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Precalculus

12.4 Derivatives

Precalculus12.4 Derivatives
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  1. Preface
  2. 1 Functions
    1. Introduction to Functions
    2. 1.1 Functions and Function Notation
    3. 1.2 Domain and Range
    4. 1.3 Rates of Change and Behavior of Graphs
    5. 1.4 Composition of Functions
    6. 1.5 Transformation of Functions
    7. 1.6 Absolute Value Functions
    8. 1.7 Inverse Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  3. 2 Linear Functions
    1. Introduction to Linear Functions
    2. 2.1 Linear Functions
    3. 2.2 Graphs of Linear Functions
    4. 2.3 Modeling with Linear Functions
    5. 2.4 Fitting Linear Models to Data
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  4. 3 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 3.1 Complex Numbers
    3. 3.2 Quadratic Functions
    4. 3.3 Power Functions and Polynomial Functions
    5. 3.4 Graphs of Polynomial Functions
    6. 3.5 Dividing Polynomials
    7. 3.6 Zeros of Polynomial Functions
    8. 3.7 Rational Functions
    9. 3.8 Inverses and Radical Functions
    10. 3.9 Modeling Using Variation
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Review Exercises
    15. Practice Test
  5. 4 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 4.1 Exponential Functions
    3. 4.2 Graphs of Exponential Functions
    4. 4.3 Logarithmic Functions
    5. 4.4 Graphs of Logarithmic Functions
    6. 4.5 Logarithmic Properties
    7. 4.6 Exponential and Logarithmic Equations
    8. 4.7 Exponential and Logarithmic Models
    9. 4.8 Fitting Exponential Models to Data
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  6. 5 Trigonometric Functions
    1. Introduction to Trigonometric Functions
    2. 5.1 Angles
    3. 5.2 Unit Circle: Sine and Cosine Functions
    4. 5.3 The Other Trigonometric Functions
    5. 5.4 Right Triangle Trigonometry
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  7. 6 Periodic Functions
    1. Introduction to Periodic Functions
    2. 6.1 Graphs of the Sine and Cosine Functions
    3. 6.2 Graphs of the Other Trigonometric Functions
    4. 6.3 Inverse Trigonometric Functions
    5. Key Terms
    6. Key Equations
    7. Key Concepts
    8. Review Exercises
    9. Practice Test
  8. 7 Trigonometric Identities and Equations
    1. Introduction to Trigonometric Identities and Equations
    2. 7.1 Solving Trigonometric Equations with Identities
    3. 7.2 Sum and Difference Identities
    4. 7.3 Double-Angle, Half-Angle, and Reduction Formulas
    5. 7.4 Sum-to-Product and Product-to-Sum Formulas
    6. 7.5 Solving Trigonometric Equations
    7. 7.6 Modeling with Trigonometric Equations
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Review Exercises
    12. Practice Test
  9. 8 Further Applications of Trigonometry
    1. Introduction to Further Applications of Trigonometry
    2. 8.1 Non-right Triangles: Law of Sines
    3. 8.2 Non-right Triangles: Law of Cosines
    4. 8.3 Polar Coordinates
    5. 8.4 Polar Coordinates: Graphs
    6. 8.5 Polar Form of Complex Numbers
    7. 8.6 Parametric Equations
    8. 8.7 Parametric Equations: Graphs
    9. 8.8 Vectors
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  10. 9 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 9.1 Systems of Linear Equations: Two Variables
    3. 9.2 Systems of Linear Equations: Three Variables
    4. 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 9.4 Partial Fractions
    6. 9.5 Matrices and Matrix Operations
    7. 9.6 Solving Systems with Gaussian Elimination
    8. 9.7 Solving Systems with Inverses
    9. 9.8 Solving Systems with Cramer's Rule
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  11. 10 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 10.1 The Ellipse
    3. 10.2 The Hyperbola
    4. 10.3 The Parabola
    5. 10.4 Rotation of Axes
    6. 10.5 Conic Sections in Polar Coordinates
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  12. 11 Sequences, Probability and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 11.1 Sequences and Their Notations
    3. 11.2 Arithmetic Sequences
    4. 11.3 Geometric Sequences
    5. 11.4 Series and Their Notations
    6. 11.5 Counting Principles
    7. 11.6 Binomial Theorem
    8. 11.7 Probability
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  13. 12 Introduction to Calculus
    1. Introduction to Calculus
    2. 12.1 Finding Limits: Numerical and Graphical Approaches
    3. 12.2 Finding Limits: Properties of Limits
    4. 12.3 Continuity
    5. 12.4 Derivatives
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Review Exercises
    10. Practice Test
  14. A | Basic Functions and Identities
  15. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
  16. Index

Learning Objectives

In this section, you will:
  • Find the derivative of a function.
  • Find instantaneous rates of change.
  • Find an equation of the tangent line to the graph of a function at a point.
  • Find the instantaneous velocity of a particle.

The average teen in the United States opens a refrigerator door an estimated 25 times per day. Supposedly, this average is up from 10 years ago when the average teenager opened a refrigerator door 20 times per day 37.

It is estimated that a television is on in a home 6.75 hours per day, whereas parents spend an estimated 5.5 minutes per day having a meaningful conversation with their children. These averages, too, are not the same as they were 10 years ago, when the television was on an estimated 6 hours per day in the typical household, and parents spent 12 minutes per day in meaningful conversation with their kids.

What do these scenarios have in common? The functions representing them have changed over time. In this section, we will consider methods of computing such changes over time.

Finding the Average Rate of Change of a Function

The functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.

A tangent line to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the slope of the curve at that point is close to the slope of the tangent line at that point.

Figure 1 represents the function f( x )= x 3 4x. f( x )= x 3 4x. We can see the slope at various points along the curve.

  • slope at x=−2 x=−2 is 8
  • slope at x=−1 x=−1 is –1
  • slope at x=2 x=2 is 8
Graph of f(x) = x^3 - 4x with tangent lines at x = -2 with a slope of 8, at x = -3 with a slope of -1, and at x=2 with a slope of 8.
Figure 1 Graph showing tangents to curve at –2, –1, and 2.

Let’s imagine a point on the curve of function f fat x=a x=a as shown in Figure 2. The coordinates of the point are ( a,f(a) ). ( a,f(a) ). Connect this point with a second point on the curve a little to the right of x=a, x=a, with an x-value increased by some small real number h. h. The coordinates of this second point are ( a+h,f(a+h) ) ( a+h,f(a+h) ) for some positive-value h. h.

Graph of an increasing function that demonstrates the rate of change of the function by drawing a line between the two points, (a, f(a)) and (a, f(a+h)).
Figure 2 Connecting point a awith a point just beyond allows us to measure a slope close to that of a tangent line at x=a. x=a.

We can calculate the slope of the line connecting the two points (a,f(a)) (a,f(a)) and (a+h,f(a+h)), (a+h,f(a+h)), called a secant line, by applying the slope formula,

slope =  change in y change in x slope =  change in y change in x

We use the notation m sec m sec to represent the slope of the secant line connecting two points.

m sec = f(a+h)f(a) (a+h)(a)        = f(a+h)f(a) a +h a m sec = f(a+h)f(a) (a+h)(a)        = f(a+h)f(a) a +h a

The slope m sec m sec equals the average rate of change between two points (a,f(a)) (a,f(a)) and (a+h,f(a+h)). (a+h,f(a+h)).

m sec = f( a+h )f( a ) h m sec = f( a+h )f( a ) h

The Average Rate of Change between Two Points on a Curve

The average rate of change (AROC) between two points (a,f(a)) (a,f(a)) and (a+h,f(a+h)) (a+h,f(a+h)) on the curve of f fis the slope of the line connecting the two points and is given by

AROC= f( a+h )f( a ) h AROC= f( a+h )f( a ) h

Example 1

Finding the Average Rate of Change

Find the average rate of change connecting the points ( 2,−6 ) ( 2,−6 ) and ( −1,5 ). ( −1,5 ).

Try It #1

Find the average rate of change connecting the points ( 5,1.5 ) ( 5,1.5 ) and (2.5,9). (2.5,9).

Understanding the Instantaneous Rate of Change

Now that we can find the average rate of change, suppose we make h hin Figure 2 smaller and smaller. Then a+h a+h will approach a aas h hgets smaller, getting closer and closer to 0. Likewise, the second point ( a+h,f(a+h) ) ( a+h,f(a+h) ) will approach the first point, ( a,f(a) ). ( a,f(a) ). As a consequence, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at x=a, x=a, and the slope of the secant line will get closer and closer to the slope of the tangent at x=a. x=a. See Figure 3.

Graph of an increasing function that contains a point, P, at (a, f(a)). At the point, there is a tangent line and two secant lines where one secant line is connected to Q1 and another secant line is connected to Q2.
Figure 3 The connecting line between two points moves closer to being a tangent line at x=a. x=a.

Because we are looking for the slope of the tangent at x=a, x=a, we can think of the measure of the slope of the curve of a function f fat a given point as the rate of change at a particular instant. We call this slope the instantaneous rate of change, or the derivative of the function at x=a. x=a. Both can be found by finding the limit of the slope of a line connecting the point at x=a x=a with a second point infinitesimally close along the curve. For a function f fboth the instantaneous rate of change of the function and the derivative of the function at x=a x=a are written as f'(a), f'(a), and we can define them as a two-sided limit that has the same value whether approached from the left or the right.

f (a)= lim h0 f( a+h )f( a ) h f (a)= lim h0 f( a+h )f( a ) h

The expression by which the limit is found is known as the difference quotient.

Definition of Instantaneous Rate of Change and Derivative

The derivative, or instantaneous rate of change, of a function f fat x=a , x=a , is given by

f'(a)= lim h0 f( a+h )f( a ) h f'(a)= lim h0 f( a+h )f( a ) h

The expression f( a+h )f( a ) h f( a+h )f( a ) h is called the difference quotient.

We use the difference quotient to evaluate the limit of the rate of change of the function as h happroaches 0.

Derivatives: Interpretations and Notation

The derivative of a function can be interpreted in different ways. It can be observed as the behavior of a graph of the function or calculated as a numerical rate of change of the function.

  • The derivative of a function f(x) f(x) at a point x=a x=a is the slope of the tangent line to the curve f(x) f(x) at x=a. x=a. The derivative of f(x) f(x) at x=a x=a is written f (a). f (a).
  • The derivative f (a) f (a) measures how the curve changes at the point ( a,f(a) ). ( a,f(a) ).
  • The derivative f (a) f (a) may be thought of as the instantaneous rate of change of the function f(x) f(x) at x=a. x=a.
  • If a function measures distance as a function of time, then the derivative measures the instantaneous velocity at time t=a. t=a.

Notations for the Derivative

The equation of the derivative of a function f( x ) f( x ) is written as y = f (x), y = f (x), where y=f(x). y=f(x). The notation f (x) f (x) is read as “ f prime of x. f prime of x. ” Alternate notations for the derivative include the following:

f (x)= y = dy dx = df dx = d dx f(x)=Df(x) f (x)= y = dy dx = df dx = d dx f(x)=Df(x)

The expression f (x) f (x) is now a function of x x; this function gives the slope of the curve y=f( x ) y=f( x ) at any value of x. x. The derivative of a function f( x ) f( x ) at a point x=a x=a is denoted f (a). f (a).

How To

Given a function f, f, find the derivative by applying the definition of the derivative.

  1. Calculate f( a+h ). f( a+h ).
  2. Calculate f( a ). f( a ).
  3. Substitute and simplify f( a+h )f( a ) h . f( a+h )f( a ) h .
  4. Evaluate the limit if it exists: f (a)= lim h0 f( a+h )f( a ) h . f (a)= lim h0 f( a+h )f( a ) h .

Example 2

Finding the Derivative of a Polynomial Function

Find the derivative of the function f(x)= x 2 3x+5 f(x)= x 2 3x+5 at x=a. x=a.

Try It #2

Find the derivative of the function f(x)=3 x 2 +7x f(x)=3 x 2 +7x at x=a. x=a.

Finding Derivatives of Rational Functions

To find the derivative of a rational function, we will sometimes simplify the expression using algebraic techniques we have already learned.

Example 3

Finding the Derivative of a Rational Function

Find the derivative of the function f(x)= 3+x 2x f(x)= 3+x 2x at x=a. x=a.

Try It #3

Find the derivative of the function f(x)= 10x+11 5x+4 f(x)= 10x+11 5x+4 at x=a. x=a.

Finding Derivatives of Functions with Roots

To find derivatives of functions with roots, we use the methods we have learned to find limits of functions with roots, including multiplying by a conjugate.

Example 4

Finding the Derivative of a Function with a Root

Find the derivative of the function f(x)=4 x f(x)=4 x at x=36. x=36.

Try It #4

Find the derivative of the function f( x )=9 x f( x )=9 x at x=9. x=9.

Finding Instantaneous Rates of Change

Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by s(t)=−16 t 2 +64t+6, s(t)=−16 t 2 +64t+6, where t tis measured in seconds and s( t ) s( t ) is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown in Figure 4 as a function of time. In physics, we call this the “s-t graph.”

Graph of a negative parabola with a vertex at (2, 70) and two points at (1, 55) and (3, 55).
Figure 4

Example 5

Finding the Instantaneous Rate of Change

Using the function above, s(t)=−16 t 2 +64t+6, s(t)=−16 t 2 +64t+6, what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

Try It #5

The position of the ball is given by s(t)=−16 t 2 +64t+6. s(t)=−16 t 2 +64t+6. What is its velocity 2 seconds into flight?

Using Graphs to Find Instantaneous Rates of Change

We can estimate an instantaneous rate of change at x=a x=a by observing the slope of the curve of the function f( x ) f( x ) at x=a. x=a. We do this by drawing a line tangent to the function at x=a x=a and finding its slope.

How To

Given a graph of a function f( x ), f( x ), find the instantaneous rate of change of the function at x=a. x=a.

  1. Locate x=a x=a on the graph of the function f( x ). f( x ).
  2. Draw a tangent line, a line that goes through x=a x=a at a aand at no other point in that section of the curve. Extend the line far enough to calculate its slope as
    change in y change in x . change in y change in x .

Example 6

Estimating the Derivative at a Point on the Graph of a Function

From the graph of the function y=f( x ) y=f( x ) presented in Figure 5, estimate each of the following:

f(0) f(0) f(2) f(2) f'(0) f'(0) f'(2) f'(2)

Graph of an odd function with multiplicity of two and with two points at (0, 1) and (2, 1).
Figure 5

Try It #6

Using the graph of the function f(x)= x 3 3x f(x)= x 3 3x shown in Figure 7, estimate: f(1), f(1), f (1), f (1), f(0), f(0), and f (0). f (0).

Graph of the function f(x) = x^3-3x with a viewing window of [-4. 4] by [-5, 7
Figure 7

Using Instantaneous Rates of Change to Solve Real-World Problems

Another way to interpret an instantaneous rate of change at x=a x=a is to observe the function in a real-world context. The unit for the derivative of a function f( x ) f( x ) is

output units   input unit  output units   input unit 

Such a unit shows by how many units the output changes for each one-unit change of input. The instantaneous rate of change at a given instant shows the same thing: the units of change of output per one-unit change of input.

One example of an instantaneous rate of change is a marginal cost. For example, suppose the production cost for a company to produce x xitems is given by C( x ), C( x ), in thousands of dollars. The derivative function tells us how the cost is changing for any value of x xin the domain of the function. In other words, C ( x ) C ( x ) is interpreted as a marginal cost, the additional cost in thousands of dollars of producing one more item when x xitems have been produced. For example, C ( 11 ) C ( 11 ) is the approximate additional cost in thousands of dollars of producing the 12th item after 11 items have been produced. C ( 11 )=2.50 C ( 11 )=2.50 means that when 11 items have been produced, producing the 12th item would increase the total cost by approximately $2,500.00.

Example 7

Finding a Marginal Cost

The cost in dollars of producing x xlaptop computers in dollars is f( x )= x 2 100x. f( x )= x 2 100x. At the point where 200 computers have been produced, what is the approximate cost of producing the 201st unit?

Example 8

Interpreting a Derivative in Context

A car leaves an intersection. The distance it travels in miles is given by the function f( t ), f( t ), where t trepresents hours. Explain the following notations:

f(0)=0 f(0)=0 f (1)=60 f (1)=60 f(1)=70 f(1)=70 f(2.5)=150 f(2.5)=150

Try It #7

A runner runs along a straight east-west road. The function f( t ) f( t ) gives how many feet eastward of her starting point she is after t t seconds. Interpret each of the following as it relates to the runner.

f( 0 )=0 f( 0 )=0 f( 10 )=150 f( 10 )=150 f ( 10 )=15 f ( 10 )=15 f ( 20 )=10 f ( 20 )=10 f( 40 )=−100 f( 40 )=−100

Finding Points Where a Function’s Derivative Does Not Exist

To understand where a function’s derivative does not exist, we need to recall what normally happens when a function f( x ) f( x ) has a derivative at x=a x=a . Suppose we use a graphing utility to zoom in on x=a x=a . If the function f( x ) f( x ) is differentiable, that is, if it is a function that can be differentiated, then the closer one zooms in, the more closely the graph approaches a straight line. This characteristic is called linearity.

Look at the graph in Figure 8. The closer we zoom in on the point, the more linear the curve appears.

Graph of a negative parabola that is zoomed in on a point to show that the curve becomes linear the closer it is zoomed in.
Figure 8

We might presume the same thing would happen with any continuous function, but that is not so. The function f(x)=| x |, f(x)=| x |, for example, is continuous at x=0, x=0, but not differentiable at x=0. x=0. As we zoom in close to 0 in Figure 9, the graph does not approach a straight line. No matter how close we zoom in, the graph maintains its sharp corner.

Graph of an absolute function.
Figure 9 Graph of the function f(x)=| x |, f(x)=| x |, with x-axis from –0.1 to 0.1 and y-axis from –0.1 to 0.1.

We zoom in closer by narrowing the range to produce Figure 10 and continue to observe the same shape. This graph does not appear linear at x=0. x=0.

Graph of an absolute function.
Figure 10 Graph of the function f(x)=| x |, f(x)=| x |, with x-axis from –0.001 to 0.001 and y-axis from—0.001 to 0.001.

What are the characteristics of a graph that is not differentiable at a point? Here are some examples in which function f( x ) f( x ) is not differentiable at x=a. x=a.

In Figure 11, we see the graph of

f(x)={ x 2 , x2 8x, x>2 . f(x)={ x 2 , x2 8x, x>2 .

Notice that, as x xapproaches 2 from the left, the left-hand limit may be observed to be 4, while as x xapproaches 2 from the right, the right-hand limit may be observed to be 6. We see that it has a discontinuity at x=2. x=2.

Graph of a piecewise function where from negative infinity to (2, 4) is a positive parabola and from (2, 6) to positive infinity is a linear line.
Figure 11 The graph of f( x ) f( x ) has a discontinuity at x=2. x=2.

In Figure 12, we see the graph of f(x)=| x |. f(x)=| x |. We see that the graph has a corner point at x=0. x=0.

Graph of an absolute function.
Figure 12 The graph of f(x)=| x | f(x)=| x | has a corner point at x=0 x=0 .

In Figure 13, we see that the graph of f(x)= x 2 3 f(x)= x 2 3 has a cusp at x=0. x=0. A cusp has a unique feature. Moving away from the cusp, both the left-hand and right-hand limits approach either infinity or negative infinity. Notice the tangent lines as x xapproaches 0 from both the left and the right appear to get increasingly steeper, but one has a negative slope, the other has a positive slope.

Graph of f(x) = x^(2/3) with a viewing window of [-3, 3] by [-2, 3].
Figure 13 The graph of f(x)= x 2 3 f(x)= x 2 3 has a cusp at x=0. x=0.

In Figure 14, we see that the graph of f(x)= x 1 3 f(x)= x 1 3 has a vertical tangent at x=0. x=0. Recall that vertical tangents are vertical lines, so where a vertical tangent exists, the slope of the line is undefined. This is why the derivative, which measures the slope, does not exist there.

Graph of f(x) = x^(1/3) with a viewing window of [-3, 3] by [-3, 3].
Figure 14 The graph of f(x)= x 1 3 f(x)= x 1 3 has a vertical tangent at x=0. x=0.

Differentiability

A function f( x ) f( x ) is differentiable at x=a x=a if the derivative exists at x=a, x=a, which means that f (a) f (a) exists.

There are four cases for which a function f( x ) f( x ) is not differentiable at a point x=a. x=a.

  1. When there is a discontinuity at x=a. x=a.
  2. When there is a corner point at x=a. x=a.
  3. When there is a cusp at x=a. x=a.
  4. Any other time when there is a vertical tangent at x=a. x=a.

Example 9

Determining Where a Function Is Continuous and Differentiable from a Graph

Using Figure 15, determine where the function is

  1. continuous
  2. discontinuous
  3. differentiable
  4. not differentiable

At the points where the graph is discontinuous or not differentiable, state why.

Graph of a piecewise function that has a removable discontinuity at (-2, -1) and is discontinuous when x =1.
Figure 15

Try It #8

Determine where the function y=f( x ) y=f( x ) shown in Figure 18 is continuous and differentiable from the graph.

Graph of a piecewise function with three pieces.
Figure 18

Finding an Equation of a Line Tangent to the Graph of a Function

The equation of a tangent line to a curve of the function f( x ) f( x ) at x=a x=a is derived from the point-slope form of a line, y=m( x x 1 )+ y 1 . y=m( x x 1 )+ y 1 . The slope of the line is the slope of the curve at x=a x=a and is therefore equal to f (a), f (a), the derivative of f( x ) f( x ) at x=a. x=a. The coordinate pair of the point on the line at x=a x=a is (a,f(a)). (a,f(a)).

If we substitute into the point-slope form, we have

The point-slope formula that demonstrates that m = f(a), x1 = a, and y_1 = f(a).

The equation of the tangent line is

y=f'(a)( xa )+f(a) y=f'(a)( xa )+f(a)

The Equation of a Line Tangent to a Curve of the Function f

The equation of a line tangent to the curve of a function f fat a point x=a x=a is

y=f'(a)( xa )+f(a) y=f'(a)( xa )+f(a)

How To

Given a function f, f, find the equation of a line tangent to the function at x=a. x=a.

  1. Find the derivative of f( x ) f( x ) at x=a x=a using f (a)= lim h0 f( a+h )f( a ) h . f (a)= lim h0 f( a+h )f( a ) h .
  2. Evaluate the function at x=a. x=a. This is f( a ). f( a ).
  3. Substitute ( a,f( a ) ) ( a,f( a ) ) and f ( a ) f ( a ) into y=f'(a)( xa )+f(a). y=f'(a)( xa )+f(a).
  4. Write the equation of the tangent line in the form y=mx+b. y=mx+b.

Example 10

Finding the Equation of a Line Tangent to a Function at a Point

Find the equation of a line tangent to the curve f(x)= x 2 4x f(x)= x 2 4x at x=3. x=3.

Analysis

We can use a graphing utility to graph the function and the tangent line. In so doing, we can observe the point of tangency at x=3 x=3 as shown in Figure 19.

Graph of f(x) = x^2-4x with a tangent line at x = 3 which has the equation of y = 2x - 9.
Figure 19 Graph confirms the point of tangency at x=3. x=3.
Try It #9

Find the equation of a tangent line to the curve of the function f(x)=5 x 2 x+4 f(x)=5 x 2 x+4 at x=2. x=2.

Finding the Instantaneous Speed of a Particle

If a function measures position versus time, the derivative measures displacement versus time, or the speed of the object. A change in speed or direction relative to a change in time is known as velocity. The velocity at a given instant is known as instantaneous velocity.

In trying to find the speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as the distance traveled divided by the elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? The use of a derivative solves this problem. A derivative allows us to say that even while the object’s velocity is constantly changing, it has a certain velocity at a given instant. That means that if the object traveled at that exact velocity for a unit of time, it would travel the specified distance.

Instantaneous Velocity

Let the function s( t ) s( t ) represent the position of an object at time t. t. The instantaneous velocity or velocity of the object at time t=a t=a is given by

s (a)= lim h0 s( a+h )s( a ) h s (a)= lim h0 s( a+h )s( a ) h

Example 11

Finding the Instantaneous Velocity

A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after t tseconds is given by s(t)=−16 t 2 +36t+200, s(t)=−16 t 2 +36t+200, find the instantaneous velocity of the ball at t=2. t=2.

Analysis

This result means that at time t=2 t=2 seconds, the ball is dropping at a rate of 28 ft/sec.

Try It #10

A fireworks rocket is shot upward out of a pit 12 ft below the ground at a velocity of 60 ft/sec. Its height in feet after t tseconds is given by s=16 t 2 +60t12. s=16 t 2 +60t12. What is its instantaneous velocity after 4 seconds?

Media

Access these online resources for additional instruction and practice with derivatives.

Visit this website for additional practice questions from Learningpod.

Footnotes

  • 37 http://www.csun.edu/science/health/docs/tv&health.html Source provided.

12.4 Section Exercises

Verbal

1.

How is the slope of a linear function similar to the derivative?

2.

What is the difference between the average rate of change of a function on the interval [ x,x+h ] [ x,x+h ] and the derivative of the function at x? x?

3.

A car traveled 110 miles during the time period from 2:00 P.M. to 4:00 P.M. What was the car's average velocity? At exactly 2:30 P.M., the speed of the car registered exactly 62 miles per hour. What is another name for the speed of the car at 2:30 P.M.? Why does this speed differ from the average velocity?

4.

Explain the concept of the slope of a curve at point x. x.

5.

Suppose water is flowing into a tank at an average rate of 45 gallons per minute. Translate this statement into the language of mathematics.

Algebraic

For the following exercises, use the definition of derivative lim h0 f(x+h)f(x) h lim h0 f(x+h)f(x) h to calculate the derivative of each function.

6.

f( x )=3x4 f( x )=3x4

7.

f( x )=2x+1 f( x )=2x+1

8.

f( x )= x 2 2x+1 f( x )= x 2 2x+1

9.

f( x )=2 x 2 +x3 f( x )=2 x 2 +x3

10.

f( x )=2 x 2 +5 f( x )=2 x 2 +5

11.

f( x )= 1 x2 f( x )= 1 x2

12.

f( x )= 2+x 1x f( x )= 2+x 1x

13.

f( x )= 52x 3+2x f( x )= 52x 3+2x

14.

f( x )= 1+3x f( x )= 1+3x

15.

f(x)=3 x 3 x 2 +2x+5 f(x)=3 x 3 x 2 +2x+5

16.

f(x)=5 f(x)=5

17.

f(x)=5π f(x)=5π

For the following exercises, find the average rate of change between the two points.

18.

( −2,0 ) ( −2,0 ) and ( −4,5 ) ( −4,5 )

19.

( 4,−3 ) ( 4,−3 ) and ( −2,−1 ) ( −2,−1 )

20.

( 0,5 ) ( 0,5 ) and ( 6,5 ) ( 6,5 )

21.

( 7,−2 ) ( 7,−2 ) and ( 7,10 ) ( 7,10 )

For the following polynomial functions, find the derivatives.

22.

f(x)= x 3 +1 f(x)= x 3 +1

23.

f(x)=3 x 2 7x=6 f(x)=3 x 2 7x=6

24.

f(x)=7 x 2 f(x)=7 x 2

25.

f(x)=3 x 3 +2 x 2 +x26 f(x)=3 x 3 +2 x 2 +x26

For the following functions, find the equation of the tangent line to the curve at the given point x xon the curve.

26.

f(x)=2 x 2 3x x=3 f(x)=2 x 2 3x x=3

27.

f(x)= x 3 +1 x=2 f(x)= x 3 +1 x=2

28.

f(x)= x x=9 f(x)= x x=9

For the following exercise, find k k such that the given line is tangent to the graph of the function.

29.

f(x)= x 2 kx, y=4x9 f(x)= x 2 kx, y=4x9

Graphical

For the following exercises, consider the graph of the function f fand determine where the function is continuous/discontinuous and differentiable/not differentiable.

30.


Graph of a piecewise function with three segments. The first segment goes from negative infinity to (-3, -2), an open point; the second segment goes from (-3, 1) to (2, 3), which are both open points; the final segment goes from (2, 2), an open point, to positive infinity.
31.


Graph of a piecewise function with three segments. The first segment goes from negative infinity to (-2, -1), an open point; the second segment goes from (-2, -4), an open point, to (0, 0), a closed point; the final segment goes from (0, 1), an open point, to positive infinity.
32.
Graph of a piecewise function with two segments and an asymptote at x = 3. The first segment, which has a removable discontinuity at x = -2, goes from negative infinity to the asymptote, and the final segment goes from the asymptote to positive infinity.
33.
Graph of a piecewise function with two segments. The first segment goes from (-4, 0), an open point to (5, -2), and the final segment goes from (5, 3), an open point, to positive infinity.

For the following exercises, use Figure 20 to estimate either the function at a given value of x xor the derivative at a given value of x ,x, as indicated.

Graph of an odd function with multiplicity of 2 with a turning point at (0, -2) and (2, -6).
Figure 20
34.

f( 1 ) f( 1 )

35.

f( 0 ) f( 0 )

36.

f( 1 ) f( 1 )

37.

f( 2 ) f( 2 )

38.

f(3) f(3)

39.

f ( 1 ) f ( 1 )

40.

f ( 0 ) f ( 0 )

41.

f (1) f (1)

42.

f ( 2 ) f ( 2 )

43.

f ( 3 ) f ( 3 )

44.

Sketch the function based on the information below:

f ( x )=2x f ( x )=2x , f( 2 )=4 f( 2 )=4

Technology

45.

Numerically evaluate the derivative. Explore the behavior of the graph of f(x)= x 2 f(x)= x 2 around x=1 x=1 by graphing the function on the following domains: [ 0.9,1.1 ] [ 0.9,1.1 ] , [ 0.99,1.01 ] [ 0.99,1.01 ] , [ 0.999,1.001 ] , [ 0.999,1.001 ] ,and [0.9999,1.0001] [0.9999,1.0001] . We can use the feature on our calculator that automatically sets Ymin and Ymax to the Xmin and Xmax values we preset. (On some of the commonly used graphing calculators, this feature may be called ZOOM FIT or ZOOM AUTO). By examining the corresponding range values for this viewing window, approximate how the curve changes at x=1, x=1, that is, approximate the derivative at x=1. x=1.

Real-World Applications

For the following exercises, explain the notation in words. The volume f(t) f(t) of a tank of gasoline, in gallons, t tminutes after noon.

46.

f(0)=600 f(0)=600

47.

f'(30)=−20 f'(30)=−20

48.

f(30)=0 f(30)=0

49.

f'(200)=30 f'(200)=30

50.

f(240)=500 f(240)=500

For the following exercises, explain the functions in words. The height, s ,s, of a projectile after t tseconds is given by s(t)=16 t 2 +80t. s(t)=16 t 2 +80t.

51.

s(2)=96 s(2)=96

52.

s'(2)=16 s'(2)=16

53.

s(3)=96 s(3)=96

54.

s'(3)=−16 s'(3)=−16

55.

s(0)=0,s(5)=0. s(0)=0,s(5)=0.

For the following exercises, the volume V Vof a sphere with respect to its radius r ris given by V= 4 3 π r 3 . V= 4 3 π r 3 .

56.

Find the average rate of change of V Vas r rchanges from 1 cm to 2 cm.

57.

Find the instantaneous rate of change of V Vwhen r=3 cm. r=3 cm.

For the following exercises, the revenue generated by selling x xitems is given by R(x)=2 x 2 +10x. R(x)=2 x 2 +10x.

58.

Find the average change of the revenue function as x xchanges from x=10 x=10 to x=20. x=20.

59.

Find R'(10) R'(10) and interpret.

60.

Find R'(15) R'(15) and interpret. Compare R'(15) R'(15) to R'(10), R'(10), and explain the difference.

For the following exercises, the cost of producing x xcellphones is described by the function C(x)= x 2 4x+1000. C(x)= x 2 4x+1000.

61.

Find the average rate of change in the total cost as x xchanges from x=10 to x=15. x=10 to x=15.

62.

Find the approximate marginal cost, when 15 cellphones have been produced, of producing the 16th cellphone.

63.

Find the approximate marginal cost, when 20 cellphones have been produced, of producing the 21st cellphone.

Extension

For the following exercises, use the definition for the derivative at a point x=a, x=a, lim xa f(x)f(a) xa , lim xa f(x)f(a) xa , to find the derivative of the functions.

64.

f(x)= 1 x 2 f(x)= 1 x 2

65.

f(x)=5 x 2 x+4 f(x)=5 x 2 x+4

66.

f(x)= x 2 +4x+7 f(x)= x 2 +4x+7

67.

f(x)= 4 3 x 2 f(x)= 4 3 x 2

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