3 ¿

¿¿

¿¿

¿

lim𝑥→−3−

𝑓 (𝑥 )− 𝑓 (−3)𝑥− (−3 )

=lim

𝑥→− 3−|𝑥+3|−|−3+3|

𝑥+3=lim𝑥→−3−

|𝑥+3|

𝑥+3

Notice that direct substitution fails for these limits as we get the indeterminate form 0/0. Instead we will utilize a table, then examine the graph.

Example 6 continued…

» Show that the function is NOT differentiable at

lim𝑥→−3+¿|𝑥+3|

𝑥+3 ¿

¿lim𝑥→− 3−

|𝑥+3|

𝑥+3

-5 -4 -3 -2 -1

?

¿−1 ¿1

Since the left and right hand limits do not agree, by the alternative definition the function is NOT differentiable at . Now let’s examine the graph.

Example 6 continued…

» Show that the function is NOT differentiable at

lim𝑥→− 3−

|𝑥+3|

𝑥+3¿−1

lim𝑥→−3+¿|𝑥+3|

𝑥+3 ¿

¿¿1

Be careful when finding these limits by using a graph. You may be inclined to say the limit is 0 since that is the -value that is approached at . However, we aren’t finding the limit of the original function, but rather the derivative, so what we are looking for is the slope from the left and right.

1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

𝑚=−1 𝑚=1

Once again this tells us that the function is NOT differentiable at . Do note that the graph IS continuous everywhere.

Example 7

» Show that the function is NOT differentiable at .

A quick look at the graph tells us the function is continuous at .

1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

Since the limit is infinite, we can conclude the tangent line is vertical at . Thus, the function is NOT differentiable at .

Summary

» Examples 6 & 7 highlighted two key situations where a function was continuous at a particular point, but was NOT differentiable at that point. We can now point out some visual cues to look for on a graph to quickly determine whether it is differentiable at a point.

1. If a graph has a sharp corner, then the function is not differentiable at that point.

2. If a graph has a vertical tangent, then it is also not differentiable at that point.