 In this video, we will provide the solution to question number 11 for practice exam number three for math 1210, in which case we're told to use the quotient rule to show that the derivative of cosecant of x is equal to cotangent of x times cosecant of x. So using the quotient rule here is critical. So basically we have to prove that this equals that. And when you try to prove that two things are equal, what you're gonna do is you're gonna provide a sequence of equalities until eventually the left-hand side arrives upon the right-hand side. So we'll start off with the left-hand side, the derivative of cosecant. Well, it tells us we should use the quotient rule. So how does the quotient rule come into play? We have the right cosecant as a quotient. So cosecant, notice, is one over sine of x. It's the reciprocal of sine. In which case now we can use the quotient rule to take the derivative. We're gonna get low d high minus high d low square the bottom, here we go. So we get a sine squared in the denominator. Some things to note, the derivative of one is zero because it's a constant. So this whole term is gonna cancel out. The derivative of sine is gonna equal a cosine. So if we apply that right there, we're gonna get zero minus cosine of x. This sits above sine squared of x. And so this gives us a negative cosine of x. Now the rest of the, sine squared of x, excuse me. So now the rest of the problem is basically just a trigonometric identity. Notice where we have to get to. We have to get just negative cotangent times cosecant. Well, cosecant is one over sine like we already observed. Cotangent is cosine over sine. So if we factor this thing, we're gonna get negative cosine x over sine x. That's a cotangent. And then you times that by one over sine of x. We can make the conclusion that this is negative cotangent of x times sine, excuse me, cosecant of x. Thus proving the identity that was required.