To find the derivative here, we need to understand the first term we're going to do the product role. So our DYDT, the derivative of the 1st just just going to be 1. So we get one times sqrt 1 -, t ^2 plus the derivative of the 2nd is going to be 1 / 2 square roots of 1 -, t ^2. By the chain rule, we have to multiply by the derivative of the inside, which would be the -2 T and then we're going to multiply it also by the first term. Then we have to remember that the derivative of sine inverse of T is 1 / sqrt 1 -, t ^2 times the derivative of T, which is really just one. So we have DYDT equaling sqrt 1 -, t ^2. Actually that's not the same term. So we're going to have that. The twos are going to cancel here. So we get a negative t ^2 / sqrt 1 -, t ^2. We can think of that as a minus in front if we wanted to. And then this is a -, 5 square roots of 1 - t ^2. Getting a common denominator. We're going to multiply this by top and bottom. When we do that, we're going to get 1 - t ^2 over the square root 1 -, t ^2 -, t ^2 over square root 1 -, t ^2 - 5 over square root 1 - t ^2. When we simplify this up, we get DYDT equaling 1 - 5 negative 4 - 2 T squared over square root 1 -, t ^2. We could factor out a negative in front if we wanted to, to make that negative, putting them in descending order. Two t ^2 + 4 square root 1 -, t ^2.