When finding the derivative of Y equal 4 to the square root of T1, method to do this is to think about Y equal E to the lane of four square roots of T. That square root of T therefore is in the exponent and can be brought down. So we get E square root of TLN 4. Remember, LN4 is a constant, so when we want to find DYDT, we're going to take the derivative of E to something which is E to that something times the derivative of that exponent or that square to TLN four. Well, we know that E squared to TLN 4 is really just the same thing as 4 sqrt t because we rewrote that as an equivalent. So now all we have to do is find the derivative of the square to TLN 4. Square to T is 1 / 2 square root of TLN of four LN4 we could think of as 2 ^2. And then this two in the exponent could come down in front and cancel with the two down in that denominator. So we'd have 4 square roots of T 1 / sqrt t lane 2 as a great final answer.