 In this video, we provide the solution to question number 24 for the practice final exam for math 1210, in which case we're asked to use implicit differentiation to show that the derivative of the inverse of cosecant is equal to negative one over x times the square root of x square minus one. So let's begin. We're going to write down our function we're trying to take the derivative of y equals cosecant inverse of x. Well, the reason we're going to do implicit differentiations because although we might not know the derivative of the inverse cosecant, we can calculate the derivative of cosecant by switching to the inverse function. So cosecant of y is equal to x. This is an observation we're going to want to remember for later on. So I'm going to put a little mark right here. But this is the moment we're going to take the derivative implicitly. We're going to take the derivative of the right hand side and the left hand side with respect to x. On the left hand side, because we're taking derivative of cosecant of y, we're going to get the derivative of cosecant, which is negative cosecant y cotangent y. But then we have to take the inner derivative, which is y prime, the derivative of course of x is going to equal a one. If you solve for y prime, you end up with one over negative one over cosecant of y times cotangent of y, for which this is now why we mark that box we did earlier. Notice that cosecant of y is equal to x. That's going to take care of this x right here. Well, how do we get rid of the cotangent? Well, the cotangent you'll notice based upon this identity that we were told about. Notice that if I subtract one, you're going to end up with cotangent squared of x, which this course we're going to replace with y. This is equal to cosecant squared of y minus one. So then if we take the square root, you're going to get that cotangent of y is equal to the square root of cosecant squared of y minus one, which that, again, for like we observed earlier, cosecant is equal to x. So this is equal to the square root of x squared minus one. And so we make that substitution in for there. We then get the derivative we were looking for. We get that y prime is equal to negative one over x times the square root of x squared minus one.