And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx.