Worksheet on Derivatives



Lesson: Deriving the Derivative

Grade Level: 12th grade, Calculus I class

Materials: Whiteboard, TI-Nspire classroom set, quadratic tangents program, and a computer projector.

Goals/Objectives: Formulate the general definition of the derivative function, use the definition to find the derivative of functions, and check our work using the quadratic tangents program on a TI-Nspire.

Standards

1) Algebra

a. Equations

b. Arithmetic with Polynomials and Rational Functions

c. Creating Equations

d. Reasoning with Equations

2) Functions

a. Interpreting Functions

b. Building Functions

c. Linear and Quadratic Functions

Assessment: Use the derivative definition to find the derivative of three functions and confirm your work using a TI-Nspire.

Lesson Plan

To begin today’s discussion I would like to review what we learned in the last section on limits with a few examples.

Find the limits.

1) [pic] = 13

2) [pic] = 6

3) [pic] =5

Now that we have refreshed our minds on that, let’s begin our new topic.

I would like to begin by reviewing a topic. Say we are given two points: (2, -3) and (5,6). How do we find the slope of the line that connects those points?

We use the fact that slope = rise/run = 6-(-3)/5-2 = 9/3 = 3. We would then say the line that connects those points two points has a slope of 3.

Now what if we were only given a function and the x coordinates of the two points, where those two points satisfy the function. For instance, what if we knew two points satisfied the function y = x3 + 2 and the x coordinates for those two points were -1 and 2. What would the slope of the line be that connects those two points?

We must first find the y coordinates of those points. How do we do that?

We use the function and find that the two points are therefore (-1, 1) and (2, 10). We then go through the same steps as we did before.

Now we will use only these tools and what we’ve recently learned about limits to find what is known as the derivative, which is nothing more than the slope of the tangent line. We begin with an arbitrary curve.

[pic]

We have derived our expression for finding the derivative function to any given function. Let’s try it out with a few examples. Starting with something we already known the answer to, let’s consider the line y = 3x. Do we know the slope of this line for all x values? Yes, it should be 3. Let’s check.

1) We know f(x) = 3x. Then f(x+h) = 3(x+h). Plugging this in we get

[pic]

Our derivative function is f’(x) = 3, so it is just a flat line.

What would happen if y = constant, say 2? Would f(x+h)=f(x)? So what is the slope of a flat line?

Let’s try another.

2) Let’s find the derivative function of f(x) = x2-x.

What is f(x+h)?

[pic]

If I plug the value x=2 into 2x-1 I get 3. What does that 3 mean?

Now that we have seen the general manner to calculate the derivative function for any given function let’s get a better feel for what is going on using our TI-Nspire.

Say we are given f(x) = x2 we want to find f’(x). We use our definition and obtain:

[pic]

Now, turn on your TI-nSpire and go to My Documents. Select the file called quadratictangents. Page 1.1 should look like the picture below.

You will see that the graph of y=x2 is graphed as well as the tangent line, T. Also note that the slope of the tangent line is provided. In this case the slope is 4.6 as m=4.6. Grab the point X and move it left and right. Describe in the space below how the slope of the tangent line changes as you move X.

Now go to page 1.2. It should look like the picture below.

We see again the function y=x2 and the tangent line T are graphed. There is another point of interest this time and that is point P. Notice that P’s x value is the x coordinate where it is located so 2.6 in this case. However, it’s y value is not it’s y coordinate. It is instead the slope of the function y=x2 at that specific x we are at so 5.2 in this case. I want you to grab point X and slide it to the right. You will notice that as you move X P moves because P depends on X. You will also see that P traces out its path. Therefore for each x value P marks what the value of the slope at that X. Once you have slid X over to at least -3 you will see a defined path formed by P. It should look like the picture below.

What is the meaning of this path?

It is indeed the derivative function of x2. Let’s check the fit by graphing 2x. Go to the double arrow in the lower left corner and click on it. F2(x) should appear. Now enter 2x and press enter. You will see the line appear.

It should look like the figure below.

How well does it fit our points?

For homework tonight I want you to repeat the same steps we did here for f(x)=x2 where we found the derivative function using our definition and then used the TI-nSpire to check our work, for three different functions. The work you need to show is the steps in the derivative definition, your derivative function and a sketch of your final graph that verifies you are correct. The three functions I want you to do are:

1) f(x) = 3x + 2

2) f(x) = x2 + 1

3) f(x) = x2 + x

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