Section 2.1: The Derivative and the Tangent Line Problem

Section 2.1: The Derivative and the Tangent Line Problem

Section 2.1 – Classwork 1

TANGENT LINE

Slope = -48

Slope = -32

Slope = -24

Slope = -17.6

Slope ≈ -16

102 16 3y x

Secant Lines

These can be

considered average slopes or average rates of change.

Secant Line

A line that passes through two points on a curve.

Tangent Line Most people believe that a tangent line only intersects a

curve once. For instance, the first time most students see a tangent line is with a circle:

Although this is true for circles, it is not true for every curve:

Every blue line intersects the pink curve only once.

Yet none are tangents.

The blue line intersects the

pink curve twice. Yet it is

a tangent.

Tangent LineAs two points of a secant line are brought together, a tangent line is formed. The slope of

which is the instantaneous rate of change











Slope of a Tangent LineIn order to find a formula for the slope of a tangent line, first look

at the slope of a secant line that contains (x1,y1) and (x2,y2):

f x2 f x1 x2 x1

yx

m

y2 y1

x2 x1

f x1 x f x1 x

f x1 x f x1 x1 x x1

(x2,y2)

(x1,y1)

Δx

In order to find the slope of the tangent line, the change in x needs to be as small as

possible.

Instantaneous Rate of ChangeTangent Line with Slope m

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

exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)).

mtan limx 0

f cx f c x

f(x)

(c, f(c))

m

Example 1Determine the best way to describe the slope of the

tangent line at each point.

A

B

C

A. Since the curve is decreasing, the slope will also be decreasing. Thus, the slope is negative.

B. The vertex is where the curve goes from increasing to decreasing. Thus, the slope must be zero.

C. Since the curve is Increasing, the slope will also be increasing. Thus the slope is Positive.

29 6 24 8 9 6

0lim x x x

xx

Example 2Find the instantaneous rate of change to at (3,-6).

c is the x-coordinate of the point on the

curve

Direct substitution

mtan limx 0

f 3x f 3 x

limx 0

3x 2 8 3x 9 32 8 3 9 x

limx 0

x 2 2xx

limx 0

x 2

Substitute into the function

0 2

limx 0

x x 2 x

Simplify in order to cancel the

denominator

f x x 2 8x 9

2

limx 0

812x6 x 2 x 3 2x 10x

Example 3Find the equation of the tangent line to at (2,10).

c is the x-coordinate of the point on the

curve

Direct substitution

mtan limx 0

f 2x f 2 x

limx 0

2x 3 2x 23 2 x

limx 0

13x6 x 2 x 3

x

limx 0

136x x 2


1360 0 2

limx 0

x 136x x 2 x


denominator

f x x 3 x

13

y 10 13 x 2

Just the slope. Now use the point-slope formula to find the

equation

A Function to Describe Slope

In the preceding notes, we considered the slope of a tangent line of a function f at a number c. Now, we change our point of view and let the number c vary by replacing it with x.

mtan limx 0

f cx f c x

The slope of a tangent line at the

point x = c.

A constant.

m x limx 0

f xx f x x

A function whose output is the slope of a tangent line at any x.

A variable.

limx 0

1x 1xx x 1xx 1x

ExampleDerive a formula for the slope of the tangent line to the graph of .


Direct substitution

m x limx 0

f xx f x x

limx 0

11xx

11x

x

limx 0

1x 1 x xx 1xx 1x

limx 0

11xx 1x

Multiply by a common

denominator

11x0 1x

limx 0

xx 1xx 1x


denominator

f x 11x

1

1x 2

A formula to find the slope of any tangent

line at x.

1xx 1x 1xx 1x

The Derivative of a FunctionThe limit used to define the slope of a tangent line is also

used to define one of the two fundamental operations of calculus:

The derivative of f at x is given by

Provided the limit exists. For all x for which this limit exists, f’ is a function of x.

0

( ) ( )'( ) lim

x

f x x f xf x

x

READ: “f prime of x.”

Other Notations for a Derivative:

dy

dx'y ( )

df x

dx

limx 0

5 xx 1 5 x 1

x

Example 1Differentiate .


Direct substitution

f ' x limx 0

f xx f x x

limx 0

5 xx 1 5 x 1 x

limx 0

5 xx 5 x

x

5 limx 0

xx x

x

5 limx 0

xx x

x xx x

limx 0

5 xx x x

Simplify in order to cancel the denominator

f x 5 x 1

5 limx 0

x

x xx x

xx x

xx x

5 limx 0

1xx x

10

5x x

52 x

Make the problem easier by factoring out common

constants

limx 0

x 3 3x 2 x3x x 2 x 3 x 3

x

Example 2Find the tangent line equation(s) for such that the tangent line has a slope of 12.

Find the derivative first since the

derivative finds the slope for an x value

Find when the derivative equals 12

f ' x limx 0

f xx f x x

3 3

0lim x x x

xx

limx 0

3x 2 x3x x 2 x 3

x

22

0lim 3 3x

x x x x

3x 2 3x0 0 2

limx 0

x 3x 2 3xx x 2 x

f x x 3

3x 2

3x 2 12

x 2 42x

Find the output of the

function for every input

2f 2f

32

32

88

8 12 2y x 8 12 2y x Use the point-slope

formula to find the equations

How Do the Function and Derivative Function compare?

f x 12x

f ' x 112x

Domain: Domain:

12 ,

12 ,

f is not differentiable at x = -½

Differentiability Justification 1In order to prove that a function is differentiable at x = c, you must show the following:

In other words, the derivative from the left side MUST EQUAL the derivative from the right side.

Common Example of a way for a derivative to fail:

limx 0

f cx f c x lim

x 0

f cx f c x

Other common examples: Corners or Cusps

Not differentiable at x = -4

limx c

f x f c


In other words, the function must be continuous.

Common Example of a way for a derivative to fail: Other common examples:

Gaps, Jumps, AsymptotesNot differentiable

at x = 0


In other words, the tangent line can not be a vertical line.

An Example of a Verticaltangent where the derivative to Fails to exist:

limx 0

f cx f c x

Not differentiable at x = 0

ExampleDetermine whether the following derivatives exist for the

graph of the function.

1. ' 8

2. ' 6

3. ' 2

4. ' 2

5. ' 3

6. ' 4

f

f

f

f

f

f

DNE

DNE

DNE

Exists

DNE

Exists

Example 2

Sketch a graph of the function with the following characteristics:

The derivative does not exist at x = -2.

The function is continuous on (-6,3)

The Range is (-7,-1]

Example 3Show that does not exist if .

f x x 3 1

f ' 3

3 2 1

0lim x

xx

First rewrite the absolute value function as a piecewise

function

Since the one-sided limits are not equal, the derivative does

not exist

2 if 3

4 if 3

x xf x

x x

limx 0

f 3x f 3 x

limx 0

xx

limx 0

1

Find the Left Hand Derivative

1

3 2 3 2

0lim x

xx

3 4 1

0lim x

xx

limx 0

f 3x f 3 x

0lim x

xx

0lim 1x

Find the Right Hand

Derivative

1

3 4 3 4

0lim x

xx

Function v DerivativeCompare and contrast the function and its derivative.

FUNCTION DERIVATIVE

-5

-5Vertex

x-intercept

Decreasing

Negative Slopes

Increasing

Positive Slopes

Function v DerivativeCompare and contrast the function and its derivative.

FUNCTION DERIVATIVE

-5 -5

Local Max

x-intercept

Decreasing

Negative Derivatives

Increasing

Positive Derivatives

5 5

Local Min

x-intercept

Positive Derivatives

Increasing

Example 1Accurately graph the derivative of the function graphed

below at left.The Derivative does not exist at a corner.

Make sure the x-value does not have a derivative

The slope from -∞ to

-7 is -2

The slope from -7 to 2

is 0

The slope from 2 to ∞

is 1

Example 2Sketch a graph of the derivative of the function graphed

below at left.

1. Find the x values where the slope of the tangent line is zero (max, mins, twists)

2. Determine whether the function is increasing or decreasing on each interval

Increasing

Positive

Decreasing

Negative

Increasing

Positive

Decreasing

Negative

Increasing

Positive


below at left.

1. Find the x values where the slope of the tangent line is zero (max, mins, twists)

2. Determine whether the function is increasing or decreasing on each interval

Decreasing

Negative

Increasing

Positive

Increasing

Positive


below.

(-3,0) (3,0)

(-2.2,-4) (2.2,-4)

(0,0)

Example 5


The derivative is only positive for -6<x<-3

and 5<x<10.

The function is differentiable on (-6,10)

Differentiability Implies Continuity

If f is differentiable at x = c, then f is continuous at x = c.

The contrapositive of this statement is true: If f is NOT continuous at x = c, then f is NOT differentiable at x = c.

The converse of this statement is not always true (be careful): If f is continuous at x = c, then f is differentiable at x = c.

The inverse of this statement is not always true: If f is NOT differentiable at x = c, then f is NOT continuous at x = c.

Example


The derivative does NOT exist at x = -2.

The derivative equals 0 at x = 1.

The derivative does NOT exist at x = 4.

The function is continuous on [-4,-2)U(-2,5]

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Section 2.1: The Derivative and the Tangent Line Problem