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## Calculus, all content (2017 edition)

### Unit 2: Lesson 30

Implicit differentiation introduction

# Worked example: Implicit differentiation

Implicit differentiation of (x-y)²=x+y-1. Created by Sal Khan.

## Video transcript

Let's get some more practice doing implicit differentiation. So let's find the derivative of y with respect to x. We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left hand side, we essentially are just going to apply the chain rule. First we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1, and the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. This right over here is the derivative of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x. We're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do. Let's make it clear. This right over here, I can rewrite as 2x minus 2y. So let me do that so I can save some space. This is 2x minus 2y If I just distribute the 2. And now I can distribute the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. So over here, we're going to subtract 2x minus 2y from that side. And then we could also subtract a dy dx from both sides, so that all of our dy dx's are on the left hand side, and all of our non dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx, or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So this is we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to-- on this side, this cancels out. We are left with 1 minus 2x plus 2y. So let me write it that way. Or we could write this as-- so negative, negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1, plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1. And we are left with-- we deserve a little bit of a drum roll at this point. As you can see, the hardest part was really the algebra to solve for dy dx. We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1.