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