We know that the derivative of inverse cotangent U DX is equal to -1 / 1 + U ^2 DU DX. So doing this problem, we're going to say our U is that sqrt 4 T So we're going to have DYDT equaling -1 / 1 plus RU in this case sqrt 4 T squared times the derivative of that sqrt 4 T Well, dy DT -1 / 1 plus the square root and the square are going to cancel. The derivative of a square root is going to be 1 / 2 times the square root times the derivative of the inside by the chain rule. So the derivative of 4T is going to give us 4. So we get -1 one plus four t * 1 / 2 sqrt 4. We know is 2 sqrt t and then the derivative of 4T is 4, so this 2 * 2 and that four is going to cancel. So our DYDT is really just -1 / sqrt t times the quantity 1 + 4 T Looks to me like in their answer they didn't simplify that sqrt 4 we know is 2, so the two and the two would cancel, leaving that sqrt 2 down in the bottom.