section 2.1: the derivative and the tangent line problem
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Section 2.1: The Derivative and the Tangent Line Problem. Section 2.1 – Classwork 1. TANGENT LINE. Slope ≈ -16. These can be considered average slopes or average rates of change. Slope = -17.6. Slope = -24. Secant Lines. Slope = -32. Slope = -48. Secant Line. - PowerPoint PPT PresentationTRANSCRIPT
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
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
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
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
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
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
Substitute into the function
1360 0 2
limx 0
x 136x x 2 x
Simplify in order to cancel the
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 .
Substitute into the function
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
Simplify in order to cancel the
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 .
Substitute into the function
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
Differentiability Justification 2In order to prove that a function is differentiable at x = c, you must show the following:
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
Differentiability Justification 3In order to prove that a function is differentiable at x = c, you must show the following:
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
Example 3Sketch 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
Decreasing
Negative
Increasing
Positive
Increasing
Positive
Example 4Sketch a graph of the derivative of the function graphed
below.
(-3,0) (3,0)
(-2.2,-4) (2.2,-4)
(0,0)
Example 5
Sketch a graph of the function with the following characteristics:
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
Sketch a graph of the function with the following characteristics:
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]