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About this Lesson
 Type: Video Tutorial
 Length: 22:25
 Media: Video/mp4
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 Download: MP4 (iPod compatible)
 Size: 242 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: Implicit Differentiation (4 lessons, $9.90)
Calculus: Applying Implicit Differentiation (2 lessons, $5.94)
Here, we will use implicit differentiation (in which you find the dy/dx of things that aren't simple functions). This will include looking at lines tangent to curves of relations (like a circle equation of x^2 + y^2 = 1). Professor Burger will show you how to find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. For instance, this is what you would use to differentiate a formula like x^2 + xy + y^2 = 3.75 or for x + x/y  y^3 = 4 + 3/8 to find dy/dx (the derivative of y with respect to x).
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
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Implicit Differentiation
Applying Implicit Differentiation
Using Implicit Differentiation Page [1 of 5]
All right, let’s take a look at that last example of ours, which was the very first example of using this notion of implicit differentiation—the ability to find dy/dx of things that aren’t even functions. The example that we worked through together was the example x squared plus y squared equals one, which is a circle centered at the origin of radius one.
Let me draw one for you right now. Again, I’m drawing circles on the fly—not bad, not bad. They’ll get progressively worse, I promise you. These are not—I’m very lucky here. I don’t want you think that I’m skilled. The radius is one. It’s centered at the origin. And we actually worked through the derivative of this. We found dy/dx and we saw that dy/dx equals negative x over y. That was the very last thing we saw in the previous discussion. And what I wanted us to do, as I said, was just to think about what is the significance of this, and to see if this answer makes sense.
Now, remember dy/dx represents – one interpretation of it is it represents the slopes of tangent lines. So let’s plot down some tangent lines and compute the slopes and see if this produces the right answer. So, let’s bring back our tangent line again, and let’s begin, maybe, way up here. Now, that point we know. That point is zero for x and one for y, because remember the circle has radius one. So, the line would be just like that.
The slope of that line, at least visually, looks like it should be zero. What happens when we plug into this formula? Namely, put in x zero, y one. Well, I see minus zero over one, which is zero. So, this is telling me that the derivative is zero there, and this shows me that the slope is, in fact, zero, and that’s consistent. That checks.
Similarly, if I plug in this point, which is x zero, y minus one, I see the same thing, because I still see zero over minus one, which is still zero. So, over here, the tangent line is zero. Great.
What about if I pick a point right in the center over here? Now, what is that point? Well, you can actually try to figure it out, but I’ll tell you what it is. It’s one over the square root of two, comma, one over the square root of two. It’s the point right in the center, and that point is one over the square root of two, comma, one over the square root of two.
You can check to see if it’s on the circle, by the way, by plugging in and seeing if you get one. One over the square root of two squared is a half. And then, another one over the square root of two squared is another half. And a half plus a half is one. Yes, that point is right in the circle, and since it’s equal, it’s right in the middle.
Well now, what is your guess as to what the slope is there? Don’t look at the formula. Just make a guess. Since it’s right in the center like that, it seems like it’s pitched downward, and the slope should be sort of just perfectly evenly coming down. So, the answer that you might come up with is negative one, right. The line that goes right through like that, slope is negative one. So, I just pulled it out to here.
So, let’s see what happens when I plug x, y into the derivative. Well, for x, I see one over the square root of two. For y, I see one over the square root of two, and what do you notice? Those things actually cancel, but I have that negative sign. And so, I see negative one, and that’s the slope.
Look how this works. This really captures it. What about this point down here? That point down there is right in the middle. That point would be – well, one over the square root of two, comma, and I’m negative – minus one over the square root of two.
What’s your guess as to the slope there? The slope there, since it’s right in the center, it looks like the slope should be just one, going right in the middle of it. So, what happens when you plug this in here? I see one over the square root of two divided by negative one over the square root of two. So those one over square roots of two cancel. But notice I have a negative down here, but then I also have a negative on top. Negative over negative makes a positive one. This has slope positive one. So we’re seeing this answer is actually consistent with the drawing and what we know the slope should be.
In fact, let’s try this one last example. What about the slope right here? This is the point one, comma, zero. Well, what’s the slope of this? Well, that line actually looks vertical. And what’s the slope of a vertical line? Well, you may remember, the slope of a vertical line is undefined, because there’s no change in x, so we’d be dividing by zero.
What happens if we plug in that point—which is one, zero—into here? Well, I get one and a zero in the denominator. That’s undefined. So, this really captures the spirit of slopes of tangent lines on the circle. And in fact, it is the derivative. So, this is really the right answer, and we got to it by implicit differentiation.
Okay, so that just gives you a sense of what all this means. Now, lets take a look at a whole bunch of examples where we use implicit differentiation to find dy/dx. So, the next example I want to look at is the following. Let’s consider the relation—it’s not a function, but it’s a relation—x squared plus xy plus y squared equals three point seven five.
Okay and the question is to find—our mission is to find dy/dx. The first thing I notice is this is not a function. I can’t just solve this for y in terms of x. So, I’ve got to differentiate the whole thing—equal sign and all—using implicit differentiation. And the way I do that is by differentiating the whole thing.
So, I’d say dy—I’m sorry—I’d say d—dx—I differentiate the whole thing with respect to x, or I take of the derivative of x squared plus xy plus y squared equals three point seven five. Now, we differentiate the entire expression.
Okay, now how do I do that? Well, the way I do that is I differentiate the lefthand side, and then I differentiate the righthand side. So, I differentiate the lefthand side, I see—well—I have to differentiate, with respect to x, x squared plus xy plus y squared. And that will equal what I get when I take the derivative of the righthand side. So, I have to differentiate with respect to x three point seven five.
I’ll make a deal with you. How about if I do the righthand side and you do the lefthand side? And the crew says no. Okay, well I’ll do both then. This side is actually pretty easy, taking the derivative with respect to x of this constant is zero. So, again, not a very hard one.
This is a little bit more involved. Let’s think about that. Well, how do you differentiate a sum? You differentiate a sum by taking the derivative of each piece. So, I’ll differentiate, but first I’ll differentiate with respect to x that first piece. And then I’ll add to it what I get when I differentiate with respect to x that second piece. And then I’ll add to that what I get when I differentiate with respect to x of that third piece. And that will equal the derivative of this side, which, again, I remind you is zero.
Okay. So now, what I do here? Well, now I’m going to differentiate each of these things. Now, I’m taking the derivative with respect to x of x squared. Well, that’s something that we know how to do, because I’m taking the derivative of something with x’s in terms of x. So that’s just good oldfashioned derivative. The answer there is two x. Okay.
But here, these have y’s in them. So, this is going to require a little bit more work. So, let’s see, this one is just going to be two x. Matter of fact, let me just continue over here, and so, what I see here is two x. I took that derivative, and then I have these other pieces. So, I have to take the derivative with respect to x of xy, and add that to the derivative with respect to x of y squared, and that equals zero. That’s where we are right now. I took the derivative of the x squared with respect to x. That was pretty easy.
Now, the derivative of this product—all I have to do here is take the derivative of this thing and take the derivative of that thing, right? I’d break it up like that. Ehhhhh, no. No, no, no, no. Derivatives of products are not the products of the derivatives. If you were fooled by that, that’s great, but now you’ve got to remember, there’s a lot of techniques that you have at your disposal, and we have to be able to recall them.
The derivative of a product is not the product of the derivative. We have to use the Product Rule. So, to break this up, we’ve got to use the Product Rule. So, let’s try that right now. I’m just going to keep writing all the other pieces that we’ve already seen, so this’ll be taken care of.
Now, the Product Rule says what? The first times the derivative of the second plus the second times the derivative of the first. So, now I’m going to have the first times the derivative of the second. And I’m going to write that in as d/dx of ythat’s the derivative of the second—plus the second times the derivative of the first. And, in fact, what is the derivative of that? That’s actually something we can do. The derivative of x with respect to x is just one.
So, I’ll write in here an invisible—or visible—one. That’s the derivative of x. Remember, anything with x in it, if I’m taking the derivative of it with respect to x is easy. We just differentiate it like we’ve learned. Okay, and that’s the Product Rule, first times the derivative of the second plus second times derivative of the first. Plus—and then I’m just going to copy this down for bookkeeping purposes—this is to take the derivative with respect to x of y squared. But I’m going to write this as I think about it, like that, and that still equals zero.
Okay, now, let’s go back and polish up, and finish all the things that we have remaining here. So, here, I still see that two x plus—and I have an x. And now, I’m told to take the derivative of y with respect to x. And what is the derivative of y with respect to x? Well, actually, there’s a name for that. It’s called dy/dx.
Remember, that dy/dx is actually the derivative of y with respect to x. The derivative of y with respect to x is just that. And here we get a plus y. And now, what do I do here? Well, now we have to use the Chain Rule, because I’m not quite sure the dependency on y with x. It’s sort of complicated. And so, what I need to do here is actually take out my play putty—which is sort of stuck in the can here.
Anyway, what I see here is blop squared. So, I take the derivative of blop squared. And what is that? Well, that’s two times the blop. And what’s the blop? The blop is y. But I’m not done, all I just did was peel off the two. Now, I’ve got to differentiate y with respect to x. Well, what is the derivative of y with respect of x? There’s a name for that. It’s dy/dx. So, this is dy/dx, and that all equals zero.
Well, I’m trying to find dy/dx. That was the question that we were asked, find dy/dx. And you’ll notice that dy/dx actually fits inside of here, in fact, in a whole bunch of places. So, we have to treat that as an unknown, but everything else is known, and we have to solve for it. So, what I do is, I look at each term, and those terms that have no dy/dx in it, I subtract it over to this side. And so that all I’m left with are terms that have dy/dx, and then I factor it out.
So, let’s do that. So, let’s keep all the dy/dx terms over on the left here. So, I’d see x dy/dx plus two y dy/dx. And that equals what? Well, I’m going to bring this term over to this side, and I see a minus two x. And then this term is sitting right here. This term has no dy/dx, so I bring it to that side as a minus y, and this term remains here. So, all terms have been accounted for.
Now, you’ll notice a common factor of dy/dx here. So, let’s factor that out. We’re trying to solve for it. If we factor that out, what do I see? I see dy/dx, that thing which I seek, and if I factor it out, I’m left with an x here plus two y here. And that still equals this side, which is minus two x minus y. And now, if I want to solve for dy/dx, there’s only one game in town, I’ll divide both sides by this thing here, and then it will be brought over to this side.
So when I divide both sides by this quantity, what do I see? Well, what I see is dy/dx equals minus two x minus y, all divided by x plus two y. And that’s the derivative. That is a machine that will generate the slopes of the tangent lines for this very, very complicated—this very, very complicated function, which I remind you of down here. That was the question. Very complicated function.
And dy/dx, we see it equals that. Notice the occurrence of both x and y in the answer, which shouldn’t be too shocking, because remember, this is a relationship. Just knowing x doesn’t tell me y. I need to know both x and y to completely identify a point on this very complicated curve, and then I can find the slope of a tangent for you.
And the process was—I want to just recap—the process was to differentiate everything with respect to x, break it up into these little pieces, and then here, I had to use the Product Rule. Here I had to use a little Chain Rule, and then just solve for dy/dx.
Okay, let’s try another one together. In fact, maybe I’ll let you try this one on your own first, and then I’ll do it to give you a chance to sort of see how all this would work. So, here we go. How about this problem? Here’s the relationship x plus x over y minus y cubed equals four and threeeighths. And what I’d like for you to try to do is find dy/dx. Give it a shot.
Any luck? Let’s do it together and see how you made out. First thing I noticed is that this, in fact, is a complicated relationship that links x and y, and it’s going to be pretty difficult, if not impossible, to solve this for x explicitly. So, what we should do, instead, is use implicit differentiation. So, what I’ll do is, I’ll just differentiate everything.
The lefthand side, x plus x over y minus y cubed, with respect to x. And I’ll differentiate with respect to x the righthand side—four and threeeighths. Again, the righthand side is not going to be too mentally challenging. This is still a constant, and the derivative of it will be zero—not a big deal.
This side will require us to do a little bit of thinking. So, in fact, what I’m going to do is I’m going to differentiate each of these terms, since I’m adding and subtracting them. So, I differentiate with respect to x the x, and then I add to that the derivative with respect to x, x over y, and then I subtract the derivative with respect to x of y cubed. And that all equals the derivative of a constant, which is zero.
Okay, well let’s go through and take each of these derivatives separately. The derivative of x with respect to x is one—not a problem. Now, here I have both x’s and y’s, so I have to be a little bit careful. And, in fact, what technique will I have to use? Well, this is a quotient of two things, so I have to use the Quotient Rule. Bottom times the derivative of the top minus the top times the derivative of the bottom all over the bottom squared.
Okay. So, let’s now compute that. So, what we have here is plus the bottom, which is y, times the derivative of the top, and the derivative of x with respect to x is just one. So, I’ll just put an invisible one there; watch me not write it in—minus—so, now where am I? I might have lost my place. So, it’s the bottom times the derivative of the top minus the top times the derivative of the bottom.
And what’s the derivative of the bottom? Well, it’s the derivative of y with respect to x. And what is the derivative of y with respect to x? Well, we have a name for that. It’s dy/dx. And where am I in the Quotient Rule? Let’s recap the chant. Bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. So, it’s all over the bottom squared.
So, that whole piece right there represents the derivative with respect to x of that. Notice, I had to use the Quotient Rule, and within the Quotient Rule, I got a little dy/dx, and I took the derivative of y with respect to x.
Now, I’m not done yet. I’ve got to deal with this term—so, minus—and then what do I do here? Well now, I’m not going to rewrite this, but I’m going to remind you, I think about this as a Chain Rule. I think of this as all this mysterious stuff cubed. So, what’s the derivative of blop cubed? The derivative of blop cubed is three blop squared. And I multiply that. I just peeled off that three, you see, and now, I’m going to multiply that by the derivative of y—the inside—the derivative of the inside with respect to x. That’s the derivative of y with respect to x, which is dy/dx. And this is dy/dx. And all of that is still equal to zero.
And now, once again, I am in a position to actually solve this for dy/dx, because dy/dx is sitting inside there. I push everything else over to one side. I’m going to be all set. How would I do this, by the way? There are a variety of ways of solving this now. The scary part, of course, is this area over here because I’ve got stuff that doesn't have dy/dx in it and stuff that does, and it’s all being divided by y squared.
How would you handle this? Let me tell you the two most standard ways of doing it. One way is to break this up into two fractions—y divided by y squared minus x dy/dx over y squared. And you’re just breaking those up into two fractions, both with the same bottom.
Another possibility would just to be multiply everything through by y squared and thus that would cancel the y squared here, and the y squared would just appear everywhere else. I’ll actually elect to do that method. That is I’ll just now multiply everything through by a y squared, both sides.
Y squared time zero remains zero, so that won’t change. But now, when I multiply through by a y squared, I’m going to get a y squared here. Here that y squared with the y squared down here cancel—which was the whole point of multiplying—and here, I get an extra y squared. So I’ve got y squared times y squared. That gives me a y to the fourth.
Okay. So, if I now record all that after I do this, what I would see is—let me keep this here so I can actually see what this is going to be. This is going to be—now, write it over here—y squared plus y minus x dy/dx minus three y to the fourth dy/dx equals zero times y squared. But zero times y squared is just zero.
Well now, I’ve cleared off that bottom that was sort of scary, and now, I can see the terms that have the dy/dx in it, and I can see the terms that don’t. And so, I can bring these over to the other side and keep these here and factor out the common factor. I’m actually going to do that in one big step there. Let me try this, and then we’ll see how I got this.
So, I have dy/dx times minus x minus three y to the fourth, and that equals—and now, I’m going to bring these two terms, which don’t have dy/dx in it over to this side. And so, I see minus y squared minus—got this squared—minus y. So, let’s make sure you see how I did that. I took these two terms, which have no dy/dx in it, and I brought them over—subtracted them over—to here. And then, in what remained, I see a common factor of dy/dx, which is the whole point. Everything that remains should have a common factor of dy/dx. And so, when I factored out that common factor—notice when I factor out of here, I’m left with a minus x. When I factor out of here, I’m left with a minus three y to the fourth.
And now, I to solve for dy/dx, I just divide both sides by that quantity so it cancels, and then it goes down here. And so, I see that dy/dx—that which I seek—is minus y squared minus y all divided by minus x minus three y to the fourth.
That’s the answer. Or if you want to be fancy, you could factor out the negative one on the top and the negative one on the bottom, and all the negative signs actually go away. And so this is the same thing as y squared plus y divided by x plus three y to the fourth. In fact, these are same thing, but that’s actually sort of nicer—no negative signs.
And so, in fact, that is the derivative of y with respect to x of this original relationship, which was quite, quite complicated. It required us to use a Quotient Rule in here and the little Chain Rule here. And then, once we’ve got this, now notice there’s a little bit of algebra that’s involved. We have to now actually go up and solve these things for the dy/dx. Again, notice that both x and y appear here.
Okay, next up, what we’re going to do is look at a couple more examples of this to really drive this home. We’ll look at a couple of applications of this, and then, I will have a very special announcement. So a special announcement after our next discussion. See you soon.
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