 Welcome back to our lecture series Math 12-10, Calculus 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 24, we are going to be starting again a two-part lecture, which we'll roll over to lecture 25, in which case we want to talk about the idea of implicit differentiation. So before we jump into the deep end of the pool about what implicit differentiation even means, we have to first kind of talk about what does it mean for a function to be expressed explicitly or implicitly. And so let me give you some examples to try to illustrate this point. So in almost all of the examples and applications you've probably seen in this calculus class and in previous mathematics classes, your functional relationships are typically given in the format y equals f of x, where f of x is some algebraic or trigonometric expression of quantities like cosine, the square root, all of that business. So we have some algebraic expression involving x and then y all by itself is on the left-hand side. This is what we mean when we say that y is given explicitly in terms of x. The relationship, because that's what a function is after all, it's a relationship between x and y. So if you take, for example, y equals f of x here, but f of x is specifically x squared, then this means that like, oh, when x is one, then y will equal one. When x is two, y equals four. When x equals three, y equals nine. So we have this connection where, okay, x equals one, two, three, y equals one, four, nine. So there's a relationship between the quantities x and y. Functions are all about those relationships. Functions are specific relationships that given a specific y or, excuse me, a specific x coordinate, there's only one y attached to it. So there's a little bit of a uniqueness in the relationship that defines a function, okay? But when you come to the study of derivatives, we're talking about dy over dx, this is supposed to be the instantaneous rate of change. How is y changing as x is changing? Okay? That's what the derivatives measure. How does y change as x changes? And so if you have a relationship between two functions, like for example, excuse me, between two variables, that doesn't necessarily make a function, like you take this graph right here, this is a so-called elliptic curve graph. This graph does not pass the vertical line test, it is not a function relationship. But even still, if we pick a specific point x, y, we can still ask questions about like, what's the tangent line at this point x, y, for which the slope of that tangent line would still be dy over dx. So how is y changing as x changes at that instant, right? We're getting to that. So our functions, function relationships are these explicit relationships, y equals f of x. And so many examples we've seen of this would be like y equals 3x minus 2, y equals x squared plus x plus 6, y equals negative x q plus 2 over x to the fourth minus one. These are all examples of these explicit functions or perhaps a better term to use here. These would be explicit equations, right? Because these are all equations after all, right? y equals something, y equals something, y equals something. The relationship on between y and x is explicitly given in the equation because y equals, et cetera, right? Now, on the other hand, some equations of x and y don't have y explicitly solved. So for example, take this equation right here, 4xy minus 3x equals 6. This is an equation. It does have some x's and some y's. And therefore, not necessarily every choice of x and y will solve this equation. So there's a relationship between x and y. What pairs of x and y coordinates satisfy that equation? That's the relationship. But we don't have a y equals whatever. Y is somewhat concealed inside the equation. Another example, y squared plus 2yx plus 4x squared equals zero. It's an equation. Yep, it's a relationship. Every time you see an equation involving the variables x and y, that forms a relationship between x and y. A little more complicated one. You take y to the fifth plus 8y cubed plus 6y squared times x squared plus 2yx cubed plus 6 equals zero. Yep, that's an equation. It's a relationship. And these are all relationships for which the relationship between x and y is said to be implicit. Y is said to be given implicitly in terms of x. And so these implicit equations still develop relationships between x and y. And it's natural to ask, at this instant, how is y changing with respect to x is x changing? That question about the derivative is still relevant. But we can't compute it the way we've usually been doing, right? Well, OK, I should mention that with some of these equations, sure, the equation is an implicit relationship. But we could switch it into an explicit one, right? I mean, after all, if you take 4 times xy minus 3x equals y, or equals 6, right? It wouldn't be too difficult to solve for y right there. Maybe we add 3x to both sides. We end up with a 4xy equals 6 plus 3x. Divide both sides of the equation by 4x, right? Do that. We then end up with y is equal to 3x plus 6 all over 4x. That would be perfectly good. And, you know, we could maybe simplify it something more, whatever, but that's beside the point. We're able to solve for y. y equals this expression, right? y equals this. And so it turns out that y is a function of x, which we discover with that function relationship is right here. So even though we had an implicit equation to start with, we were able to solve for y to get an explicit equation, so y is explicitly defined in terms of x. On the other hand, consider the equation y squared plus 2yx plus 4x squared equals 0. This one is a quadratic equation with respect to the y. We see it right here. So we could try to solve the quadratic equation. Maybe we write factoring. That one seems a little bit difficult, not just because of the x's that are present. And so then we could try maybe using the quadratic formula or completing the square. The two forms are the equivalent to each other. But if you were to solve for y explicitly, y equals, you know, the negative b plus or minus the square root of, you know, et cetera, the quadratic formula. If we use that, we're going to get a negative 2x plus or minus the square root of, we're then going to get a 2x squared minus 4 times 1 times 4x squared. And this all sits above 2 times 1. And so the quadratic formula would tell us that y would solve in this manner right here, which we could simplify that a little bit more. Y equals negative x plus or minus the square root. We're going to get a 4x squared, which you can actually factor out from the group there. You can take out the 4x squared, that leaves behind a 1 minus 4, something like that, all over 2. Okay, we could keep on going. And of course, there's some concerns, of course, going on right here. When we take, you know, the square root to take square of 4, take square of x, that's all that nice and dandy, we're going to get the square root of this negative 3. So it's got like imaginary relationship, but this again, this is beside the point. The thing here, we could solve for y explicitly. Of course, though, this is not a function relationship. You'd be have this plus or minus going on right here, right? There's not a single x that's attached to y. You actually have to break up the problem into two pieces. It's kind of like this picture we were looking at a moment ago that I drew on the previous slide, this idea of this elliptic curve. It fails the vertical line test. What you could do is you could break it up in half. So you have this upper half, and then you have like this lower half. So if we segregate the graph, we could treat like this part is a function relationship. This other part's a function relationship. So if you're willing to treat us like two separate problems, you can solve for it explicitly. So even though we solved for it explicitly, it still wasn't a function relationship. But with that caveat, we can deal with it. But then look at this dreaded beast right here. y to the fifth plus 8yq plus 6y squared x squared plus 2yxq plus 6 equals 0. We don't have a quote unquote quadratic formula for this type of expression, and you would be hard-fetched if it's even possible to solve y equals whatever, even if it's not even a function relationship. So when you have these implicit equations, it does become necessary to consider, can I solve for y? And even if the answer there is yes, you have to ask yourselves, is it worth solving for y when you're studying these these relationships between x and y? And the short answer is oftentimes, if your goal is to find the derivative, no, it's not worth solving for y even if you can. But sometimes you can't. And so now we're ready to get to this idea of implicit differentiation. So even though the equations may or may not correspond to an explicit function relationship or some explicit equation, it doesn't matter. Because when we take an equation like this, y squared equals x cubed plus 17, this is one of these elliptic curves I keep on drawing, its graph would look something like the following if we were to put it on the xy axis. We could still ask ourselves, what's the derivative at the point x equals negative 2, y equals 3? So that might be like some point around here. We could ask about this tangent line right here. Well, the tangent line would still have the equation where you're going to get y minus, well the y coordinate which is 3, equals the slope, well the slope should still be y prime, right? The derivative times x plus 2. So if you want to find the equation with this tangent line, the only concern is how do you find the derivative? How do you find the slope of the tangent line which is this dy over dx? The slope is still going to be a rise over run, a small infinitesimal change of y with respect to a small infinitesimal change of x. And the strategy of doing that is going to be the same. What we're going to do is we're going to take the derivative with respect to x of both sides of the equation, are written in a slightly better manner. We're going to take the derivative of y squared with respect to x and we're going to take the derivative with respect to x of the right-hand side as well, x cubed plus 17. And so we take the derivatives. Now the right-hand side, it's not so crazy here, right? Because this derivative will look like the normal sense. We're going to take the derivative of x cubed. We're going to take the derivative of 17. 17 is a constant, so its derivative will just be 0. By the usual power rule, the derivative of x cubed with respect to x should be 3x squared. That's our right-hand side. But what do you do with this derivative of y squared? We take the derivative with respect to x. We know that the derivative of y squared with respect to y, you know, if we were to do that one, the derivative of y squared with respect to y, that would be a 2y. But we're not taking the derivative with respect to x. We're not taking the derivative with respect to y. We're taking the derivative with respect to x. So how does that change affect things? y is not constant with respect to x because the relationship between x and y is a dependent relationship. y does depend on what you choose for x. The two are not freely independent of each other, so we can't treat it like a constant against 0. What do we do? Well, it turns out the way we're going to save ourselves is to use the chain rule because the chain rule tells us that if you take the derivative of any expression u with respect to x, you can always change this into be like the derivative of u with respect to v times the derivative of v with respect to x. You can insert any new variable into the situation if you want to. Now in this situation, that's going to be extremely useful. We can take the derivative of y squared with respect to y. Ah, that we know how to do. And then we're going to take the derivative of y with respect to x, for which the derivative of y squared with respect to y, that's a 2y. And the derivative of y with respect to x, I'm just going to abbreviate that as a y prime here for short. This is equal to 3x squared. Notice if I divide both sides of the equation by 2y, I end up with y prime equals 3x squared over 2y. More specifically here, I've now discovered that dy over dx. The derivative of y with respect to x is equal to 3x squared over 2y. Now this is a little bit different than how we've seen things previously, right? Because now the derivative is expressed as a function of both x and y. This happens in this so-called technique of implicit differentiation. Can we calculate the derivative when we have this implicit equation, as we saw in this example, using the chain rule, we exactly can do that. So we have a formula for the derivative. What we need to compute, right, from our tangent line right here, we need to find the derivative at the point negative 2 comma 3. So we can evaluate the derivative dy over dx at the point, at the point negative 2 comma 3. So that would mean x is a negative 2. That would mean the y-coordinate is 3, like so. So simplifying that, you see that there's going to be a 2 that cancels on top and bottom. So it just gets rid of one of the 2s. The 3s will cancel. The negative sign is still going to be squared. So in the end, this turns out to be a positive 2. So this way, we have a little more careful here with our cancellation here. The 3s cancel out simple enough. The negative 2 squared, because of the negative sign, you can just cancel out the negative there, because, sorry, because you're squaring a negative will be positive. And so 2 goes into there once. So it turns out to be 2 when you're done with this thing here. The slope is 2. And so then we come back up here and we put that into our equation for the tangent line. The slope turned out to be a 2. So y minus 3 equals 2x plus 2, or 2x plus 4. If you distribute it, add 3 to both sides, you get y equals 2x plus 7. This is the equation of that tangent line that we sketched up right here. And we found this completely in this implicit manner. We don't have to solve for x. And so I should mention it's often useful to use this symbol dy over dx, as opposed to the prime notation, because we use the prime notation. If you write something like y prime, am I taking the derivative with respect to y or with respect to x? It matters a lot. So sometimes it's better to write this dy over dx, but it can get a little bit cumbersome. So we sometimes abbreviate. Just if you do that abbreviation, make sure it's clear from context which variable you're taking the derivative with respect to, because as you switch, you know, the derivative with respect to y versus the derivative with respect to x, that does give you a different outcome here.