 Welcome back to our lecture series Math 12-10, calculus one for students at Southern Utah University. As usual, be your professor today, Dr. Andrew Misledine. If you've been in our lecture series before, then you'll probably know that lecture 25 here is the second part of our two-part lecture about implicit differentiation. We first started that in lecture 24, like I mentioned a moment ago. We've learned that we can compute derivatives that is dy over dx without necessarily having an explicit relationship y equals f of x. This video, I want to do another example of such a calculation. That is, I want to use implicit differentiation to compute the equation of a tangent line of a curve that looks like the following that you see right here. This curve, which clearly is not a function, you'll notice that it violates in many locations the vertical line test. It's not the graph of a function. This is what's known as a fully of Descartes. That's some geometric jargon that you don't necessarily memorize. But the equation that determines this relationship is x cubed plus y cubed is equal to 3xy. You can throw in this extra parameter a into there, and you can fluctuate and get these little loopy loops. It's like a function for a roller coaster or what have you. But we want to find a tangent line, the equation of tangent line for this fully of that we see here on the screen. In particular, we want to find the tangent line at the point 2 comma 4, for which you can see that this is a legitimate point on the graph, right? If you take 2 cubed plus 4 cubed, you'll notice you get 8 plus 64, which is equal to, of course, that's going to be 72, sorry. Then on the other hand, you get over here 9 times 2 times 4, 4 times 2 is 8, 9 times 8 is likewise 72. This is a point on there. We want to find the tangent line at that place, for which the equation of the tangent line always has the same basic format, you're going to end up with y minus the y-coordinate. Give me some point on the line. In this case, we know that's going to be 4. This equals the slope, which we don't know at the moment, times x minus the point, the x-coordinate of that point, which is going to be here too. The question that we don't know because we know the x and y coordinates, we don't know the slope of the line, which the slope of the line is going to be the derivative. What we're looking for here is dy over dx at the point 2 comma 4. That's where calculus comes into play here. In order to compute the derivative, we're going to do this implicitly. We're going to take the equation x cubed plus y cubed is equal to 9 x y, and we're going to take the derivative of both sides. We're going to take the derivative with respect to x with both sides. When we're trying to find the slope of the tangent, it always needs to look like dy over dx. It needs to look like rise over run. This dy is a small change of y, so it's an infinitesimal rise, and then this dx right here, this is going to be a small change, an infinitesimal change of the x. It's a small, small run right there, but we still need to have dy over dx. We're going to apply the operator d over dx to both sides of the equation, because what's good for the goose is good for the gander. As long as we do the same thing to both sides, we're going to be just fine. Applying our derivative rules on the left-hand side, because we have a sum of functions inside, we're going to take their derivative separately. We're going to take x cubed derivative plus y cubed derivative, and notice this prime notation is just an abbreviation of this d over dx, so we don't have to write it all over and over again. On the right-hand side, because we have this coefficient of 9, we can factor it out, so we get 9 times x y prime. We now have to investigate how do these derivatives proceed from here? Well, this x cubed prime, we're looking for dx cubed over dx. We want the derivative of x cubed with respect to x by the usual power rule. That's going to be a 3x squared. No big deal there. So we get 3x squared. Now for the y cubed prime, notice this is where we need to emphasize where implicit differentiation is happening. Dy cubed, we're taking the derivative with respect to x. So this is the derivative of y cubed with respect to x, but because the relationship between x and y is implicit, we don't have a function formula that says y equals yada yada yada. We have to take the derivative using the chain rule. Instead, we're going to take the derivative dy cubed with respect to y, and then we times that by dy over dx. R in more compact form, the derivative of y cubed with respect to y, that is going to be 3y squared, just like we saw a moment ago, but the thing you don't want to forget is we need the inner derivative. We're going to get this y prime. Y prime, of course, is abbreviation for dy over dx. On the right-hand side, when we take the derivative of x times y, we're going to use the product rule. We take x prime y plus xy prime by the usual product rule. Now this can be simplified because like we mentioned before, what is x prime in this situation? It's dx over dx, which that's just a one. So we could simplify this expression, just putting a one in right there instead of the x prime. Let's do that. But what about the y prime? Well, y prime is an abbreviation for dy over dx. We don't know yet. That's actually what we're looking for, isn't it? So let's leave it alone. And so in fact, I'm going to kind of emphasize, right, that we have this y prime right here. So the y prime is what we want to solve for. So there's two things we could do at this moment. We could proceed to solve for the y prime, and I'm going to illustrate how to do that right now. So we want to solve for y prime. What I'm going to suggest is distribute this nine through so that we get a 3x squared plus a 3y squared y prime is equal to the nine y, because that's just a one right there, nine times one is nine. And then we have a nine xy prime. So what we want to do is combine all the y primes together. So move them to the same side of the equation. It doesn't matter where you go, just bring them together. However you want to do it, no big deal. So what we're going to do is we're going to move this one to the other side of the equation. That is we're going to subtract nine xy prime for both sides. And we're going to take the 3x squared and move it to the other side. Again, we're subtracting 3x squared from both sides of the equation. This then gives us a 3y squared y prime minus nine xy prime. This will equal nine y minus 3x squared. Now you'll notice that on the left hand side, everyone's divisible by y prime. You can factor it out. So we end up with this y prime times 3y squared minus nine x. This equals nine y minus 3x squared. We then can solve for y prime by dividing by its coefficient, so to speak, 3y squared minus nine. We do that to both sides, 3y squared minus nine x, like so. And so then in the end we see that dy dx, because that's what y prime even means, can be computed by the formula nine y minus 3x squared over 3y squared minus nine x. If we want to simplify the derivative, we can do that a little bit, because you'll notice the numerator, everything's divisible by three. We can factor out the three right there. I'll leave behind the three with the y and with the x square we're going to take out the three as well, so that leaves three times 3y minus x squared. We can do that same thing on the denominator. Take out the three, we'll factor it out here, factor it out here, and then that'll leave a y squared minus 3x, like so. The reason that could be of advantage to us is that the threes cancel on top of the bottom. That simplifies the derivative. So this gives us the derivative which we found implicitly. It'll be a function of both x and y. Now what we have to do is what we care about, remember what we care about is we need the derivative of y with respect to x when x equals two and y equals four. So that means plug these things in here. That means we're going to plug in a four right here, plug in a two right here, we plug a four in right here, we get a two right here. In which case we can then proceed to simplify this thing. What we can end up with a three times two is a 12 minus a four, minus a four is going to give us an eight in the numerator. In the denominator we have four squared which is 16 minus three times two, which is a six that gives us a 10. So we end up with eight tenths or four fifths that gives us the derivative. We can plug that into the equation of the tangent line just a second. What I want to show you slightly, a different approach to the same situation, turns out we did a lot more calculation than we needed. We didn't actually need the derivative, okay? Now I want you to think about that for a second. We're trying to find a tangent line. We need to know it's slope, right? Why don't we need to know the derivative? We don't need to know a formula for the derivative. What we need to know is actually a specific derivative. So if we actually come back up to this moment right here in our calculation, at this moment, all of the calculus was done. Going through this problem again, you'll notice that all of this business, this is all just algebra at this moment. And then there's all this other stuff that I erased. That's just arithmetic, okay? It turns out the algebra in this situation is not necessary because at this moment, once the calculus is complete, we actually could insert the values for X and Y, right? X is two, Y is three. If we do that, we end up with three times two squared plus three times four squared Y prime. We don't know what Y prime is, we're gonna figure that out. I mean, we do, it's four-fifths, don't tell anyone. We're gonna get nine times Y, Y is four, plus two times Y prime. So you'll notice if you plug this in, we have far fewer variables, we just have Y prime here. We can start combining and solving this linear equation. I think this will be a much easier chore for many of us. So we can say this one right here, two squared is four times three is 12, that 12 sounds familiar. You're gonna get four squared, which is 16 times three, that's a 48, Y prime. And the next one over here, you're gonna get nine times four, which is 36. And then you're also gonna get nine times two, which is 18 Y prime. So combining like terms, we're going to subtract 18 Y prime from both sides. And we're also gonna subtract 12 from both sides. What happens when we do that? Well, 48 take away 18 Y, that's going to give us a 30 Y prime. And then on the other side, 36 take away 12 gives us a 24, for which you can divide both sides by 30. We end up with Y prime equals 24 over 30, which if you cancel out the factor of two, you end up with eight tenths, which we already know is four fifths. So you get the exact same value, but I think probably in comparison, you'll find what we did here much simpler. So you can plug in the numbers X equals two, Y equals four, only after the calculus is complete. Once you've reached the algebraic stage, you can actually just plug the numbers and then solve it for there. So once we figure out the slope is one fifth, that means we come back up here, we erase this thing, and we record the, excuse me, the four fifths, not one fifth, the four fifths slope we found before, distribute the four fifths through, we're going to get four fifths X minus eight fifths. And so then if we add four to both sides, well, four is the same thing as 20 fifths. We're going to take 20 minus eight, which is 12, and this then gives us the equation of this tangent line to our Fulium of Descartes, Y equals four fifths X plus 12 over five. So I think there's two important takeaways from this example. One, if you're trying to find the tangent line of some implicit curve, that is it's not a function, don't worry about taking the derivative of Y equals F of X, just calculate the derivative implicitly. It's probably a lot easier. Implicit differentiation is our good friend here. And then the second thing I want you to take away is that if you're trying to find the slope of the tangent line, you don't necessarily need the prettiest, simplest, bestest formula for the derivative, you're looking for a specific number, the slope of a line. Once the calculus is complete, you can plug in the point of tangency for X and Y and then solve for Y prime. And that'll probably be a much easier task than our first attempt at this problem.