Find the derivative of Y with respect to Theta. So if we let F of X equal hyperbolic cosecant X, then F inverse of X is hyperbolic cosecant inverse of X. So the derivative of that is going to be -1 divided by the derivative of that hyperbolic cosecant is hyperbolic cosecant of the F inverse of X times hyperbolic cotangent of the F inverse of X. That was negative because it starts with AC. It's a Co function, so the hyperbolic cosecant of the hyperbolic cosecant of inverse of X is just X. We know that the hyperbolic cotangent based off of our identity of the hyperbolic cosine squared X minus the hyperbolic sine squared X equal 1. If we divided them all by hyperbolic sine squared, we'd get hyperbolic cotangent X -, 1 equal hyperbolic cosecant squared X. So we're going to replace this hyperbolic cotangent with the square root of hyperbolic cosecant squared X + 1. And remember that X is really this whole hyperbolic cosecant inverse of X. So now these are going to cancel, leaving us an X ^2 + 1. So our formula for the derivative of just plain old hyperbolic cosecant inverse of X is -1 / X sqrt X ^2 + 1. Now we're going to have to do the chain rule for the actual problem. So we're going to have -1 over this 1/8 to the Theta square root, 1/8 to the two Theta plus one times DD Theta of 1/8 to the Theta. Well, the derivative of 1/8 to the Theta is the negative natural log of 1/8 tie. Actually the negative came from there times 1/8 to the Theta. So now if we simplify this up, we can think of 1/8 as 8th to the -1 and the negative and the negative there can cancel. If I get a common denominator in the bottom, we'd get one to the two Theta, which is 1 plus 8 to the two Theta, all over 8 to the two Theta, and the 1 / 8 to the thetas top and bottom are going to cancel. So now we have the square root of this 8 to the two Theta, the square root and the two are going to cancel. So that's going to leave us 8 Theta, 8 to the Theta outside. And because it was in the denominator, we're going to move it up to the numerator. So 2 lane 3 * 8 to the Theta over the square root, 1 + 8 to the two Theta. Or if we had wanted to, we could actually think of that as lane of eight. Oh wait, I did this one backwards right here. 8 is 2 ^3, so that would have been 3-2. Or we could have thought of it as lane 8 * 8 to the Theta all over the square root 1 + 8 to the two Theta.