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Implicit differentiation, what's going on here? | Chapter 6, Essence of calculus
3Blue1Brown
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May 12, 2026
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Transcript
0:10
Let me share with you something I found particularly
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weird when I was a student first learning calculus.
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Let's say you have a circle with radius 5 centered at the origin of the xy plane.
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This is something defined with the equation x2 plus y2 equals 5 squared,
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that is, all the points on the circle are a distance 5 from the origin as
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encapsulated by the Pythagorean theorem, where the sum of the squares of
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the two legs on this triangle equals the square of the hypotenuse, 5 squared.
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And suppose you want to find the slope of a tangent line to the circle,
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maybe at the point xy equals 3,4.
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Now if you're savvy with geometry, you might already know that this
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tangent line is perpendicular to the radius touching it at that point.
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But let's say you don't already know that, or maybe you want a
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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,
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the key thought here is to zoom in close enough that the curve basically
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looks just like its own tangent line, and then ask about a tiny step along that curve.
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The y component of that little step is what you might call dy,
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and the x component is dx, so the slope we want is the rise over run, dy divided by dx.
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But unlike other tangent slope problems in calculus,
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this curve is not the graph of a function, so we can't just take a simple derivative,
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asking about the size of some tiny nudge to the output of a function caused by some
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tiny nudge to the input.
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x is not an input, and y is not an output, they're both
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just interdependent values related by some equation.
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This is what's called an implicit curve, it's just the set of all points x,
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y that satisfy some property written in terms of the two variables, x and y.
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The procedure for how you actually find dy, dx for curves like
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this is the thing I found very weird as a calculus student.
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You take the derivative of both sides like this, for x squared you write 2x times dx,
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and similarly y squared becomes 2y times dy, and then the derivative
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of that constant 5 squared on the right is just 0.
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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
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variables in it, and why is it that we're tacking on dy and dx in this way?
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But if you just blindly move forward with what you get,
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you can rearrange this equation and find an expression for dy divided by dx,
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which in this case comes out to be negative x divided by y.
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So at the point with coordinates x, y equals 3, 4,
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that slope would be negative 3 divided by 4, evidently.
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This strange process is called implicit differentiation.
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Don't worry, I have an explanation for how you can interpret
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taking a derivative of an expression with two variables like this.
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But first I want to set aside this particular problem and show how it's connected
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to a different type of calculus problem, something called a related rates problem.
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Imagine a 5 meter long ladder held up against a wall where the top
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of the ladder starts 4 meters above the ground,
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which by the Pythagorean theorem means that the bottom is 3 meters away from the wall.
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And let's say it's slipping down in such a way that the top
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of the ladder is dropping at a rate of 1 meter per second.
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The question is, in that initial moment, what's the rate at
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which the bottom of the ladder is moving away from the wall?
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It's interesting, right?
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That distance from the bottom of the ladder to the wall is 100%
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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
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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, it's always nice to give names to the quantities that we care about,
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so let's label that distance from the top of the ladder to the ground y of t,
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written as a function of time because it's changing.
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Likewise, label the distance between the bottom of the ladder and the wall x of t.
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The key equation that relates these terms is the Pythagorean theorem,
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x of t squared plus y of t squared equals 5 squared.
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What makes that a powerful equation to use is that it's true at all points of time.
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One way that you could solve this would be to isolate x of t,
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and then figure out what y of t has to be based on that 1 m per second drop rate,
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and you could take the derivative of the resulting function dx dt,
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the rate at which x is changing with respect to time.
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That's fine, it involves a couple layers of using the chain rule,
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and it'll definitely work for you, but I want to show a different way that you can think
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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 doesn't change
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while time passes, but it's still written as an expression dependent on time,
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which means we can manipulate it like any other function that has t 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 if I let a little bit of time pass, some small dt,
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which causes y to slightly decrease and x to slightly increase,
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how much does this expression change?
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On the one hand, we know that the derivative should be 0,
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since the expression is a constant, and constants don't care about your tiny nudges in
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time, 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 x of t squared is 2 times x of t times the derivative of x.
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That's the chain rule that I talked about last video.
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2x dx represents the size of a change to x squared caused by some change to x,
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and then we're dividing out by dt.
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Likewise, the rate at which y of t squared is
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changing is 2 times y of t times the derivative of y.
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Now evidently, this whole expression must be 0,
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and that's an equivalent way of saying that x squared plus y squared must not
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change while the ladder moves.
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At the very start, time t equals 0, the height, y of t,
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is 4 meters, and that distance x of t is 3 meters.
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And since the top of the ladder is dropping at a rate of 1 meter per second,
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that derivative, dy dt, is negative 1 meters per second.
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Now, this gives us enough information to isolate the derivative, dx dt,
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and when you work it out, it comes out to be 4 thirds meters per second.
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The reason I bring up this ladder problem is that I want you to compare
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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 x squared plus y squared equals 5 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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it's 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 a small
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amount of time dt has passed, which causes x and y to change,
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the derivative just has these tiny nudges dx and dy just floating free,
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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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Let's give this expression x squared plus y squared a name, maybe s.
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s is essentially a function of two variables.
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It takes every point xy on the plane and associates it with a number.
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For points on the circle, that number happens to be 25.
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If you stepped off the circle away from the center, that value would be bigger.
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For other points xy 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 s,
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is to consider a tiny change to both of these variables, some tiny change dx to x,
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and some tiny change dy to y, and not necessarily one that keeps you on the circle,
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by the way, it's just any tiny step in any direction of the xy plane.
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From there you ask how much does the value of s change?
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That difference, the difference in the value of s before the nudge and after the nudge,
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is what I'm writing as ds.
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For example, in this picture we're starting off at a point where
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x equals 3 and where y equals 4, and let's just say that the
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step I drew has dx at negative 0.02 and dy at negative 0.01.
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Then the decrease in s, the amount that x squared plus y squared changes over that step,
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would be about 2 times 3 times negative 0.02 plus 2 times 4 times negative 0.01.
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That's what this derivative expression, 2x dx plus 2y dy, actually means.
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It's a recipe for telling you how much the value x squared plus y squared changes as
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determined by the point xy where you start and the tiny step dx dy that you take. And
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As with all things derivative, this is only an approximation,
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but it's one that gets truer and truer for smaller and smaller choices of dx and dy.
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The key point here is that when you restrict yourself to steps along the circle,
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you're essentially saying you want to ensure that this value of s doesn't change.
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It starts at a value of 25 and you want to keep it at a value of 25.
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That is, ds should be 0.
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So setting the expression 2x dx plus 2y dy equal to 0 is the condition
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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
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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, there's nothing special about the expression
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x squared plus y squared equals 5 squared.
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It's always nice to think through more examples,
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so let's consider this expression sin of x times y2 equals x.
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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 xy where the
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value of sine of x times y squared happens to equal the value of x.
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Now imagine taking some tiny step with components dx and dy,
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and not necessarily one that keeps you on the curve.
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Taking the derivative of each side of this equation will tell
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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
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video tells us that this should be left d right plus right d left.
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That is sine of x times the change to y squared, which is 2y times dy,
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plus y squared times the change to sine of x, which is cosine of x times dx.
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The right side is simply x, so the size of a change to that value is exactly dx,
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right? Now
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setting these two sides equal to each other is a way of saying whatever your tiny
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step with coordinates dx and dy is, if it's going to keep us on the curve,
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the values of both the left hand side and the right hand side must change by the
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same amount.
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That's the only way this top equation can remain true.
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From there, depending on what problem you're trying to solve,
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you have something to work with algebraically,
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and maybe the most common goal is to try to figure out what dy divided by dx is.
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As a final example here, I want to show you how you can use this technique
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of implicit differentiation to figure out new derivative formulas.
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I've mentioned that the derivative of e to the x is itself,
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but what about the derivative of its inverse function, the natural log of x?
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Well the graph of the natural log of x can be thought of as an implicit curve.
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It's all of the points xy on the plane where y happens to equal ln of x.
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It just happens to be the case that the x's and y's of this
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equation aren't as intermingled as they were in our other examples.
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The slope of this graph, dy divided by dx, should be the derivative of ln of x, right?
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Well to find that, first rearrange this equation
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y equals ln of x to be e to the y equals x.
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This is exactly what the natural log of x means, it's saying e to the what equals x.
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Since we know the derivative of e to the y, we can take the derivative
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of both sides here, effectively asking how a tiny step with
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components dx and dy 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 the left side of the equation,
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which is e to the y times dy, must equal the change to the right side,
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which in this case is just dx.
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Rearranging, that means dy divided by dx, the slope of our graph,
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equals 1 divided by e to the y.
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When we're on the curve, e to the y is by definition the same thing as x,
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so evidently this slope is 1 divided by x.
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And of course, an expression for the slope of a graph of a function
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written in terms of x like this is the derivative of that function,
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so evidently the derivative of ln of x is 1 divided by x.
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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
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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
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tiny nudges are at play, and how exactly they depend on each other.
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Next up, I'm going to be talking about limits,
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and how they're used to formalize the idea of a derivative.
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Thank you.
— end of transcript —
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