Let's scroll down a little bit. The derivative of y with respect to x is equal to, is equal to, well, the twos cancel out. We are left with negative x, negative x over, over y. So this is interesting. We didn't have to 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, 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 two over two, comma, the square root of two over two.