Find the derivative of y = 6 to the square root of S with respect to S First thing we should know is there's actually a formula of DDX equaling DDX of A to EU. Taking the derivative in terms of X of A to some function, U is equal to A to EU times lane of a DUDX. Now I don't memorize things well, and that might be a formula that you don't remember, but we can rewrite this in terms of our ES. So if we think about Y equaling 6 to the square root of S, that's really the same thing as E to the lane of six to the square root of S. Now we can bring that sqrt 6 down and or square root of S down in front. So we get E square root of SLN 6. Now when we take the derivative, we know that the derivative of E to anything is E to the anything times the derivative of that anything. So we get E to the square root of SLN 6 times the derivative of that exponent. Well, LN6 is just a constant. So all we really have to do then is figure out the derivative of that square root of S. And that square root of S is 1 / 2 sqrt s, And then E to the square root of SLN 6 is really just six to the square root of S. So we take this out and we replace it with what it was. If we look up here, the formula said it's going to be our A to our U 6 to the square root of S, which is right here. The lane of a, in this case, our a was six. So there's my lane of six times the derivative of U in terms of X. So the derivative of this exponent of square root of S in terms of X sqrt s would be 1 / 2 sqrt s.