Unit 2 - Differentiation Lesson 1: The Derivative

Unit 2 - Differentiation Lesson 1: The Derivative

AP Warmup #2

1) For which value of k is the following function continuous at x = 4?

f

(x)

sin

x

,

x4

k

x, 2

x4

(A) k = 2 (B) k = 1 (C) k = -1 (D) k = 1 (E) k = 1

2

2

2) Suppose f (a) 2 and f (b) 3 . Which of the following statements is always true?

(A) If f is continuous on [a, b] then the maximum value of f on [a, b] is less than 2.

(B) f takes on a maximum value on [a, b]. (C) If f is continuous, then f has a zero on [a, b]. (D) If f is continuous on [a, b] then f may take on a minimum value on [a, b]. (E) If f is not continuous on [a, b] then f does not take on a minimum value on [a, b].

3) The graph of a function f is shown below and describes the position of a particle as it moves along the x-axis.

(A) Describe the movement of the particle on the interval [-1, 2]. (B) Describe the movement of the particle as t approaches infinity. (C) Can we use Intermediate Value Theorem on the interval [-3, 0] to show

that f has a zero in that interval? On the interval [0, 3]? Explain your reasoning.

Unit 2 - Differentiation Lesson 1: The Derivative

What is a Derivative? Let's DISCOVER IT!

The definition of the derivative states that f (x) lim f (x h) f (x) . You may notice that this looks

h0

h

really similar to the average rate of change of a function from point (x, f (x)) to a point (x h, f (x h)) .

The definition of average rate of change is f (x h) f (x) . Do you see how this is hiding in our h

definition of the derivative?

(1) Let f (x) x2 . In the calculator place f (x) in y1 . Graph the function in a Zoom 4: Decimal

window. Your picture should match the one I have shown here. Turn

off y1 , and then enter the following equation in

y2

:

y1 (

x

0.1) 0.1

y1 ( x)

.

This

new

function

is

calculating

a

set

of

average rates of change along the graph of f (x) x2 . For each slope

calculated, the x values are 0.1 units apart.

(a) Look at a set of table values associated with the new function in y2 . Can you write a function to

represent these table values?

(b) Now, graph y2 to see what it looks like. Were you right? Hopefully, your equation matches the

graph you see!

(2)

Change the value of h to a smaller value. Let

h 0.01. Change

y2

:

y1 (

x

0.01) 0.01

y1

(

x)

.

Recheck the table values and graph. Write an equation that is

represented by the new set of table values for y2 .

Unit 2 - Differentiation Lesson 1: The Derivative

(3) One last time, change the value of h to an even smaller value. Let h 0.001. Change

y2

:

y1

(

x

0.001) 0.001

y1 (

x)

.

Recheck

the

table

values

and

graph.

Write

an

equation

that

is

represented by the new set of table values for y2 .

(4) If we were to continue decreasing the value of h, predict what the equation will be if

y2

y1(x 0.000001) 0.000001

y1 ( x)

.

(5) What would happen to the equation if we continued to let h get very small (or in other words,

approach 0)? Write an equation that would represent y2 as h approaches 0.

We would represent this, mathematically, as

y2

lim h0

y1(x h) y1(x) h

Bonus ? Try to find some other derivatives using the same techniques described above. Use both the graph

and the set of table values to predict the equation represented by y2 for:

(a) y 1 x2 4

(b) y ln x

(c)

y ex

Differentiation Defined

Recall that the slope of something represents its rate of change. We've expressed this in the past as y2 y1 x2 x1

But that really only helps us when dealing with linear concepts. Enter....."differentiation"....or the process of finding a derivative.....OR the process of finding the rate of change of ANY function.

The derivative is defined mathematically as

f ' (x) lim

x 0

f (x x) f (x) x

follows:

Other common notations for the derivative include: Example 1

y', dy , d [ f (x)] dx dx

Use the definition of the derivative to find f (x) if f (x) 2x2 x .

Unit 2 - Differentiation Lesson 1: The Derivative

Finding the Value of a Derivative at a Point ..........What is this, the DISCOVERY CHANNEL?

We know that f (a) lim f (x) f (a) . We can use the calculator to help us determine the value of xa x a

f (a) given f (x) .

Let

f (x) sin x

and let

x 0 . Enter

y1 sin x

and

y2

y1(x) y1(0) x0

or

sin x x

. Create a table to

study the values near x 0 .

(a) What do you notice about the x and y values in the table on the left? So, we can conclude that

sin x must equal _____!!

(b)

What does

y 2

represent? The ___________ of

sin x x

!!! What is this value near 0? _______

Look carefully!

(c) Change to y1 sin(2x) . What is the slope now?

We know that the value at x 0 is undefined. But looking at the two tables we see values get very close to 1 as we get very close to x 0 . Therefore I will make the conjecture that y2 approaches 1 as x approaches

a.

Use the table feature of the calculator to approximate the value of f (a) on the function f (x) .

1)

y ex @ x 1

2)

y 1 @x 1

x

2)

y x @x 1

4)

y x2 4 @ x 3

x2

Unit 2 - Differentiation Lesson 1: The Derivative

In order to find the slope of a tangent line at a specific point, we use this:

Example 2

Use this alternate form of the derivative to find f (2) if f (x) x2 3 .

f '(c) lim

xc

f (x) f (c) xc

Discovering the Derivative Graphically and Numerically

After completing this part, you should be able to:

Define the slope of a function at a point by zooming in on that point. Develop the definition of the derivatives of a function at a point by examining slopes of secant lines. Understand situations in which the derivative will fail to exist.

Remember from algebra that the slope of a line may be expressed and calculated by y y2 y1 , where x x2 x1

(x1, y1) and (x2 , y2 ) are any two points on the line. Most functions we see in calculus have the property

that if we pick a point on the graph of the function and zoom in, we will see a straight line.

Part I

(1) Graph the function f (x) x3 6x 3 for 3 x 4 .

(2) Is the slope of this function constant or is it changing? (3) Since the slope is not constant let's look at the function in a different window to see how it behaves.

Zoom in on the point (1, -2). Try to get the curve to appear to have a constant slope. Record a copy of your window below.

Slope = __________

Equation in y2 = ______________________________

(4) Select two points on this part of the graph, label them in the drawing and calculate the slope of a straight line between these two points.

Unit 2 - Differentiation Lesson 1: The Derivative

(5) Enter an equation y2 in point-slope form for the line passing through (1, -2) with the calculated

slope. Graph this line in the same window you are working with. Can you see two lines or one line?

(6) If the two lines are distinguishable then continue to zoom in on the point (1, -2) again and repeat steps 4 and 5.

(7) Once you have zoomed in on a point and the function appears to have a constant slope, you have a good approximation of the slope of the function near this point (1, -2), even though the function did not have constant slope. You will also notice that when you constructed the tangent line at that point (using the point-slope form) the tangent and the function appeared to be the same line.

(8) Zoom in a little more and repeat steps 4 and 5. Calculate and record a new slope three more times. Slope = ____________

Slope = ____________

Slope = ____________ (9) Are the slopes all the same? Do the numbers indicate that slope is approaching a particular value?

(10)The numbers you calculated are approximations to the slope of the function f (x) x3 6x 3 at the point (1, -2). The number that the slopes are approaching would be called the derivative of f at x = 1, and is denoted by f (1) .

Part II

Repeat this step for the slope at two other points on this graph: (0, 3) and (-3, -6)

Unit 2 - Differentiation Lesson 1: The Derivative

Part III

In parts I and II you learned how to use the graph of a function to estimate the value of the slope of a function at a point or the derivative of the function at that point. We did this by creating a window that was smaller and smaller. This caused the x-values to be very close to each other. We can also do this numerically using the table feature of the graphing calculator.

(1) Set the table minimum (TblStart) to 1 and the steps of the table (Tbl) to 0.1. Create the table for

the given function.

X

y 1

0.9

1

1.1

(2) Calculate the slope of a line through x = 0.9 and x = 1. Calculate the slope between x = 1 and x = 1.1. Calculate the slope again between x = 1 and the two x- values near x = 1.

(3) Set the table minimum (TblStart) to 1 and the steps of the table (Tbl) to 0.01. Create a table for

the given function.

X

y 1

1

(4) Calculate the slope of a line again through two points using this new chart.

(5) Repeat this step three more times:

X

y 1

1

X

y 1

X

y 1

(6) What do you observe about the slopes?

(7) Again the value that you calculated are approximations to the slope of the function

f (x) x3 6x 3 at the point (1, -2). The number that the slopes are approaching would be called the derivative of f at x = 1, and is denoted by f (1) .

Unit 2 - Differentiation Lesson 1: The Derivative

(8) Create tables to approximate the slope of the function at (0, 3) and (-3, -6).

(9) The calculator can determine the derivative of the function at these points. To do this start on the

home screen and press Math, 8: nDeriv( and enter this line:

nDeriv (Y1, x,1, 0.001) . This will

find an approximation for the derivative of the function at x = 1 using two points that are 0.001

apart on the x-axis. How do these values compare to what you had?

(10)Repeat this command for the other two points (0, 3) and (-3, -6).

Part IV

Study these three functions geometrically via zooming, analytically via use of limits, and directly via your calculator's derivative command:

(1) f (x) x4 x2 3 at (1, 3) (2) f (x) sin x at (n, 0) (3) f (x) tan x at (0, 0)

Part V

Here are two interesting functions. Study them the same way. What do you notice about the values? Can you use these values to predict how you will describe the derivative at the value of a?

4

(1) f (x) (x 1) 3 at a = 1 (2) f (x) x2 4 at a general point a. Try some different values of a. (3) For what values of a does f (a) exist and for what values does it fail to exist?

Further Exploration

(4) Explain in your own words how calculating the slope of a function at the point (a, f (a)) by repeated

zooming is related to the definition of the derivative

f (a) lim

f (a h)

f (a) .

h0

h

 ................
................

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