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About this Lesson
 Type: Video Tutorial
 Length: 12:13
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 131 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: Inverse and Hyperbolic Functions (14 lessons, $19.80)
Calculus: Inverse Functions & Logarithmic Diff (6 lessons, $10.89)
In this lesson, we will review the properties of inverse functions and learn about finding the derivative of an inverse function using a formula. For invertible functions, the inverse of f(x) is continuous if f(x) is continuous, is differentiable if f(x) is differentiable, is increasing if f(x) is monotonically increasing, and is decreasing if f(x) is monotonically decreasing. There is a technique that you can use to find the derivative of the inverse of a function without even having to find the inverse of its function. Once you find the derivative of a function's inverse using the provided formula, you will also be able to find the inverse function from the derivative you've calculated.
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
 Joined:
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
 derivative of the inverse function
 01/04/2013

the video never played.when i clicked the mouse to play the lesson nothing happended.After paying $2.97 for the lesson I received nothing in return.I'm very disapointed! jim.
 Helpful
 05/16/2012

This was well explained, but I would need at least one more example.
Elementary Functions and Their Inverses
Calculus of Inverse Functions
Derivatives of Inverse Functions Page [1 of 2]
Okay, so now let’s take a look at the calculus of inverse functions. What do we tend to look at with calculus, taking derivatives and so forth and so on? So if I give you a function, a basic question is how can you find the derivative of its inverse? So suppose you had a function and you know you can take the inverse of it, it’s invertible, how can you take the derivative of its inverse? Here are some basic facts about a function that has an inverse.
So suppose you have a function and it actually has an inverse. Well then, if the function is continuous, then the inverse will be continuous. If the function is very, very smooth, there’s no bumps in it, then, in fact, the inverse will be smooth. Because remember all we’re doing, when you think of the graph, is just flipping it over the y = x line. If a function is increasing, then its inverse will be increasing. If a function is decreasing, its inverse will be decreasing. So these are all properties that are sort of easy to see just be thinking about that flip along the diagonal.
But what about taking derivatives? Let’s think about this. What I really want to find out here, my mission, is what is the derivative? So given an invertible function, f(x), I want to find the derivative of its inverse. So I’ll write it this way: . Now, why am I writing it that way? Well, if you use the prime notation, how would you do that? So you’d have f1’(x). So, yes, you could do that. So if you like that, that’s good, but there’s so many little things dangling there. It looks like maybe f11, and I was a little bit naughty. So I’ll just use the d, dx to mean to take the derivative of the inverse function. If you like the prime, use the prime, but I won’t.
So how would you do that? So you’re given a function, you know it has an inverse and you want to find the derivative of its inverse. I don’t know. There’s only one thing I know, and that is the connection that links a function to its inverse, and that’s this: f f1(x) = x. That’s the only relationship I know that has f1 in it, for sure, because that’s the definition. That’s one of the properties that’s required to be an inverse. So what if I took this whole thing and differentiated it?
So let’s differentiate this whole thing with respect to x. So now, if I differentiate this with respect to x, what’s going to happen? I have to differentiate this side with respect to x, and then differentiate this side with respect to x. Well, this side with respect to x is pretty easy. The derivative of x = 1, so that’s not a problem. But what about this? This looks so complicated. Well, actually, it’s nothing more than a chain rule, because I have an inside, namely, that, and then I have an outside, namely, f of stuff. So what I’ve got to do is use a chain rule to take the derivative of this side, and then the derivative of this side is just going to be 1, taking the derivative with respect to x. So let’s see what we get.
If we try this, what I see is – well, what’s the derivative of f of junk? It’s f’ of junk, using the chain rule now, so the junk is f1(x). But I’m not done yet. I’ve got to multiply that by the derivative of the inside and that is the derivative of the inverse function. So that’s the derivative of the inside. Now this may look a little bit weird, so let’s just pause and make sure that we’re all okay on this. I’m taking the derivative of this thing with respect to x. But we have to realize that, in fact, this has an inside and outside. There’s all this junk inside here, all this blob, and then we have the outside, the f. So I've got an inside and then an outside, so I use the chain rule. The chain rule says take the derivative of something like this, I take the derivative of the outside with respect to the blob, and then multiply that by the derivative of the inside with respect to x. So if I’ve got f of junk, the derivative of that is f’ of junk and I just copy the stuff there. Multiply that by the derivative of the inside, the derivative of the junk, with respect to x. So that’s just the derivative of the inside. And what does that equal? Well, it equals the derivative of this, which is 1. And this is fantastic, because now I can actually solve for this. Remember I was asked to find this, that was my mission in life, and now I see it. It’s dangling right in there. All I’ve got to do, in fact, is divide by this. So if I divide by this, I can find out what the derivative of an inverse is. So what do I see? What I see is . And there’s the formula.
So how do you find the derivative of the inverse function? It’s just the reciprocal, one over, the derivative of the original function. This is f’, so the derivative of the good oldfashioned function would then plug in the inverse function. Just take the inverse function and plug it in. And that gives you the derivative of the inverse function.
Now, there’s something that you should be cautious of here. When is this thing actually undefined? It’s undefined whenever this thing is zero. So I have to make sure that this thing is never zero; otherwise, that derivative doesn’t make sense. So I have to make sure that f’ f1(x) ? 0, because I can’t divide by 0, but that’s clear. So as long as we make sure that the bottom is not zero, then this is the derivative.
So let’s actually try a little example so you can see this thing in action. So here’s a good little example. Let’s find the derivative of f1, and let’s find the derivative at this point: x = ?, where – and I’ll tell you a few things. First of all, I’ll tell you what f is. I’ve got to tell you something. But I’ll tell you what f is. f(x) = 2x + cos x. And I’ll even tell you something else. I’ll tell you what f1 is at ?. I’ll tell you it’s . So this is also given. So let’s make sure we understand the question. Here’s this function and what I’d like for you to do is I’d like for you to find the derivative not of the function, but I want you to find the derivative of the inverse, evaluated with x = ?. And I’ll help you out and tell you that the value of the inverse at ? is . So I’ll give you that much, but I want to now find the derivative, I want to find the slope of the tangent at that point.
Well now, how can we proceed? Well, first of all, there’s one thing we should check. Does this function have an inverse? Is this an invertible function? Well, to check that, we’ve got to take a derivative. So let’s do a little sidebar. So we take the derivative of the function – I see 2, and then what’s the derivative of cosine? It’s sine. And so I see that f’(x) = 2 – sin x. Now, what are the values for sin x? Well, sin x wiggles. It wiggles between1 and –1. So the biggest this thing could be is 1 and the smallest it could be is –1. If I make this thing as big as it can be, namely, 1, this is 2 – 1, which is 1. If I make this thing as small as it can be, then it’s –1 and I see 2 minus –1, which is 3. No matter what, this number is going to be between 1 and 3, namely, it’s always positive. This is an increasing function. If it’s an increasing function, we know it has an inverse. And, in fact, we could even do more. We could actually check to make sure that, in fact, this point makes sense. How could we check to see if this really is a true statement? Well, all we have to do is reverse the roles, because if f1(?) = , then f( ) = ?. They must undo each other. So what is f( )? If I plug in , this would be 2( ), that’s just ?, and I’d see cos ( ). And the cos ( ) = 0. So all I see is just ?, and that’s what I have here. So everything is okay.
Now, how could I actually now find the derivative? Well, I’ll just use this fact here. So if I use that fact, I could immediately report the news. I see that that the derivative of f1(x) = one over – and now I've got to take the derivative, which I already – oh, I don’t have it anymore, but I can do it for you live. Here we go, here’s the derivative. It’s 2 +, and now the derivative, cosine is –sine, but I don’t plug in x, I plug in f1. So that’s the derivative and I want to know the value at this point. So let’s plug in ?. And if I plug in ?, what do I see? Well, I see that at ?, this equals . But what is f1(?)? We know that’s . So we’re given that fact, so let’s put that in right now. So this equals , and what’s the sin ( )? You think back to the graph. That’s actually a high point, so that’s actually 1, so what I see is . So, in fact, this equals 1. So the derivative of the inverse function of this at ? turns out to equal 1. So the slope of the tangent line of the inverse function at ? turns out to be 1.
So, using this formula, you can actually find inverses of functions and the derivative of the inverse of the function. Okay, I’ll see you at the next lecture.
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the video never played.when i clicked the mouse to play the lesson nothing happended.After paying $2.97 for the lesson I received nothing in return.I'm very disapointed! jim.
This was well explained, but I would need at least one more example.