When looking at this problem, we want to do the derivative of Y in terms of X. So we're going to take DY. We're going to take the derivative in terms of X of four y ^2, and that's going to equal the derivative in terms of X of the three X -, 5 over three X + 5 S. The derivative in terms of X of four y ^2 is going to be 8Y using the power rule, and it's the derivative of Y in terms of X because our variable was AY over here on the right. We're going to use the quotient rule. So we're going to have the derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared. So we're going to have the 8Y DYDX equaling the derivative of the top was 3 and it's derivative of X in terms of X. So we could just think of that as one, but it's 3D XDX times three X + 5 minus the derivative of the bottom. The derivative of the bottom is 3 again, but it's the derivative of X in terms of X. So DXDX, which once again is going to be one times the top all over that bottom square SO8Y DYDX is going to equal. I'm going to distribute out the three. So nine X + 15. I'm going to distribute out the second three. So negative nine X + 15 all over three X + 5 ^2. So DYDX is going to equal 30 over three X + 5 ^2, and we're going to take that eight Y to the other side. So we're going to divide so 30 / 8 Y times three X + 5 ^2, and then thirty and eight Y can reduce to 15 / 4 Y three X + 5 quantity squared. So that's our final answer.