When doing this problem we are trying to find the derivative so DYDX we also or DYDTI guess. In this case we also know that T is going to be positive. It's going to be in between zero and sqrt 1 over root 3. So we have one over EU, which in this case is 1 / 9 T to the 4th square root of our U ^2. So 1 / 81 T to the 4th -1. That's all times the derivative of the inside. Well, we could think of that derivative of the inside as one ninth T to the -4. So we're going to bring down the exponent and multiply by 1 less power. So we get -4 / 9 T to the fifth times. If I take the reciprocal of this 1 / 9 T to the 4th, we get 9 T to the 4th in the top. Then I'm going to get a common denominator in the bottom 1 -, 81. Oops, I didn't square this one. Did I-81 T to the 8th because nine T to the fourth times 9 T to the 4th. So 81 T to the 8th over 81 T to the 8th. Simplifying this up, the nines cancel and the T to the 4th over T to the five is just going to leave us a -4 on top square root 1 -, 81 T to the 8th. But then sqrt 81 T to the 8th is 9 T to the 4th. So simplifying it up so we don't have a complex fraction, we get negative. Oh, we had a 1T still left down here, didn't we -36 T cubed over square root 1 -, 81 T to the 8th, so the T to the 4th over T to the fifth left a single T down here. This 19th to the 4th would cancel with that T to get us a nine t ^3. When we take that up to the top, we get -36 T cubed over square root 1 -, 81 T to the 8.