### Video Transcript

If π¦ is equal to the inverse csc of π to the power of three π₯, which of the following represents dπ¦ by dπ₯? Is it a), negative three π to the three π₯ times cot of π to the power of three π₯ times csc of π to the power of three π₯? Is it b), negative three π to the power of three π₯ times sin squared π¦ over cos π¦? Is it c), three π to the power of three π₯ times tan π¦? Or d), three times cos π¦ time sec π¦?

Letβs go right back to our original function, π¦ is equal to the inverse csc of π to the power of three π₯. What makes this function really difficult to differentiate is that itβs an inverse function. So ideally, weβll find a different way to write it and integrate from there. Now, the first thing weβre going to do is take the cosecant of both sides of this equation. On the left, that gives us csc of π¦. And then, csc of the inverse of csc of π to the power of three π₯ is simply π to the power of three π₯.

Our job is now to differentiate both sides of this equation with respect to π₯. But of course, the expression on the left-hand side of our equation is in terms of π¦. So, how do we differentiate that with respect to π₯? Well, we use a special version of the chain rule, which is called implicit differentiation. This says that the derivative with respect to π₯ of some function in π¦ is equal to the derivative of that function of π¦ with respect to π¦ times dπ¦ by dπ₯. And so, this means if weβre going to differentiate csc of π¦ with respect to π₯, weβre going to differentiate it with respect to π¦ and then multiply that result by dπ¦ by dπ₯. But what is the derivative of csc of π¦ with respect to π¦?

Well, csc of π¦ is one over sin of π¦ or sin of π¦ to the power of negative one. And itβs important we donβt confuse this power of negative one with the inverse. Well, we can use the general power rule, itβs a special case of the chain rule, to differentiate this with respect to π¦. What we do is multiply the entire term by the exponent and then reduce that exponent by one. So, we get negative one times sin π¦ to the power of negative two. And then, we multiply that by the derivative of the inner function. Well, the derivative of sin π¦ is cos π¦. Of course, sin π¦ to the power of negative two is the same as one over sin squared π¦. So, we find that the derivative of sin of π¦ to the power of negative one, and therefore csc of π¦ with respect to π¦, is negative cos π¦ over sin squared π¦.

And now, we can say that the derivative of csc of π¦ with respect to π₯ is this multiplied by dπ¦ by dπ₯. Itβs negative cos π¦ over sin squared π¦ dπ¦ by dπ₯. We can now differentiate the right-hand side of our equation with respect to π₯, and thatβs much more straightforward. Itβs three times π to the power of three π₯. And thatβs just a result that we can quote, the derivative of π to the power of ππ₯ with respect to π₯ is π times π to the power of ππ₯ for real constants π. Now, of course, weβre looking to find an expression for dπ¦ by dπ₯. So, weβre going to make dπ¦ by dπ₯ the subject by dividing through by negative cos of π¦ over sin squared of π¦.

Well, dividing by our fraction is the same as multiplying by the reciprocal of that fraction. So, we find that dπ¦ by dπ₯ is equal to three π to the power of three π₯ times negative sin squared π¦ over cos π¦. By writing three to the power of three π₯ as three to the power of three π₯ over one, we see we can multiply these fractions quite easily. We multiply the two numerators and then multiply the two denominators. And we get negative three π to the power of three π₯ times sin squared π¦ over cos π¦. And in doing so, we see that the correct answer is b). The derivative of the inverse csc of π to the power of three π₯ is negative three π to the power of three π₯ sin squared π¦ over cos π¦.