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Calculus: Derivative of the Natural Log Function

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About this Lesson

  • Type: Video Tutorial
  • Length: 13:24
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 144 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: Special Functions (10 lessons, $15.84)
Calculus: Logarithmic Functions (3 lessons, $6.93)

In this lesson, you will learn how to find the derivative of the natural log function. To start with, we will review exponential functions and the natural logarithmic function, and then you move on to learning about composite functions that include natural log functions. To find the derivative of one of these composite functions, like e^(3x^2), you must combine the fact that the derivative of e^x is e^x and the chain rule. You will also see how to use the natural exponential function to prove why the derivative of f(x) = ln x is equal to 1/x. Last, this lesson will cover how to use the chain rule or the properties of logs to find the derivative of the composition of a natural log function and another function, like finding the derivative of ln(x^3).

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 UT-Austin 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, p-adic 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
Thinkwell
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Founded in 1997, Thinkwell has succeeded in creating "next-generation" 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 technology-based 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 star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...

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Recent Reviews

Nopic_blu
Excellent job of explaining a tricky topic of C...
01/24/2009
~ travis5818

I was getting the differentiation and integration of exponential functions mixed up pretty bad. After viewing this view, the differentiation part is definitely easy. Mr. Burger doesn't go over the integration of exponential functions in this one, but the differentiation sets a foundation for that.

-Travis

P.S. It's a shame thinkwell is such a scam. Mindbites has the right idea with their pricing.

Nachan_homepage
Thanks!
01/19/2009
~ nachan

Professor Burger explains the questions what is the fundamental derivative of the log function? He helps us understand log function by first explaining the exponential function and shows us how to solve both types of problems and explains the meaning of inverse functions.

Nopic_blu
Excellent job of explaining a tricky topic of C...
01/24/2009
~ travis5818

I was getting the differentiation and integration of exponential functions mixed up pretty bad. After viewing this view, the differentiation part is definitely easy. Mr. Burger doesn't go over the integration of exponential functions in this one, but the differentiation sets a foundation for that.

-Travis

P.S. It's a shame thinkwell is such a scam. Mindbites has the right idea with their pricing.

Nachan_homepage
Thanks!
01/19/2009
~ nachan

Professor Burger explains the questions what is the fundamental derivative of the log function? He helps us understand log function by first explaining the exponential function and shows us how to solve both types of problems and explains the meaning of inverse functions.

Special Functions
Logarithmic Functions
The Derivative of the Natural Log Function Page [1 of 1]
Well, we’ve taken a look at logs, and now the fundamental question remains. What is the derivative of the log function, and in particular, what’s the derivative of the natural log. So, here’s the question. If the function f of x equals the natural log of x—which means, that’s just the log base e—that special number, 2.71 something—of x. What’s the derivative? So, what is that equal to?
Okay. Well, to answer that question, I actually want to warm up by reviewing the exponential stuff that we already saw, do a couple of examples, and then we can actually see what the answer to this question is. Not surprisingly, we can use the exponential function to help us with this, because the exponential function and the log function sort of interact so well together. They sort of interplay so nicely.
In particular, let me just remind you that if I take e and raise it to the natural log of x power, that just gives me x. There’s a relationship that links the exponential function and the log function together. They’re inverse functions.
Okay. Well, let’s warm up. So, let’s warm up, always good to warm up whether you’re doing physical or mental exercise. Okay, so in this mental exercise, let’s do the following problem. Suppose I give you a function g of x. Let’s start off very simple, simply g of x equals e to the x. What is the derivative of that? G prime of x equals—well, it equals e to the x. That’s basically how we define the number e. It’s that base for which e to the x has, as its derivative itself, e to the x. Really neat function. Its derivative is equal to itself. It’s really easy. So, if you didn’t like the exponential function stuff before, you should really like it now. Because taking derivatives of e to the x is easy. E to the x. Derivative—e to the x.
But now, let’s try to actually look at some more elaborate exponential functions. For example, how about this one. H of x equals e to the three x squared. How would you find the derivative of this? Well, this is a little bit trickier, until you realize that you know to find the derivative of e to the x. And therefore, you know how to find the derivative of e to the blop. And in particular, if you treat –whoops, excuse me, I just – did you miss me? If you treat that whole thing there as a blop, I see e to the blop. This cries out for the Chain Rule.
So, the derivative of e to the blop is what? Well, it’s itself—e to the blop. And what was the blop? It was three x squared. And then, what does the Chain Rule say to do after that? You take that answer, and you multiply it by the derivative of the blop. So, I just peeled off that e, and now I’m just looking that and have to take the derivative of that; derivative of that is easily seen to be six x.
So, this answer turns out to be—if I write the six x first—six x e to the three x squared. Let me review that again with you. So, the idea is if I see e to something very complicated, since I know how to take the derivative of e to the x—it’s just itself. Then I treat this whole thing as a blop, and I say, “Well, what’s the derivative of e to the blop?” Well, the derivative of e to the blop is e to the blop. So, it equals e to the blop.
But then, the Chain Rule says I’ve got to multiply that by the six x. Remember the onions. Remember the onions. I’m peeling off layers, but now, what I’m peeling off is the e. I peel off the e, and then I have the three x squared—that’s the inside—and take the derivative of that and get to six x.
Let me try another one to really drive this issue—uh—notion home. So, let’s call this function k. K of x equals e to the four x cubed plus three x minus one. Why don’t you, right now, try to take the derivative of this function. See if you can get the right answer. Okay.
Well, how’d you make out? Let’s do it together and see what we get. The first thing I notice is this is very, very complicated, because I see to a very complicated exponent. So, I just take my time, take it in stages, and try to peel things away. First, I’m going to peel off that exponent thing. So, what I’m going to do is I’m going to say, “Well, this is just e to some big blop.” And what’s the derivative of e to the blop? Well, it’s itself. It’s e to the blop.
E has the great feature that the derivative of e to the x is itself. The derivative of e to the blop is itself. So, the derivative of e to the blop is e to the blop. Now, what’s the blop? The blop was four x cubed plus three x minus one. But now, I’ve got to multiply that whole thing by the derivative of the blop. So, I just peeled away that exponent, and now I’m taking the derivative of the inside. I take the derivative of that, what do I see? I see twelve x squared plus three.
And so, the answer, if I were to write it on one line, would be k prime of x—the derivative—would equal twelve x squared plus three times e to the power four x cubed plus three x minus one. Sort of cramped all that in there. Sorry about that. But, this represents the derivative of this function. And you can see how I used the Chain Rule. Notice how the Chain Rule looks when you have an exponential. It’s e to the blop. The derivative is e to the blop times the derivative of the blop.
Okay. Well now, armed with all of this as a warm-up, we can now find the derivative of the natural log function, and all I’m going to use is this fact that we’ve already seen. Let me show you how we could actually execute that.
So, what I want to do is I want to find the derivative of that function. And all I know is that the e raised to the natural log of x equals x. Now, watch this really sneaky trick. This is a little bit of magic. I’m going to define a new function. I’m going to call it m for the magical function. Now, you’ve got to watch this. This is going to be sneaky. I’m going to define this to be e to the natural log of x. Okay? Now, that magical function, actually, looks pretty impressive until you remember this fact, and then you realize, that magical function is just a very complicated way of just saying x. Right. Remember. Just say x!
Okay. Now, what’s the derivative of the magical function? Well, watch this. Well, on the one hand, we know the magical function is x. And the derivative of x, we know, is one – not that big of a stretch. On the other hand, this is a very complicated way of writing x. We can take the derivative of this, though. Now, watch what happens when we take the derivative of this. I see e to the blop. And what’s the derivative of e to the blop? It’s e to the blop. I just peeled away that e, and now, I have to take the derivative of that.
Well, what is the derivative of that? Well, that’s what I’m trying to find out. That’s the derivative of the natural log, agreed? That’s what I’ve got to figure out. The derivative of this is the derivative of the natural log. That’s what I was trying to find. So that’s actually f prime of x. It’s the derivative of the natural log.
Let me say that again. This was the function m. It equals e to the natural log of x, which also is the same thing as x. So, if I take the derivative, on the one hand I know that the derivative of x is just one. On the other hand, since it also equals this, I can take the derivative of this using the Chain Rule. E to the blop—the derivative is e to the blop—multiplied now by the derivative of the blop. But the derivative of the natural log is what I’m trying to find. It’s the derivative of the natural log—f prime of x. Do you see it?
Okay. Well therefore, I’ve got this nice little fact. What is e to the natural log of x? We know it’s x. And so, what I see here is that one equals—and this is just a very fancy way of saying x—multiplied by, well, f prime of x. So, I can solve this. I want to find out what the derivative is. So, if I divide through by x, what do I discover? I discover that f prime of x equals one over x.
And so, what we’ve just established is the fact that the derivative of the natural log of x turns out to be one over x. Isn’t that a surprise? If you take the derivative of a natural log, it’s just one over x—just one over x. Derivative of a natural log is one over x.
Let’s try one quick example to indicate how you can actually combine this idea—this fact—with, say, the Chain Rule. So, here’s the last example I’ll take a look at here. Suppose the f of x equals the natural log of x cubed. How would you take the derivative of this? Well, you could use the Chain Rule. Let me show you how you use the Chain Rule.
I see natural log of blop. And so, what’s the derivative of the natural log of blop? Well, we’ve just discovered that to take the derivative of the log, you just take the reciprocal of whatever is in there. So, the natural log of blop equals one over the blop. That’s what we just discovered—take the derivative of the natural log, it’s just one over.
But, I’ve got to now multiply that—I just peeled off that natural log—I’ve got to multiply that by the derivative of x cubed, which we know is three x squared. And I can actually cancel, you’ll notice. I have x squared on top and x cubed on the bottom. I can cancel my x squared with two of my x’s down here, leaving me with just one x there on the bottom. And so, this gives me the answer of three divided by x. So, the derivative of this turns out to be three over x.
Now, if you were just a little bit sneaky, you could have actually solved this question in a different way without ever using the Chain Rule. And let me close with that. You could remember one of the properties that we’ve learned about logs. That’s why these logs properties are so handy. You might have recalled that if you have an exponent inside here, you can bring that out in front. If you do that, you take this problem that required a Chain Rule, and turned it into an even easier problem. Because that’s now just a constant. And so, how would you take the derivative of this? Well remember, when you just have a constant number in front, you just put that in a holding pattern, and you multiply that by the derivative of this. And the derivative of the natural log, we saw, was one over x. And so, we get three over x once again.
So, the point is if you just take this problem and try to do it using the Chain Rule, you’ll get the right answer after doing a little bit of extra work. If you use the basic properties of logarithms to simplify the function a little bit before taking the derivative, then the derivative actually turns out to be a fairly easier thing to find, and you get the same answer.
Anyway, what we’re doing next, though, is a number of examples to really illustrate the logarithm function, the sine function, the cosine function, the exponential function, the tangent function. The list goes on and on. But now, we’re finally empowered to find derivatives of much more elaborate functions and really the functions that capture the activity of the world around us and in nature.
And I’ve got some good news for you. This is basically the end of all the functions that we’re going to have to figure out the derivatives for. So, in some sense, we’re bringing closure right now, in that there will be no other functions that we’re going to have to figure out how to take the derivatives of. We now are empowered to be able to take the derivatives of all the functions we’re interested in.
So, in some sense, this is a milestone. But up next, we’ll try some examples, and then we’ll move on from there. Okay. I’ll see you in a bit.

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