Current time:0:00Total duration:5:58
0 energy points
Sal finds dy/dx for e^(xy²)=x-y using implicit differentiation. Created by Sal Khan.
Video transcript
So let's take another implicit derivative of the somewhat crazy relationship. And I've graphed the relationship here. As you can see, it is actually quite bizarre. e to the x times y squared is equal to x minus y. This is some at least in the range that's depicted here, the x's and the y's that satisfy this relationship. Well let's take the derivative of both sides. So we'll apply our derivative operator to both sides. And actually this is a good chance to maybe explore some different notation. We tend to be using that notation, but oftentimes you'll see a derivative operator that just looks like a big capital D. So maybe we'll do that right over here. So let me make it clear. This is the equivalent of saying d over dx. I'll just use this big capital D operator so that you are familiar with the notation. And instead of using dy dx for the derivative of y with respect to x, I'm just going to write that as y prime. So we get a little bit of practice with different notation. So let's take the derivative of this thing right over here. Well we're going to apply the chain rule. Actually, we're going to apply the chain rule multiple times here. The derivative of e to the something with respect to that something is going to be e to the something times the derivative of that something with respect to x. So times the derivative of xy squared. So that's our left-hand side. We aren't done taking the derivative yet. And on our right-hand side, the derivative of x is just 1. And the derivative with respect to x of y is just going to be minus-- or I could write-- negative dy dx. But instead of writing dy dx, I'm going to write y prime. As you can tell, I like this notation and this notation more because it makes it explicit that I'm taking the derivative with respect to x. Here, we just have to assume that we're taking the derivative with respect to x. Here, we have to assume that's the derivative of y with respect to x. But anyway let's stick with this notation right over here. Actually, let me make all of my y primes, all my derivatives of y with respect to x, let me make them pink so I keep track of them. So once again, this is going to be equal to e to the xy squared times the derivative of this. Well the derivative of this, we can just use the product and actually a little bit of the chain rule here. So the derivative of x is just 1 times the second function. So it's going to be times y squared. And then to that, we're going to add the product of the first function which is this x times the derivative of y squared with respect to x. Well that's going to be the derivative of y squared with respect to y, which is just going to be 2y times the derivative of y with respect to x, which we are now writing as y prime. And then that's going to be equal to 1 minus y prime. And like we've been doing, we now have to just solve for y prime. So let's distribute this exponential, this e to the xy squared. And we get e, or maybe I should say y squared times e to the xy squared. So that's that. Plus 2xye to the xy squared. y prime, the derivative of y with respect to x, is equal to 1 minus the derivative of y with respect to x. Now let's get all of our y primes on one side. So let's add y prime to both sides. So let's add-- and just to be clear, I'm adding one y prime to both sides. So let's add a y prime to both sides. And let's subtract this business from both sides. So let's subtract y squared e to the xy squared subtracting from both sides. So we're going to subtract y squared e to the xy squared. And we are left with 2xye to the xy squared plus 1 times y prime. We had this many y primes and then we add another 1y prime so we have this many plus 1 y primes. That's going to be equal to-- well, I purposely added y prime to both sides and so we are left with 1 minus-- this is kind of a crazy expression-- y squared times e to the xy squared. And now we just have to divide both sides by this. And we're left with the derivative of y with respect to x is equal to this, which I will just copy and paste. Actually, let me just rewrite it. Scroll down a little bit. It's equal to 1 minus y squared e to the xy squared over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but fundamentally no different than what we've been doing in the last few examples.