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Calculus: Derivatives of Inverse Trig Functions


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:31
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 124 MB
  • Posted: 07/02/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Inverse and Hyperbolic Functions (14 lessons, $19.80)
Calculus: Calculus of Inverse Trig Functions (2 lessons, $2.97)

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 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 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.

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Elementary Functions and Their Inverses
Calculus of Inverse Trig Functions
Derivatives of Inverse Trigonometric Functions Page [1 of 3]
Okay, so now that we're becoming world's experts on these inverse trig functions, you see this question coming down the horizon like a train barreling down. What about the calculus of inverse trig functions? How do you take derivatives of these really exotic functions? The graphs are sort of weird, because they sort of start and stop at weird places. And so how can you actually find slopes of tangents, how can you actually find sort of what the derivative is? Well, what I want to do is show you a way of thinking about looking for the derivatives. And that way will actually work for every single inverse trig function. So I'm going to show you that way with one particular example, and then I want you on your own to try it, and rederive and figure out all the derivatives of all the inverse trig functions. And you'll see that, in fact, these things get really messy. In fact, really good people can make mistakes on finding these things. So it's good to know the method for figuring them out, so you can always check. You don't have to actually memorize it.
So while I want to do is take a look at the function y = arcsin x. And I want to show you a way of thinking about finding the derivative. So let's now think about finding the derivative for y = arcsin x. All right, so what I want here is I want to find, here's my goal, . So what do I do? Well, what I'm going to do is I'm going to say, "Look, let's be honest, let's just admit it up front. We don't like the arcs. The arcs are too confusing. Arcsin, who knows what that means? It's something that we're not very familiar with." So when you're faced with something that's really hard and a little scary, the thing is avoid it and instead try to convert it into something that you can tackle. And so what can I tackle? Well, I'll just think about the fact that arcsin x is the inverse function of sin x. Which means if I switch the roles of x and y, I can replace it with a sine. In particular, if you think about this says y = the angle whose sine is x. So what that means is sin y = x. So, in fact, this expression right here is identical to the following: x = sin y. Let's just see. y = arcsin x. And notice that x = sin y. That's the same statement. Now, why do I like this statement better, since they're identical? The reason why I like this statement better is there's no arcs. Let's get rid of the arcs. I don't want to have anymore arcing. Let's just look at this. Now, all I'm trying to do is find . So what I'll do is I'll take this while expression and differentiate it with respect to x.
Now, one thing that's sort of worth thinking about is - well, what does this look like? Can you visualize this somehow? And the answer is yes. Let's just pretend for now that y is an angle that's between 0 and 90 degrees, just for argument sake. Because if I do that, then I'm allowed to convert this into a little triangle. And so y is the angle, so let me call this y. And since it's my triangle, I can make it any length I want. I'll make the hypotenuse 1. Now, once I make the hypotenuse 1, all these other things are going to be fixed, because this angle is actually given to me to be y. Now, what is this length right here? Sin y = opposite/hypotenuse. So it's this length divided by 1. So sin y is actually just this length, because dividing by 1 is just doing nothing. So, in fact, sin y is this. But look, sin y, we already know, is x. So, in fact, this is just x. That's pretty neat. We'll able to take this sort of algebraic expression and convert it to a picture expression. And as you know, a picture is worth a thousand words. So this saves me a lot of talking. So there's the picture.
Now, what I'll do is differentiate. So I want to differentiate this whole expression now with respect to x. So how would that look? Well, if I differentiate with respect to x, I have to differentiate the left-hand side. That's not going to be too hard. And then I differentiate the right-hand side. That's going to be harder. So it looks like that. Now, the derivative of x with respect to x, that's just 1. So that's pretty easy. But what about this thing right here? Well, actually, it's sort of threatening because you've got some y's in there. What do you do with the y's? Well, the y's depend upon x in some very complicated way. And so what I'm going to do here is use the chain rule. I'm going to think about this as a big blop in there, and so I'm going to view this as taking the derivative of sin blop. So to use the chain rule, what do I do? I say, well the derivative of sin blop = cosin blop. And then I've got to multiply it by the derivative of the blop, the derivative of the inside. So let's think about this now. So the derivative of sin blop = cos blop, and the blop is just a y. And then I have to multiply all that by the derivative of the inside. So that's the derivative of y with respect to x. And what is the derivative of y with respect to x? Well, we actually have a name for that. That's called . So, in fact, that's the derivative. And this is fantastic news, because look, that thing which I'm trying to find now has appeared. This is terrific. In fact, now I can solve for by just dividing by cos y. So, in fact, if I divide by cos y, what do I see? I see =. So that's the answer. Well, you don't hear trumpets. Well, why? Because it's very unsatisfying to have a y in a derivative when you're taking a derivative with respect to x. You take a derivative of x^2. It's 2x. There's no y's there. So somehow having a y here is very unfulfilling. So even though this is the correct answer, it's not a very satisfying answer. Could I have just a derivative that just has x's in it? Well, that's asking the question can I write cos y just in terms of x? Well, how can I do that? Well, actually, I can do that if I return to this picture, because if I return to this picture, what I see is - well, what is cos y? cos y is adjacent/hypotenuse. So that's this length divided by 1. So, in fact, cos y is nothing more than that length. If I could find that length, that's cos y. So this thing here is exactly this. And see, it matches up perfectly. So all I've got to do is figure out what that is in terms of x. How can I do that? Well, this is a great application of the Pythagorean theorem, because, you see, I've got a right triangle. So if I've got a right triangle, then I've got a 1 here, and I've got an x here, and now I just don't know what this is.
So then what do you do? What you do is you say, "Okay, well, I know that ??^2 + x^2 = 1." And now I can solve for x. So what would it be to solve for x, I'm sorry for question mark. x I am already given. It's the question mark that's a question mark. So ?? = 1 - x^2. Now what about the square there? Well, I'll just take square roots of both sides - well, technically, I've got to take plus or minus the square root, but the question marks represent a length, and length is always positive, so I'll just take the positive square root. And that equals . And so that equals this length right here. So that length, in fact, now becomes clear to the Pythagorean theorem. It's . But remember that, in fact, that length is exactly cos y, because cos y is adjacent/hypotenuse. So it's that thing over 1. So, in fact, this thing right down here is really equal to . And so, in fact, we just found the derivative of arcsin. The derivative of arcsin turns out to be a very strange-looking thing, . And you say, "Gee, that's sort of a funny thing," because you'd think maybe the derivative of arcsin would have all sorts of other trig things in it, but it doesn't. And the reason is, I think, believable: first of all, when we reconvert that statement into here and write the triangle, now it's the Pythagorean theorem that kicks in and that's what gives us that square root stuff. And then we use a little bit of chain rule to actually untangle this expression and we figure out exactly what the derivative is. So, in fact, you can use this exact same procedure with all the other inverse trig functions. In fact, the only difference is where you put the x and stuff. And you should actually try that and verify for yourself that you can take any arc expression, turn it into an expression without arcs, then you can build a little teeny right triangle, where this angle will always be y. And then where the x gets placed depends on which trig function you use. But if you do that what you'll see is the following. Well, the first one we already saw. So the derivative of arcsin = . But what about the other ones? If you do this for arccos, you'll see that its derivative = . And I'm just going to start flashing these up here real fast. The derivative of arctan = . Now, where's the 1 + x^2 and how comes there's no square root? Well, because now what's going to happen is you're going to be solving for the hypotenuse. arccsc = . Notice there's a little difference between the arcsin and the arccsc. The arcsec = . Again, notice the difference in order. And the last one is arccot = .
So, in fact, you can see all these derivatives here, and there are an awful lot of them. You can see there's tons of them. And the thing to do is not to try to memorize them, because there are just so many and it's so easy to make a mistake. So you try to understand the method. If you're taking a test or doing something, you can always reconstruct it and know exactly whether it s 1 - x^2 or x^2 - 1, should there be a negative sign or a square root or whatever. You can always figure out. So think about the idea of finding it, and then you'll always have it.
All right, we'll take a look at some examples of how to actually take derivatives of more complicated-looking things involving these new formulas up next. I'll see you there.

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