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Calculus: Finding the Derivative Implicitly

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  • Type: Video Tutorial
  • Length: 12:15
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 133 MB
  • Posted: 06/26/2009

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: Implicit Differentiation (4 lessons, $9.90)
Calculus: Implicit Differentiation Basics (2 lessons, $4.95)

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 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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Implicit Differentiation
Implicit Differentiation Basics
Finding the Derivative Implicitly Page [1 of 3]
Right. Let's get going and take a look at how to take derivatives of things that aren't even functions. Let's recap here. First, a little vocabulary lesson and remind you that dy/dx--that's the derivative of y with respect to x. That is a noun. That is an object. That differs dramatically from d/dx, which is an action, which is a verb, which means take the derivative with respect to x of whatever I write after that. So, this is a command, whereas this is an object.
Okay, now armed with that notion, we were actually able to find the derivative of functions without ever using the fact that they were a function. Let me do a very simple example to remind you of what we've just discovered. So, for example, if I see that y equals x squared and I want to differentiate it, instead of just cavalierly saying, "Well, dy/dx equals two x," instead, what I can do is the following.
Put in some intermediate steps. First, say, "Okay, I want you to differentiate everything here with respect to x." Remember, this is a commandment. Differentiate with respect to x the entire thing. How do you do that? Well, you differentiate the left-hand side with respect to x. Whoops, I didn't want to do that. You know what I wanted to do? I wanted to have the y be in a purple font. And then, this would equal the derivative--take the derivative--of the right-hand side with respect to x. And that would be x squared.
So these are still two commandments. Now, let's execute. Well, what's the derivative with respect to x of y? Well, what's the derivative of y with respect to x? Well, that's something, and we have a name for it. What do we call the derivative of y with respect to x? We call that dy/dx. So, this thing produces dy/dx, and on this side, what's the derivative of x squared with respect to x? That's just two x. Not a big deal, and there's the answer.
So, in fact, there is a way of taking the derivative--I admit, it's so much longer than just looking at this and just immediately reporting the answer. However, notice that I never used the fact that this was just a function. In particular, this method--the method of putting the verb here, this action--actually works for anything, even things that aren't functions.
So, keeping these two little vocabulary things in mind, let me now introduce the last of our fancy methods for taking derivatives of exotic things. And this method is called implicit differentiation, because the idea is we can now take derivatives of things that just aren't y equals stuff that explicitly has x's in it. Now, the dependency on x and y can be interwoven, and y and x can depend on each other in sort of implicitly by a very complicated formula, which, in fact, might not be a function.
So, that's the idea of implicit differentiation. Let me show you, now, how you actually execute that by looking at some examples. Once you understand this basic idea, there's nothing to it. Let me illustrate with an example. So, this is now what we think of as implicit differentiation.
Let's consider the circle example. The equation of a circle is x squared plus y squared equals one. That's the circle that's centered right at the origin and has radius one. And you'll notice that, as we saw before, is not a function, because it fails the vertical line test. You can take a vertical line and put it right through there and see it pierces the curve at two different points. So, in fact, this is not a function.
Another way of seeing it's not a function is by looking at this and noticing that you just can't solve this for y. But you might say, "Oh, yeah. I can solve this for y. I'd bring the x over to this side and then square root both sides." You got to be careful, when you square root, you've got to take plus or minus the square root, and that corresponds to the plus wing and the minus wing of the circle.
Okay. So now, how do we proceed? The way we proceed is--well, what's the question? I want to find the slope of tangent lines on the circle, so the question is find dy/dx. The question remains the same. How do you find dy/dx? Now, how do we proceed?
The thing to do is to proceed just as we did before. Namely, let's just differentiate everything here with respect to x. So, what I'm going to now say is, "Okay. I'll just differentiate both sides with respect to x". So, I'll differentiate the left-hand side, which is x squared plus y squared with respect to x, and that will equal differentiating the right-hand side with respect to x.
By the way, one of these sides is very easy to differentiate. Can you guess which one it is? This side's not too hard. The derivative of a constant is zero. So, that's not going to be the big brainteaser here. So, in fact, why don't we just do that right now? The derivative of this one--the derivative of a constant--is zero. So, in fact this right-hand side equals zero, not a big deal.
Now, how do we differentiate the sum of two things? Well, we already know the rule. The rule is if you want to take the derivative, as this is telling us to do, take the derivative of a sum, you just take the derivative of each piece separately. All right, we combine them. So, in fact, I could write this--and I'm not going to write the colorful notation anymore--d/dx of x squared plus d/dx of y squared.
So, that's where I am now. I took this expression. I want to find the derivative of y with respect to x. So, I've differentiated both sides with respect to x. This side's actually pretty easy to do. It's just zero. And this side's the sum. So, when I take the derivative of a sum, I just get the sum of the derivative.
And now, I've got to differentiate each of these separately. Well, how would I differentiate this one? Well, we know how to do that. What's the derivative with respect to x of x squared? Well, we know what the answer is to that. That's just two x. So, this, we can actually do that command and report two x.
Now, I'm going to have to take the derivative of y squared with respect to x. Now, this is the place where something new has to happen, because I just can't say two y. You might be tempted to say, "Well, the derivative is two y." But that would be the derivative of this with respect to y. I'm asked to take the derivative of this with respect to x. There are no x's there. If there were x's there, no problem. I'm done. But there's no x's. Somehow y implicitly depends upon x, but I don't know how.
So, how do I do this? How do I do this? Well, let me show you how I think about that. I'm going to move this over here, then we'll bring it right back. Let's just concentrate on that piece. Let's just take a look at the derivative with respect to x of y squared. Now, let me show you how I write that. This is the key, which I think helps me a lot. I'm going to write this exactly as I see it, and it looks like this to me.
Take the derivative with respect to x of this thing. And you want to know what that thing is? It's this. Actually, it's the same thing as that, isn't it? I just notationally wrote it a different way. But, the way I wrote it captures the spirit of what's going on. Really, this is going to be a Chain Rule. There's an inside and an outside--something squared. So, we're actually going to use the Chain Rule here.
And let's use the Chain Rule, be really careful, and we'll see what the answer is. So, I'm going to take the derivative right now. First, I've got blop squared. What's the derivative of blop squared? The derivative is two blop. And what's the blop? The blop is y. I just peeled off the two, and now, what do I have to do? Now, I have to take the derivative with respect to x of y. What's the derivative of y with respect to x?
We have a name for that, and what's the name for the derivative of y with respect to x? It's dy/dx. So, I multiply this by dy/dx. And that is the derivative of this piece. So, it's actually a little Chain Rule. I look at this as blop squared, and I write that the derivative of that is two blop times the derivative of the blop with respect to x.
You see, I don't know exactly how y depends upon x, so I just write that in as dy/dx. If I now insert this piece, which I'll just momentarily write there, back to the original thing that we had--that original. Let me just show you where we are here. So, now that's no longer a question mark. I know what that is. The answer to that is it's two y dy/dx, and it still all does equal zero.
Then, if I write this out--let me just finish this problem up for you--I see that two x plus two y dy/dx equals zero. Well, the question originally was find dy/dx. Well, notice something that we've never seen before. Dy/dx is in front of us, but it's entangled. It's entangled inside this formula. But I can solve for it by bring this term over to this side, and then dividing by this co-efficient.
So, if I bring that term over, I see two y dy/dx equals minus two x. And if I, now, divide through by the two y on both sides, what do I see? Well, I see that the two y's now cancel, and on the left-hand side, I'm just left with dy/dx. So, I'm left with dy/dx equals--and here, I can cancel the two. So, let's see, minus x over y. So, it would be dy/dx--the derivative of y with respect to x--for the circle turns out to be minus x over y. Remember the original equation we were given was the circle centered at the origin of radius one. The question was find dy/dx, and we just did that. We just did that.
Notice that the derivative of this relation is not just stuff with x's in it. There's x's and y's in there. Not surprising when you consider that this thing is implicitly involved with x and y's. So, you can't separate the x and the y's If you could separate the x from the y, you would have some sort of function if you had y equal stuff, but you don't.
So, it shouldn't be a great shock to see something we've never seen before, namely, the slope of the tangent is given in terms of both x and y, because just knowing x doesn't tell me y.
Well, next up, what we're going to do is we're going to return to this example. And we're going to take a look at the significance and the meaning of this in terms of slopes of tangents of circles, make sure this answer actually makes sense, and then we'll take a look at a whole bunch of extra examples to really solidify this new notion of implicit differentiation, which you'll notice is just using the Chain Rule in a very creative way.
All right. Up next, we'll take a look at what this means, and then look at some more examples. Thanks, and I'll see you soon.

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