Let me share with you something I found particularly weird when I was a student first learning calculus. Let's say you have a circle with radius 5 centered at the origin of the xy plane. This is something defined with the equation x2 plus y2 equals 5 squared, that is, all the points on the circle are a distance 5 from the origin as encapsulated by the Pythagorean theorem, where the sum of the squares of the two legs on this triangle equals the square of the hypotenuse, 5 squared. And suppose you want to find the slope of a tangent line to the circle, maybe at the point xy equals 3,4. Now if you're savvy with geometry, you might already know that this tangent line is perpendicular to the radius touching it at that point. But let's say you don't already know that, or maybe you want a technique that generalizes to curves other than just circles. As with other problems about the slopes of tangent lines to curves, the key thought here is to zoom in close enough that the curve basically looks just like its own tangent line, and then ask about a tiny step along that curve. The y component of that little step is what you might call dy, and the x component is dx, so the slope we want is the rise over run, dy divided by dx. But unlike other tangent slope problems in calculus, this curve is not the graph of a function, so we can't just take a simple derivative, asking about the size of some tiny nudge to the output of a function caused by some tiny nudge to the input. x is not an input, and y is not an output, they're both just interdependent values related by some equation. This is what's called an implicit curve, it's just the set of all points x, y that satisfy some property written in terms of the two variables, x and y. The procedure for how you actually find dy, dx for curves like this is the thing I found very weird as a calculus student. You take the derivative of both sides like this, for x squared you write 2x times dx, and similarly y squared becomes 2y times dy, and then the derivative of that constant 5 squared on the right is just 0. Now you can see why this feels a little strange, right? What does it mean to take the derivative of an expression that has multiple variables in it, and why is it that we're tacking on dy and dx in this way? But if you just blindly move forward with what you get, you can rearrange this equation and find an expression for dy divided by dx, which in this case comes out to be negative x divided by y. So at the point with coordinates x, y equals 3, 4, that slope would be negative 3 divided by 4, evidently. This strange process is called implicit differentiation. Don't worry, I have an explanation for how you can interpret taking a derivative of an expression with two variables like this. But first I want to set aside this particular problem and show how it's connected to a different type of calculus problem, something called a related rates problem. Imagine a 5 meter long ladder held up against a wall where the top of the ladder starts 4 meters above the ground, which by the Pythagorean theorem means that the bottom is 3 meters away from the wall. And let's say it's slipping down in such a way that the top of the ladder is dropping at a rate of 1 meter per second. The question is, in that initial moment, what's the rate at which the bottom of the ladder is moving away from the wall? It's interesting, right? That distance from the bottom of the ladder to the wall is 100% determined by the distance from the top of the ladder to the floor. So we should have enough information to figure out how the rates of change for each of those values actually depend on each other, but it might not be entirely clear how exactly you relate those two. First things first, it's always nice to give names to the quantities that we care about, so let's label that distance from the top of the ladder to the ground y of t, written as a function of time because it's changing. Likewise, label the distance between the bottom of the ladder and the wall x of t. The key equation that relates these terms is the Pythagorean theorem, x of t squared plus y of t squared equals 5 squared. What makes that a powerful equation to use is that it's true at all points of time. One way that you could solve this would be to isolate x of t, and then figure out what y of t has to be based on that 1 m per second drop rate, and you could take the derivative of the resulting function dx dt, the rate at which x is changing with respect to time. That's fine, it involves a couple layers of using the chain rule, and it'll definitely work for you, but I want to show a different way that you can think about the same problem. This left hand side of the equation is a function of time, right? It just so happens to equal a constant, meaning the value evidently doesn't change while time passes, but it's still written as an expression dependent on time, which means we can manipulate it like any other function that has t as an input. In particular, we can take a derivative of this left hand side, which is a way of saying if I let a little bit of time pass, some small dt, which causes y to slightly decrease and x to slightly increase, how much does this expression change? On the one hand, we know that the derivative should be 0, since the expression is a constant, and constants don't care about your tiny nudges in time, they just remain unchanged. But on the other hand, what do you get when you compute this derivative? Well, the derivative of x of t squared is 2 times x of t times the derivative of x. That's the chain rule I talked about in the last video. 2x dx represents the size of a change to x squared caused by some change to x, and then we're dividing out by dt. Likewise, the rate at which y of t squared is changing is 2 times y of t times the derivative of y. Now evidently, this whole expression must be 0, and that's an equivalent way of saying that x squared plus y squared must not change while the ladder moves. At the very start, time t equals 0, the height, y of t, is 4 meters, and that distance x of t is 3 meters. And since the top of the ladder is dropping at a rate of 1 meter per second, that derivative, dy dt, is negative 1 meters per second. Now, this gives us enough information to isolate the derivative, dx dt, and when you work it out, it comes out to be 4 thirds meters per second. The reason I bring up this ladder problem is that I want you to compare it to the problem of finding the slope of a tangent line to the circle. In both cases, we had the equation x squared plus y squared equals 5 squared, and in both cases we ended up taking the derivative of each side of this expression. But for the ladder question, these expressions were functions of time, so taking the derivative has a clear meaning, it's the rate at which the expression changes as time changes. But what makes the circle situation strange is that rather than saying that a small amount of time dt has passed, which causes x and y to change, the derivative just has these tiny nudges dx and dy just floating free, not tied to some other common variable, like time. Let me show you a nice way to think about this. Let's give this expression x squared plus y squared a name, maybe s. s is essentially a function of two variables. It takes every point xy on the plane and associates it with a number. For points on the circle, that number happens to be 25. If you stepped off the circle away from the center, that value would be bigger. For other points xy closer to the origin, that value would be smaller. Now what it means to take a derivative of this expression, a derivative of s, is to consider a tiny change to both of these variables, some tiny change dx to x, and some tiny change dy to y, and not necessarily one that keeps you on the circle, by the way, it's just any tiny step in any direction of the xy plane. From there you ask how much does the value of s change? That difference, the difference in the value of s before the nudge and after the nudge, is what I'm writing as ds. For example, in this picture we're starting off at a point where x equals 3 and where y equals 4, and let's just say that the step I drew has dx at negative 0.02 and dy at negative 0.01. Then the decrease in s, the amount that x squared plus y squared changes over that step, would be about 2 times 3 times negative 0.02 plus 2 times 4 times negative 0.01. That's what this derivative expression, 2x dx plus 2y dy, actually means. It's a recipe for telling you how much the value x squared plus y squared changes as determined by the point xy where you start and the tiny step dx dy that you take. And As with all things derivative, this is only an approximation, but it's one that gets truer and truer for smaller and smaller choices of dx and dy. The key point here is that when you restrict yourself to steps along the circle, you're essentially saying you want to ensure that this value of s doesn't change. It starts at a value of 25 and you want to keep it at a value of 25. That is, ds should be 0. So setting the expression 2x dx plus 2y dy equal to 0 is the condition under which one of these tiny steps actually stays on the circle. Again, this is only an approximation. Speaking more precisely, that condition is what keeps you on the tangent line of the circle, not the circle itself. But for tiny enough steps, those are essentially the same thing. Of course, there's nothing special about the expression x squared plus y squared equals 5 squared. It's always nice to think through more examples, so let's consider this expression sin of x times y2 equals x. This corresponds to a whole bunch of u-shaped curves on the plane. And those curves, remember, represent all of the points xy where the value of sine of x times y squared happens to equal the value of x. Now imagine taking some tiny step with components dx and dy, and not necessarily one that keeps you on the curve. Taking the derivative of each side of this equation will tell us how much the value of that side changes during the step. On the left side, the product rule that we talked through last video tells us that this should be left d right plus right d left. That is sine of x times the change to y squared, which is 2y times dy, plus y squared times the change to sine of x, which is cosine of x times dx. The right side is simply x, so the size of a change to that value is exactly dx, right? Now setting these two sides equal to each other is a way of saying whatever your tiny step with coordinates dx and dy is, if it's going to keep us on the curve, the values of both the left hand side and the right hand side must change by the same amount. That's the only way this top equation can remain true. From there, depending on what problem you're trying to solve, you have something to work with algebraically, and maybe the most common goal is to try to figure out what dy divided by dx is. As a final example here, I want to show you how you can use this technique of implicit differentiation to figure out new derivative formulas. I've mentioned that the derivative of e to the x is itself, but what about the derivative of its inverse function, the natural log of x? Well the graph of the natural log of x can be thought of as an implicit curve. It's all of the points xy on the plane where y happens to equal ln of x. It just happens to be the case that the x's and y's of this equation aren't as intermingled as they were in our other examples. The slope of this graph, dy divided by dx, should be the derivative of ln of x, right? Well to find that, first rearrange this equation y equals ln of x to be e to the y equals x. This is exactly what the natural log of x means, it's saying e to the what equals x. Since we know the derivative of e to the y, we can take the derivative of both sides here, effectively asking how a tiny step with components dx and dy changes the value of each one of these sides. To ensure that a step stays on the curve, the change to the left side of the equation, which is e to the y times dy, must equal the change to the right side, which in this case is just dx. Rearranging, that means dy divided by dx, the slope of our graph, equals 1 divided by e to the y. When we're on the curve, e to the y is by definition the same thing as x, so evidently this slope is 1 divided by x. And of course, an expression for the slope of a graph of a function written in terms of x like this is the derivative of that function, so evidently the derivative of ln of x is 1 divided by x. By the way, all of this is a little sneak peek into multivariable calculus, where you consider functions that have multiple inputs and how they change as you tweak those multiple inputs. The key, as always, is to have a clear image in your head of what tiny nudges are at play, and how exactly they depend on each other. Next up, I'm going to be talking about limits, and how they're used to formalize the idea of a derivative. Thank you.
