 This lesson is on the derivatives of implicit relations. Implicit relations sounds very strange, I know, but the idea has to do with the concept of relations. What is an implicit relation? Well, they are circles, parabolas, hyperbolas, ellipses, all of those relations that you worked with before. They can even be functions of some sort. But the important thing about them is that many times you cannot solve for y. We could also have relations in the form of x squared plus y squared is equal to 4, as you know, to be the circle. Or x plus xy squared plus y cubed is equal to cosine y. And as you can see, it would be very, very difficult to solve for y, maybe even impossible in this particular relation. Or we could have cosine xy plus y cubed is equal to secant x. And again, it would be just about impossible to solve for y in this case. So the first problem we're going to work with is our x squared plus y squared is equal to 4. And we're actually going to do that in two ways, because it can be done in two ways. We're first going to do it with the implicit relation idea. And then we're going to solve for y and show you how that matches the way of doing it with implicit relations. So we're going to take derivatives. In this first example, we're going to not solve for y and we're going to determine the derivative of x squared plus y squared is equal to 4. And we are going to determine the derivative dy dx. That's very important because we could also determine dx dy on these. So how do we do that? Well, we already know the derivative of x squared is 2x. And what we haven't talked about in the past when we were using chain rule is what is that final step? Well, the final step is to take the derivative of x with respect to x and it's dx dx. And normally we never saw this because it was equal to 1. So when we do our derivatives implicitly, we have to take this into account. What is that final derivative on the variable we are trying to take the derivative on? So that leads us to plus the derivative of y squared is 2y. And since the final derivative of y is dy dx, we write that in. And of course we go on and do the rest of this normally. The derivative of 4 is 0. Again, dx dx becomes 1 so we can just cross that out and leave it as a 1. Now we're going to solve for dy dx. So the way I normally do this, I put everything with the dy dx on one side. And everything that doesn't have dy dx goes on the other side, in this case the negative 2x. And then we finally divide in order to get dy dx by itself. And we get negative 2x over 2y, which simplifies to negative x over y. This is the way we take the derivative implicitly. Now let's go on and show how to do it with solving for y. If we solve for y, we have two cases. We're going to have y is equal to plus or minus the square root of 4 minus x squared. Now we have to take these derivatives separately. So we get the first dy dx to be, if you remember how to do this using chain rule, you'll have one half times the square root of 4 minus x squared, all in the denominator, times the derivative of 4 minus x squared, which is negative 2x. And that becomes, once we reduce it, negative x over the square root of 4 minus x squared. And you will see if you go back to the original, we had y is equal to the square root of 4 minus x squared. And of course we can change that to negative x over y, which was the same thing we got when we did it implicitly. Well let's try it with the negative. This time we're going to say dy dx is equal to negative one half square root of 4 minus x squared, times the derivative of 4 minus x squared, which again is negative 2x. Again we're going to reduce this and this time we're going to get x over the square root of 4 minus x squared. Now because we use y is equal to negative square root of 4 minus x squared, that means that this square root of 4 minus x squared is equal to x over negative y, or to simplify it, negative x over y. So now we have both of these coming back to our implicit form of negative x over y. And you can do them either way, get nice results, but many times solving for y again is impossible. And so we have to do our derivatives in an implicit way. Let's go on with another example. Determine dy dx of x plus x times y squared plus y cubed equals cosine y. We cannot solve for y in this. I would not try in a million years to do that. But we can determine the derivative. Now we can go along. This is normal, no problems there. But when we come to this term here, we see that we have to use product rule. So that's very important when we get here. And then this is pretty normal for the y and then the cosine y is not bad either. So let's try this. So the derivative of x we know to be 1 times dx dx, which we're not going to write in because as we become more sophisticated, we know that. Plus, now let's use that product rule. So we're going to have x times the derivative of y squared, which is 2y dy dx. Plus y squared times the derivative of x, which is 1, which we can write in or not. Y cubed becomes 3y squared dy dx. That equals the derivative of cosine y. Well, the derivative of cosine is negative sine y. And of course, then we go in and take the derivative on the y. We get dy dx. Now my rule of thumb is to put everything with the dy dx on the left-hand side. Everything else goes on the right-hand side. We see that this 2xy is a dy dx. We see the 3y squared is a dy dx. And when we bring the negative sine over, we get plus sine y. All of these are dy dx. And that equals everything that doesn't have a dy dx goes to the other side. So we have a negative 1 minus y squared. So now we can solve for dy dx. We can pull out that negative 1 plus y squared. And that's all over everything that was on the left-hand side, which is 2xy plus 3y squared plus sine y. As you can see, this was not that difficult to solve as long as we remembered to use product rule on our xy squared. Let's try another one. Again, we look at this problem. We see x squared, no problem there. We have e to the y. We're going to have to put that dy dx in. We have product rule here. And of course, the derivative of 5 will be 0. So let's try this one. So we have 2x, then dx dx, which is 1, plus e to the y. The derivative of e to the y is e to the y dy dx. So we put e to the y, then our dy dx. And then we have minus. And again, we have product rule. So I put the minus in front because we're going to have to distribute that. So we have x times the derivative of y, which we write as dy dx, plus y times the derivative of x, which we know to be 1. And that's going to equal 0. So let's distribute an x. So we'll have 2x plus e to the y dy dx minus x dy dx minus y equals 0. Again, put everything that has dy dx on one side and everything else goes on the other side, all the other terms. So we have e to the y minus x times dy dx equals negative 2x plus y. And the final step to solve for dy dx, and that's negative 2x plus y over e to the y minus x. And that is the final answer. If you are like most mathematicians, you do not like to see this written that way. You like to see it y minus 2x over e to the y minus x, because the negative in front can be a little bit confusing. Now this one has the product rule mixed in with a cosine function. So this becomes a little bit more difficult. But again, if you go through it slowly, don't skip steps. You will be able to get through this kind of a problem too. And again, we see that it is impossible to solve for y. One is embedded in the cosine and the other one is a y cubed. So the best way to do this problem is with implicit relations. So first we take the derivative of cosine, which we know to be negative sine of xy. Then we have to use our train rule and go in and take the derivative of x times y. So we know that is x dy dx plus y. And make sure you distribute this before you go on and do anything else. And that's what I'm going to do is I continue the problem. So the derivative of y cubed is 3y squared dy dx. And the derivative of secant x is secant x tan x. Now we have to distribute that negative sine y. So we have negative x sine of xy dy dx minus y sine xy plus 3y squared dy dx equals secant x tan x. Again, all the dy dx's go on the left-hand side. So we're going to have, if I put the first one positive, we're going to have the 3y squared. And then we're going to have minus x sine xy. And all of those would be dy dx. And that equals our secant x tan x, which is already on the right-hand side. And then plus y sine xy. And the last thing we do is divide by the 3y squared minus x sine xy. So we'll have dy dx is equal to secant x tan x plus y sine xy over 3y squared minus x sine xy. I think we have done enough of the basic implicit relations so that you get an idea of how to do these. So let's go on and do something else with these problems. The next thing we have to look at is how do you take second derivatives? We know first derivatives not too bad. We can get those. But how do we do a second derivative? Now if you remember when we did the problem before, we got dy dx on this problem to equal negative x over y. Now we're looking at the second derivative. And the second derivative is d squared y over dx squared. So the left-hand side is our normal second derivative. Now when we take the derivative on the right-hand side, we see we have a y in there. And we have to use quotient rule. So as usual, I'm going to leave that negative out and do my quotient rule. Square the bottom. Then we do that denominator times the derivative of the numerator, which is 1 minus the numerator times the derivative of the denominator and the derivative of y is dy dx. Well, we're not finished. We cannot leave this dy dx in that numerator. We have to substitute in. And of course we get that substitution from the previous step. So now we'll have negative y minus x times negative x over y all over y squared. Again, remember the negative is out front. All right, so let's clean this up. Negative times the negative becomes a positive. And we get this to be negative y plus x squared over y over y squared. All right, we don't like to have this kind of a fraction, so we have to clean this up some more. So we're going to get negative y squared plus x squared over y cubed. We're not quite finished. This equals something. Now if we think back, we know we have y squared plus x squared somewhere or x squared plus y squared somewhere. And it happens to be up here in the original problem. So we see that x squared plus y squared is equal to 4. So now we can substitute that 4 in and have negative 4 over y cubed as our final answer. And this is what you're supposed to look for anytime you do second derivatives on implicit relations. But other kind of problem can we do? Well, there are lots of application problems, but one type is determining tangent lines. So the problem this time reads, determine the equations of the tangent lines which are parallel and perpendicular to the x axis of the relation x squared plus y squared equals 4. Again, I'm using that same circle, so we know dy dx is equal to negative x over y. Now when we have a tangent line that's parallel to the x axis, we know that we want negative x to equal 0. So we say parallel to the x axis means we want that horizontal tangent line negative x to equal 0 or x is equal to 0. So when x is equal to 0, we see y is equal to plus or minus 2. And those are our tangent lines. Remember, this is a circle centered at 0, radius of 2. The horizontal tangent lines will be here and here. If we want it perpendicular, that means we will take the y portion of this. So that means y is equal to 0. And again, substituting in, we find x is equal to plus or minus 2. And those would be our perpendicular ones, which would be here and here. This concludes our lesson on implicit relations.