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About this Lesson
 Type: Video Tutorial
 Length: 14:42
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 159 MB
 Posted: 06/27/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Special Functions (10 lessons, $15.84)
Calculus: Logarithmic Functions (3 lessons, $6.93)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. 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'Hopital'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
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Special Functions
Logarithmic Functions
Using the Derivative Rules with Transcendental Functions Page [1 of 4]
Okay. Welcome back. We've got a great show tonight. We've got some derivatives we're going to take in a second. But right now, first, in my right handladies and gentlemenis tonight's top ten listtop ten list. You want to hear it? Okay. Tonight's top ten list. The top ten most important derivatives and why you should love them. Do you love them, Paul? I know you do.
Okay. Here we go. Number ten, the derivative of sine x is cosine x. Number nine; the derivative of cosine x is minus sine x, so watch it. Okay, number eight, they don't leave a sticky residue. Number seven, the derivative of tangent x is secant squared x. Secant squared x. Number six  asymptotes, asymptotes, asymptotes. Number sixthat was number six, Paul. Number five, the derivative of e to the x is e to the x. Little bit selfreferential, but what the heck. Number four, they stay crunchy in milk. Number three, the derivative of the natural log of x is one over x. Number two; never again will you be left speechless at a party conversation about natural logs. And the number one answer why you should love these exotic derivatives  yes, they will be on the test.
Okay, well. Thank you so much. Okay, great. Well, I thought we'd now try some examples and actually see these things in action. So, let's use those top ten list to try and solve some of these problems. For the first one, let's let f of x equal sine squared of two x cubed plus one. And the question for each of these problems is to find the derivative.
Okay. Well now, you look at this thing, and there's a lot of things going on here. The first thing we have to do is figure out what is sort of a laundry list of how we're going to attack this problem. So, in particular, we need to figure out what is it that's holding this whole thing together. And the answer is, it's this square. That's the most outside part of this function. In fact, if you rewrote this as sine of all that stuff quantity squared, you would really see that two on the outside.
Really confusing point for a lot of folks when they first get introduced to this notation, because it looks like the squared sort ofyou knowin the middle there. So, you might think, "Well, gee. The sign's on the outside, or this is x." But remember, this equals sine of all that stuff squared. So, it's that squared that is the outside thing. That's the last thing that you would do if you were to actually compute this on a calculator.
And so, in fact, what we've got to do is peel away that squared first. I mean, think about these things as the onions now. What we've got to do is peel off the layers. So, this is going to be a Chain Rule problem. In fact, it's going to be a multiple Chain Rule problem. How many times am I going to have to peel off things? How many times am I going to have to use the Chain Rule?
Well, let's see. I'm going to have to peel that off. And let's visualize right now, in our minds, while we peel that off. Then I've got to take the derivative of that. But that's also a Chain Rule, because I've got sine of stuff. So, now I've got to peel off the sine. That's another Chain Rule. So, in fact, I'm going to have twotwo Chain Rules in this one problem.
Okay. Let's try to execute. So, let's take the derivative here, f prime of x. I'm going to first peel off that outside two. How do I do that? Well, I see that it's just a big old blop squared. What's the derivative of blop squared? It's two blop to the first power. And what's the blop? Well, that was sine two x cubed plus one. And that's all to the first power.
Now, I just peeled off that two, and I've got to take the derivative of what's left. Well, that's a sine of blop. So, I look at that and see that as sine of blop. So what's the derivative of sine of blop? Well, derivative of sine is cosine of blop. So, it's cosine of blop, and the blop was two x cubed plus one.
Now, we just peeled off the sine, so all we're left with is that thing, and we have to multiply by the derivative of that. The derivative of that is easily seen just to be six x. So, this long answer is the derivative. You could rewrite this a little bit, I guess, if you want. And remember we're multiplying everything, so that two times that six could be written as a twelve. I could put that x in front. I'm just sort of cleaning things up a little bit here. And then, we have sine of two x cubed plus one times cosine two x cubed plus one. So, there's the answer to this first question. That's the derivative. And that required a double Chain Rulea double Chain Rule on that one.
Okay, great. Let's try another one. How about this one? How about g of x equals cosine x times e to the power sine x. So, this is sort of a weird one. It's cosine x times e to the sine x. Okay, what are we going to do here? You might want to actually make a little laundry list of what's holding this thing together, and what we're going to have to do to actually pull this thing apart.
Well, if you look at this and think about it, the last thing you'd have to do is what? Is it that power? No, if you were actually computing this on a calculator, the very last thing you'd do would be press the multiplication keythis product right here. You'd have to figure out that number first. You'd have to figure out that number first, and the last thing you do is press product. What is that product?
So, this is going to be a Product Rule. I'm going to make a little list hereProduct Rule. By the way, I do this. And maybe you would like to is too. I actually make a listan outlineto make sure you're not forgetting anything. Within the Product Rule, I'll have to take the derivative of that, which is not a big deal, but I'll have to take the derivative of that.
What technique am I going to use to take the derivative of that? Well, that's going to require a little Chain Rule right there, in and of itself, because I see e to the blop. And I only know how to take the derivative of e to the x. And so e to the sine x is going to require another little Chain Rule. So, inside the Product Rule, we're going to have a Chain Rule.
Okay. Let's execute this and see how it goes. So first, we're going to execute the Product Rule. So, that's the firstI'm going to chant itso it's cosine of x times the derivative of the second. So now, I come to the derivative of that, and so here I am at the Chain Rule. So now, I'm about to do the Chain Rule, and so, what do I see? I see e to the blop. And what's the derivative of e to the blop, it's just e to the blop.
And then, I have to multiply that by the derivative of the blop. I just peeled away this, and now, I'm going to take the derivative of sine, and the derivative of sinewe've already seenis cosine.
So that's the derivative of this whole thing. The derivative of e to the blop is e to the blop. I peel that off. Take the derivative of sine and get cosine. So, there I am, and I just took the derivative of that.
Where am I in the Product Rule? AnswerI don't know. So, I've got to rechant the Product Rule. First times the derivative of the secondokay, now I know where I amplus the secondso e to the sine x powermultiplied by the derivative of the first. And what's the derivative of cosine? It's not sine. The derivative of cosine is minus sine. So, minus sine of x.
So, I could rewrite this as how? Well, I could see cosine times cosine. That would be cosine squared. So, I could write this as cosine squared x, e to the power sine x minus e to the sine x times sine x. So, that's a perfectly fine answer. Or you might notice that I could factor out the common factor of sine x if you wanted to do a little bit more cleaning. Any of these answers are all correct, by the way, but I'm just trying to show you can simplify this.
A lot of times, you look in the back of the book, if you do an odd problem, and you see an answer, and the answer is different than what you said. It actually might be equivalent but just simplified a little bit, and so it looks different. So, don't be faked out. If you think your answer's right, stick with it. A lot of times there's also typos in the back of the book.
So, we factor that out. I would see cosine squared x minus sine x. So, any of those answers are correct, and I'll just box in this one here.
So, great. There's the derivative of this complicated function. Notice how I used the Product Rule, and within the Product Rule, I used the Chain Rule. It was a nice example. Okay? Let's try another one. In fact, how about if I let you try this one? How about if I tell you that y equals x times the natural log of x squared plus one. And the question is to find the derivative of thisyou want to find dy dx. Remember, that's just the notational way of saying take the derivative of this. Why don't you try right now, and see if you can do it? Make a laundry list first, and then see if you can execute the laundry list.
How'd it go? Sometimes these problems are a little involved. Let's take a look. What's holding this thing together? Well, is it this term here? No. Is it this? No. Is it this? No. It's actuallyit's invisible multiplication right here. If you want to compute this on a calculator, you wouldn't lastly do this. You wouldn't lastly do this. You'd have to compute all that stuff first. You'd have to figure out what that is, and then the last thing you do is take the product.
So, this is a Product Rule, and the Product Rule's going to require me to take the derivative of that, which is not going to be that difficult, but then take the derivative of that. That's going to be a little more challenging, because the derivative of that is the natural log of blop. And so, to take the derivative of that function, I'm going to have to actually use a Chain Rule to unlock that. So, inside this, I'm going to have the Chain Rule.
Okay, let's execute and see how we make out. So, to do this, I'm first going to start off dy dx equalsand I'm going to do the Product Rule. That's the first times the derivative of the second. Well, here I am already arriving at the Chain Rule. So, how do I take the derivative of that? Well, notice that's just the natural log of blop. So what's the derivative of natural log of blop? It's one over the blop.
Remember, the derivative of natural log is just the reciprocal of whatever the natural log thing is. So, the derivative of the natural log of blop is actually one over the blop, which in this case is x squared plus one. But I'm not done yet; I just peeled away the natural log by doing that. I still have to take the derivative of this. The derivative of that is two x.
So, that's now the derivative of this whole thing. Do you see it? The derivative of natural log of blop is one over blop. Peel that away. Multiply that by the derivative of the inside, which is two x. So, there we are. And so, that takes care of the derivative of this. Where am I in my Product Rule? I don't know. I've got to rechant.
First times the derivative of the secondokay, now I know where I amplus the secondyou see how I rechant to make sure I don't forget anythingmultiplied by the derivative of the first. Well, the derivative of x is just one, so I'll write an invisible one. Can you not see that? Good.
Well, you could simplify this, I guess, a little bit. You see an x times two x. That's going to be a two x squared, all divided by x squared plus one, plus this natural log of x squared plus one. I guess you could combine those if you want to and make them into one big fraction, but I'm not going to do that. I'll look at that as the answer. There's the answer. Required a Product Rule, again, and then another Chain Rule.
Great. Here's the last one we're going to try. And, again, let me give you a shot at this. It's sort of fun to try them yourselvesgood practice. Suppose that y equals tangent of x divided by one plus the natural log of x. Why don't you find the derivativefind dy dx? Right now, give it a shot.
Success? Let's see. To find this, what I'm going to do is first, I have to think to myself what is holding this thing together. And what's holding this thing together is that division. So, this is actually going to be a Quotient Rule problem. And within the Quotient Rule, am I going to have to do anything funny? Well, let's see. I've got to take the derivative of that, which I can just do, and I have to take the derivative of that, which I can just do. So, actually, I don't think there's anything going on here. There's no Chain Rule or anything else here. So, this is going to be a Quotient Rule.
So, let's take a look at the Quotient Rule and see how this is going to work. So, dy dx equalsso, what's the Quotient Rule chant? I'll chant it for you right now. Bottom times the derivative of the top minus the top times the derivative of the bottom all divided by the bottom squared. Okay. Let's execute it.
So, it's the bottomso, one plus natural log of xtimes the derivative of the top. So, I have to take the derivative of tangent of x, which we already saw is secant squared x minus the toptangent xtimes the derivative of the bottom. And what's the derivative of the bottom? Well, the derivative of one is zero, and the derivative of natural log of x is just one over x. And that's all divided by the bottomone plus the natural log of x squared. And that's the answer.
I'm just going to make a little point here. A classic mistakeby the wayis to somehow in the denominator, you get so wrapped up in taking derivatives, people sometimes take the derivative of the bottom and square it. Don't do that. We just divide by the bottom squared. Just divide by the bottom squared. So, it's the bottom times the derivative of the top minus the top times the derivative of the bottom, all divided by the bottom squared.
Okay, so there's some elaborate examples where we're using all these complicated transmittal functions. And you can see, once you're empowered with the Product Rule, Quotient Rule and Chain Rule, then no matter what function we throw at youas long as you, sort of, break it up into small, little, manageable, bitesized piecesyou can actually bite it off. So, in fact, you can now bite off any derivative that you want.
Next up, we're going to take a look at taking derivatives of things that aren't even functions. What does it mean not to even be a function? Well, stay tuned. We'll see. See you in a bit.
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