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About this Lesson
 Type: Video Tutorial
 Length: 11:17
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 121 MB
 Posted: 11/18/2008
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Intro to Derivatives (10 lessons, $17.82)
Calculus: Using the Derivative (4 lessons, $8.91)
In this lesson, we will review tangent lines, learn how to find the derivative, and learn how to use the derivative once we find it. We begin by finding the slope the tangent line f(x) = 2x^2 at x=3. We find the slope by taking the derivative of f(x). We compute this derivative by evaluating the limit as delta x approaches 0 of [f(x + delta x)  f(x)]/delta x or, in this case, the limit as delta x approaches zero of [2(x+ delta x)^2  2x^2]/delta x.
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
 2174 lessons
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11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
 The Slope of a Tangent Line
 12/29/2010

This is an excellent stepbystep demonstration. I am so happy I found it. I'm sure I will be purchasing more of these lessons as I progress through my Calculus Class.
 Calculus: The Slope of a Tangent Line
 10/01/2009

Another great hit by the Thinkwell staff but not perfect. I was going to give you a 5 but when the professor kept saying..."look at the graph or look at the other side"... no window appeared so I looked at nothing else but the teacher. Maybe this is a postproduction error but it would be nice if it was corrected.
An Introduction to Derivatives
Using the Derivative
The Slope of a Tangent Line Page [1 of 1]
Okay folks. It’s time to get our feet wet in this whole derivative business. So let’s take off our shoes and get walking. Okay, so what I want to take a look at now are a whole bunch of examples. We’re still walking, just like that little pink bunny. Okay, oops, the pink bunny bit the dust.
Okay, let’s take a look at some examples now. The first example I want to us to take a look at is the following. The question is – Find the slope of the tangent line to the function f of x equals two x squared at the point x equals three. Okay, so the first reaction you may have is “Huh? What does that mean?” Well, just take it slowly and think about what’s going on here.
First of all, what does that look like? That’s actually something we can graph. So I’m going to graph that one. It’s going to be rare in these problems when we can actually graph the functions because they can get really complicated, really quickly. But here’s one that we can actually graph and so what the heck, let’s graph it. This is a parabola. You’ll notice that it’s centered right here at the origin and instead of being the good old fashioned parabola that comes out like this; this is going to be a tighter one because that number of course, is bigger than two, bigger than one, plus two. So in fact, our parabola will look something like this.
And I’m interested in the action at three. So let’s look at three; one, two, three. Let’s take a look at what’s going on there at three. Well, look at three and zoom up, boom, I come out here. By the way, what is this value right, just for fun? Well, that’s just going to be three plugged into the function. That’s going to be three squared times two. That’s going to be two times nine, which is eighteen. So that number actually is eighteen.
Okay and now what’s the question. I want to find the slope of the tangent line. So what I'm looking after is the slope of the line that just grazes. How do I find the slope of the line that just grazes? I remind you of what we just discovered. The derivative at a particular point represents the slope of the tangent line, at that point. So this question actually requires us to first find the derivative.
Sort of interesting here now, we’re making an intellectual leap here. There’s a question and the question doesn’t even talk about derivatives but knowing that the derivative represents this object, we now see we need the derivative to answer this question. All we have to do is find the derivative. Okay, so how do we find the derivative? Well, I remind you of the definition of the derivative. The definition of the derivative is given by this beautiful fact. Can you see that? Can you see that fact right there? Look at that fact. That is the definition of the derivative. Remember it’s just the limit of a quotient and that quotient is just a slope. It’s the change in y over the change in x. Really cool. Bring this back down to earth.
Okay, now what I’d like to do is actually compute the derivative. Now what do we do to compute the derivative? I have to evaluate that limit. Okay, so let’s now evaluate the limit. I’m going to take this picture and I’m going to put it over there right now. I’m going to clean the slate off and we’re going to use this definition and find the derivative of this function. Okay, let’s do it right, right now, chop, chop. Here we go.
So the first step is to write down the definition of the derivative, since that’s the only thing we know about the derivative. The definition of the derivative I remind you, is this. Okay, and the function we’re looking at, I also remind you – it’s just up here. I’ll put it way up there and now here, whoops, it fell. And here’s the definition. So now, what I’m going to do is I’m going to actually insert all this particulars of this function into here.
Let’s see if we could do that right now. So f prime of x –remember that’s just the name for the derivative – that equals the limit. We have the limit – as delta x approaches zero of this long quotient. So how do I figure out this long quotient? We have to take it easy, take it slowly and make sure we do this correctly. Remember what happens. What does f of x plus delta x mean? Well, it means wherever I see an x in my original function I have to replace that x by this whole quantity there. So wherever I see an x, I put in this whole thing. This is always a sticky point for people when they first see this stuff. That’s understandable. It’s sort of a complicated looking thing. There’s a delta x; there’s an x, there’s that x. There’s all this stuff going on here. There’s x’s here and what not. This is where you have to remember the chicken. This is where you have to remember the quarters. This is where you have to remember all that ridiculous stuff when we were looking at function. It’s just a matter of plugging in. This is a placeholder so wherever I see that symbol I’m now going to replace it by that entire symbol.
Let’s do that. What is f of x plus delta x? It’ll be two multiplied by x plus delta x, all squared. I just plugged in that wherever I saw the x. With me? Okay, now I’m not done yet. I have to now subtract off f of x. Well, f of x is just a function itself. Now to divide all that thing by delta x. Okay, now that is the limit I have to compute. It looks long and unwieldy. Let’s compute it.
What’s the very first thing I try when computing a limit? I let this thing approach whatever that says. Let’s let delta x equal zero. If we let delta x equal zero, what do we get here? If delta x equals zero, that term drops out. So I’m just left with an x squared. So I see a two x squared minus two x squared, that’s zero over zero. Zero over zero, indeterminate form, needs more work. What should I do here? What techniques should I think about? Well, the technique that I think I’m going to use is just expand out this whole thing here. I see that parentheses, I’m squaring this whole thing.
Let me FOIL and see what I get. So what do we have? This equals – and don’t forget I always wish you will write limit delta x approaches zero – of and then what do we have here. Two times – and then I’ve got to square all this out. Let’s square that out. You can go off and try that FOILing thing and you’ll see you get an x squared plus two x delta x plus delta x squared. That’s what you get when you FOIL. Then you’d subtract off the two x squared. That’s from this term.
Now I want you to look at that. I’m going to divide the whole thing by delta x. I’m going to want you to look at that and see if you followed all that and if that’s really correct. And now let me ask and lead the witness a little bit. Tell me what’s wrong with this. It looks so close to being correct, in fact, I bet that maybe you even thought it was correct. And that’s completely reasonable but actually, it’s not. Because in fact you were being very stingy. You didn’t want to share that two with everybody. That two gets multiplied by every single person here, but instead you decide to hoard it all for the x square here and so what do I have to do. Well, what I’ve got to do is I’ve got to multiply that entire thing through by the two.
Now, let’s do it. Now we’re on the road correctness here. Limit is delta x approaches zero of – and if I distribute, I see, and this always fun. By the way, this is, I think the greatest thing about these problems, look how long this gets. Two x squared plus four x delta x plus two delta x squared minus two x squared. Look at all that stuff. Have you ever seen such a long thing in your life? Look at that thing. It’s incredible. Looks really, really, really, really hard until you realize a spectacular coincidence has occurred. Notice there’s a two x squared and minus two x squared. Those things actually drop out. And what’s neat about that? What’s neat about that is that everything that remains has a delta x in it. I could actually factor out that delta x; thus giving me the zero over zero term.
Let’s take this and I’m going to take this exact thing, I’m going to push it over there so you can see it. I’ll just keep going. What does that equal? So if I now factor, I’m going to put the limit as delta x approaches zero of and I take out that delta x factor on the top. I’m left with four x plus two delta x. See I took out the – look over there – I took out the one delta x and I took out one of the delta x’s here, but I had another one, so I have that one. Divided by delta x and look there it is. There’s the zero over zero term. That was the thing that made this indeterminate. Do you see how we actually grab that term there by expanding  and it looked really, really awful. Look at thing over there. It’s awful, it’s long, but the really cool thing is it all collapses down. In these kind of questions, it’s always wonderful because there will always be collapse.
Well, now we can cancel these things away as long as you promise me one thing and of course, that promise is that delta x doesn’t equal zero. Is that a problem? Is that a problem, camera people? No! No, that’s right. It’s no because I’m just approaching zero. I never equal zero. And now if I take the limit, what do I do? I let delta x go to zero. If I let delta x go to zero, this term gets really, really small, shrinks down to zero. What am I left with? I’m left with four x and so, I see the derivative, our very first derivative that we computed is equal to four x. And remember that the derivative at a particular point represents the slope of the tangent line at that point.
So if I want to find the slope of the tangent line, which was our mission at the point x, equals three what I do is I go to this machine. This is a slopeproducing machine and I plug in three here. And if I plug in three here, into the derivative, that gives me twelve and that twelve represents the slope of the tangent line of this function at x equals three. Take a look at the graph that’s now there and you’ll see that slope of that tangent line is in fact, equal to twelve. It’s fairly steep, notice how steep it is. It’s twelve.
So in fact, we’re now empowered to find slopes of tangent lines. The way we do it is we first go through this process of finding the derivative and then that derivative is a little machine that produces the slope, once you plug in the value your interested in. In this question, the value is three. I plug in three into this derivative machine; this slope machine and I get twelve. The slope is twelve. We found our very first derivative and we found the answer to this question. One down, about a thousand more to go. See you in a bit.
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This is an excellent stepbystep demonstration. I am so happy I found it. I'm sure I will be purchasing more of these lessons as I progress through my Calculus Class.
Another great hit by the Thinkwell staff but not perfect. I was going to give you a 5 but when the professor kept saying..."look at the graph or look at the other side"... no window appeared so I looked at nothing else but the teacher. Maybe this is a postproduction error but it would be nice if it was corrected.