[00:10] Let me share with you something I found particularly
[00:13] weird when I was a student first learning calculus.
[00:16] Let's say you have a circle with radius 5 centered at the origin of the xy plane.
[00:22] This is something defined with the equation x2 plus y2 equals 5 squared,
[00:26] that is, all the points on the circle are a distance 5 from the origin as
[00:30] encapsulated by the Pythagorean theorem, where the sum of the squares of
[00:34] the two legs on this triangle equals the square of the hypotenuse, 5 squared.
[00:40] And suppose you want to find the slope of a tangent line to the circle,
[00:44] maybe at the point xy equals 3,4.
[00:48] Now if you're savvy with geometry, you might already know that this
[00:51] tangent line is perpendicular to the radius touching it at that point.
[00:56] But let's say you don't already know that, or maybe you want a
[00:59] technique that generalizes to curves other than just circles.
[01:03] As with other problems about the slopes of tangent lines to curves,
[01:07] the key thought here is to zoom in close enough that the curve basically
[01:11] looks just like its own tangent line, and then ask about a tiny step along that curve.
[01:17] The y component of that little step is what you might call dy,
[01:21] and the x component is dx, so the slope we want is the rise over run, dy divided by dx.
[01:28] But unlike other tangent slope problems in calculus,
[01:31] this curve is not the graph of a function, so we can't just take a simple derivative,
[01:36] asking about the size of some tiny nudge to the output of a function caused by some
[01:41] tiny nudge to the input.
[01:44] x is not an input, and y is not an output, they're both
[01:47] just interdependent values related by some equation.
[01:52] This is what's called an implicit curve, it's just the set of all points x,
[01:58] y that satisfy some property written in terms of the two variables, x and y.
[02:04] The procedure for how you actually find dy, dx for curves like
[02:08] this is the thing I found very weird as a calculus student.
[02:12] You take the derivative of both sides like this, for x squared you write 2x times dx,
[02:19] and similarly y squared becomes 2y times dy, and then the derivative
[02:24] of that constant 5 squared on the right is just 0.
[02:29] Now you can see why this feels a little strange, right?
[02:32] What does it mean to take the derivative of an expression that has multiple
[02:37] variables in it, and why is it that we're tacking on dy and dx in this way?
[02:42] But if you just blindly move forward with what you get,
[02:46] you can rearrange this equation and find an expression for dy divided by dx,
[02:51] which in this case comes out to be negative x divided by y.
[02:56] So at the point with coordinates x, y equals 3, 4,
[02:59] that slope would be negative 3 divided by 4, evidently.
[03:05] This strange process is called implicit differentiation.
[03:09] Don't worry, I have an explanation for how you can interpret
[03:12] taking a derivative of an expression with two variables like this.
[03:16] But first I want to set aside this particular problem and show how it's connected
[03:20] to a different type of calculus problem, something called a related rates problem.
[03:26] Imagine a 5 meter long ladder held up against a wall where the top
[03:30] of the ladder starts 4 meters above the ground,
[03:33] which by the Pythagorean theorem means that the bottom is 3 meters away from the wall.
[03:39] And let's say it's slipping down in such a way that the top
[03:42] of the ladder is dropping at a rate of 1 meter per second.
[03:46] The question is, in that initial moment, what's the rate at
[03:50] which the bottom of the ladder is moving away from the wall?
[03:55] It's interesting, right?
[03:56] That distance from the bottom of the ladder to the wall is 100%
[04:00] determined by the distance from the top of the ladder to the floor.
[04:05] So we should have enough information to figure out how the rates of
[04:08] change for each of those values actually depend on each other,
[04:12] but it might not be entirely clear how exactly you relate those two.
[04:16] First things first, it's always nice to give names to the quantities that we care about,
[04:21] so let's label that distance from the top of the ladder to the ground y of t,
[04:25] written as a function of time because it's changing.
[04:29] Likewise, label the distance between the bottom of the ladder and the wall x of t.
[04:34] The key equation that relates these terms is the Pythagorean theorem,
[04:39] x of t squared plus y of t squared equals 5 squared.
[04:43] What makes that a powerful equation to use is that it's true at all points of time.
[04:50] One way that you could solve this would be to isolate x of t,
[04:54] and then figure out what y of t has to be based on that 1 m per second drop rate,
[04:59] and you could take the derivative of the resulting function dx dt,
[05:03] the rate at which x is changing with respect to time.
[05:07] That's fine, it involves a couple layers of using the chain rule,
[05:10] and it'll definitely work for you, but I want to show a different way that you can think
[05:15] about the same problem.
[05:17] This left hand side of the equation is a function of time, right?
[05:21] It just so happens to equal a constant, meaning the value evidently doesn't change
[05:26] while time passes, but it's still written as an expression dependent on time,
[05:30] which means we can manipulate it like any other function that has t as an input.
[05:36] In particular, we can take a derivative of this left hand side,
[05:40] which is a way of saying if I let a little bit of time pass, some small dt,
[05:45] which causes y to slightly decrease and x to slightly increase,
[05:49] how much does this expression change?
[05:53] On the one hand, we know that the derivative should be 0,
[05:55] since the expression is a constant, and constants don't care about your tiny nudges in
[06:00] time, they just remain unchanged.
[06:03] But on the other hand, what do you get when you compute this derivative?
[06:08] Well, the derivative of x of t squared is 2 times x of t times the derivative of x.
[06:14] That's the chain rule that I talked about last video.
[06:17] 2x dx represents the size of a change to x squared caused by some change to x,
[06:23] and then we're dividing out by dt.
[06:27] Likewise, the rate at which y of t squared is
[06:30] changing is 2 times y of t times the derivative of y.
[06:35] Now evidently, this whole expression must be 0,
[06:38] and that's an equivalent way of saying that x squared plus y squared must not
[06:42] change while the ladder moves.
[06:45] At the very start, time t equals 0, the height, y of t,
[06:49] is 4 meters, and that distance x of t is 3 meters.
[06:54] And since the top of the ladder is dropping at a rate of 1 meter per second,
[06:59] that derivative, dy dt, is negative 1 meters per second.
[07:04] Now, this gives us enough information to isolate the derivative, dx dt,
[07:08] and when you work it out, it comes out to be 4 thirds meters per second.
[07:14] The reason I bring up this ladder problem is that I want you to compare
[07:17] it to the problem of finding the slope of a tangent line to the circle.
[07:22] In both cases, we had the equation x squared plus y squared equals 5 squared,
[07:26] and in both cases we ended up taking the derivative of each side of this expression.
[07:32] But for the ladder question, these expressions were functions of time,
[07:36] so taking the derivative has a clear meaning,
[07:38] it's the rate at which the expression changes as time changes.
[07:43] But what makes the circle situation strange is that rather than saying that a small
[07:48] amount of time dt has passed, which causes x and y to change,
[07:52] the derivative just has these tiny nudges dx and dy just floating free,
[07:56] not tied to some other common variable, like time.
[08:01] Let me show you a nice way to think about this.
[08:03] Let's give this expression x squared plus y squared a name, maybe s.
[08:08] s is essentially a function of two variables.
[08:11] It takes every point xy on the plane and associates it with a number.
[08:16] For points on the circle, that number happens to be 25.
[08:20] If you stepped off the circle away from the center, that value would be bigger.
[08:25] For other points xy closer to the origin, that value would be smaller.
[08:29] Now what it means to take a derivative of this expression, a derivative of s,
[08:34] is to consider a tiny change to both of these variables, some tiny change dx to x,
[08:40] and some tiny change dy to y, and not necessarily one that keeps you on the circle,
[08:45] by the way, it's just any tiny step in any direction of the xy plane.
[08:51] From there you ask how much does the value of s change?
[08:56] That difference, the difference in the value of s before the nudge and after the nudge,
[09:01] is what I'm writing as ds.
[09:04] For example, in this picture we're starting off at a point where
[09:09] x equals 3 and where y equals 4, and let's just say that the
[09:15] step I drew has dx at negative 0.02 and dy at negative 0.01.
[09:21] Then the decrease in s, the amount that x squared plus y squared changes over that step,
[09:28] would be about 2 times 3 times negative 0.02 plus 2 times 4 times negative 0.01.
[09:35] That's what this derivative expression, 2x dx plus 2y dy, actually means.
[09:41] It's a recipe for telling you how much the value x squared plus y squared changes as
[09:46] determined by the point xy where you start and the tiny step dx dy that you take. And
[09:53] As with all things derivative, this is only an approximation,
[09:56] but it's one that gets truer and truer for smaller and smaller choices of dx and dy.
[10:02] The key point here is that when you restrict yourself to steps along the circle,
[10:07] you're essentially saying you want to ensure that this value of s doesn't change.
[10:12] It starts at a value of 25 and you want to keep it at a value of 25.
[10:17] That is, ds should be 0.
[10:20] So setting the expression 2x dx plus 2y dy equal to 0 is the condition
[10:25] under which one of these tiny steps actually stays on the circle.
[10:30] Again, this is only an approximation.
[10:33] Speaking more precisely, that condition is what keeps you
[10:36] on the tangent line of the circle, not the circle itself.
[10:40] But for tiny enough steps, those are essentially the same thing.
[10:45] Of course, there's nothing special about the expression
[10:47] x squared plus y squared equals 5 squared.
[10:50] It's always nice to think through more examples,
[10:53] so let's consider this expression sin of x times y2 equals x.
[10:58] This corresponds to a whole bunch of u-shaped curves on the plane.
[11:02] And those curves, remember, represent all of the points xy where the
[11:06] value of sine of x times y squared happens to equal the value of x.
[11:16] Now imagine taking some tiny step with components dx and dy,
[11:19] and not necessarily one that keeps you on the curve.
[11:23] Taking the derivative of each side of this equation will tell
[11:27] us how much the value of that side changes during the step.
[11:32] On the left side, the product rule that we talked through last
[11:35] video tells us that this should be left d right plus right d left.
[11:39] That is sine of x times the change to y squared, which is 2y times dy,
[11:44] plus y squared times the change to sine of x, which is cosine of x times dx.
[11:52] The right side is simply x, so the size of a change to that value is exactly dx,
[11:56] right? Now
[11:59] setting these two sides equal to each other is a way of saying whatever your tiny
[12:04] step with coordinates dx and dy is, if it's going to keep us on the curve,
[12:09] the values of both the left hand side and the right hand side must change by the
[12:14] same amount.
[12:15] That's the only way this top equation can remain true.
[12:20] From there, depending on what problem you're trying to solve,
[12:23] you have something to work with algebraically,
[12:26] and maybe the most common goal is to try to figure out what dy divided by dx is.
[12:33] As a final example here, I want to show you how you can use this technique
[12:37] of implicit differentiation to figure out new derivative formulas.
[12:42] I've mentioned that the derivative of e to the x is itself,
[12:46] but what about the derivative of its inverse function, the natural log of x?
[12:50] Well the graph of the natural log of x can be thought of as an implicit curve.
[12:56] It's all of the points xy on the plane where y happens to equal ln of x.
[13:01] It just happens to be the case that the x's and y's of this
[13:04] equation aren't as intermingled as they were in our other examples.
[13:09] The slope of this graph, dy divided by dx, should be the derivative of ln of x, right?
[13:16] Well to find that, first rearrange this equation
[13:20] y equals ln of x to be e to the y equals x.
[13:24] This is exactly what the natural log of x means, it's saying e to the what equals x.
[13:31] Since we know the derivative of e to the y, we can take the derivative
[13:35] of both sides here, effectively asking how a tiny step with
[13:39] components dx and dy changes the value of each one of these sides.
[13:44] To ensure that a step stays on the curve, the change to the left side of the equation,
[13:50] which is e to the y times dy, must equal the change to the right side,
[13:54] which in this case is just dx.
[13:57] Rearranging, that means dy divided by dx, the slope of our graph,
[14:03] equals 1 divided by e to the y.
[14:06] When we're on the curve, e to the y is by definition the same thing as x,
[14:11] so evidently this slope is 1 divided by x.
[14:15] And of course, an expression for the slope of a graph of a function
[14:19] written in terms of x like this is the derivative of that function,
[14:24] so evidently the derivative of ln of x is 1 divided by x.
[14:32] By the way, all of this is a little sneak peek into multivariable calculus,
[14:37] where you consider functions that have multiple inputs and how
[14:40] they change as you tweak those multiple inputs.
[14:44] The key, as always, is to have a clear image in your head of what
[14:48] tiny nudges are at play, and how exactly they depend on each other.
[14:54] Next up, I'm going to be talking about limits,
[14:56] and how they're used to formalize the idea of a derivative.
[15:17] Thank you.