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# Implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)

## Video transcript

So we've got the equation x squared plus y squared is equal to 1. I guess we could call it a relationship. And if we were to graph all of the points x and y that satisfied this relationship, we get a unit circle like this. And what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle. And what immediately might be jumping out in your brain is, well a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value we actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of 1 minus x squared. And you could say y is equal to the negative square root of 1 minus x squared. Take the derivatives of each of these separately. And you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point. But what I want to do in this video is literally leverage the chain rule to take the derivative implicitly. So that I don't have to explicitly define y is a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation. And then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x, and explicitly getting y is equal to f prime of x, they call this-- which is really just an application of the chain rule-- we call it implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, on the left hand side of our equation. And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. So let's see. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power. We could just say 2x. Now what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative-- let me make it clear-- we're just going to take the derivative of our something. The derivative of y squared-- that's what we're taking, you can kind of view that as a function-- with respect to y and then multiply that times the derivative of y with respect to x. We're assuming that y does change with respect to x. y is not some type of a constant that we're writing just an abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y with respect to x. It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y, as a function of x. This might be, or y is a function of x squared, which is essentially another way of writing what we have here. This might be a little bit clearer in terms of the chain rule. The derivative of y is a function of x squared with respect to y of x. So the derivative of something squared with respect to that something, times the derivative of that something, with respect to x. This is just the chain rule. I want to say it over and over again. This is just the chain rule. So let's do that. What do we get on the right hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y, is just going to be 2 times y. 2 times y, just an application of the chain rule. And the derivative of y with respect to x? Well, we don't know what that is. So we're just going to leave that as times the derivative of y with respect to x. So let's just write this down over here. So we have is 2x plus the derivative of something squared, with respect to that something, is 2 times the something. In this case, the something is y, so 2 times y. And then times the derivative of y with respect to x. And this is all going to be equal to 0. Now that was interesting. Now we have an equation that has the derivative of a y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x. Solve this equation. So let's do that. And actually just so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off. And let's continue there. So let's say let's subtract 2x from both sides. So we're left with 2y times the derivative of y, with respect to x, is equal to-- we're subtracting 2x from both sides-- so it's equal to negative 2x. And if we really want to solve for the derivative of y with respect to x, we can just divide both sides by 2y. And we're left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, well the 2s cancel out. We we're left with negative x over y. So this is interesting. We didn't have to us explicitly define y as a function of x here. But we got our derivative in terms of an x and a y. Not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point right over here. Which, if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of 2 over 2 comma the square root of 2 over 2. What is the slope of the tangent line there? Well, we figured it out. It's going to be negative x over y. So the slope of the tangent line here is going to be equal to negative x. So negative square root of 2 over 2 over y. Over square root of 2 over 2, which is equal to negative 1. And that looks just about right.
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