# Video: Using Implicit Differentiation to Find the Derivative of an Inverse Trigonometric Function

If π¦ = cscβ»ΒΉ (π^(3π₯)), which of the following represents dπ¦/dπ₯? [A] β3π^(3π₯) cot (π^(3π₯)) cosec (π^(3π₯)) [B] β(3π^(3π₯) sinΒ² π¦)/(cos π¦) [C] 3π^(3π₯) tan π¦ [D] 3 cos π¦ sec π¦

03:18

### 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¦?

Letβs 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 itβs an inverse function. So ideally, weβll find a different way to write it and integrate from there. Now, the first thing weβre 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 weβre going to differentiate csc of π¦ with respect to π₯, weβre 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 itβs important we donβt confuse this power of negative one with the inverse. Well, we can use the general power rule, itβs 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₯. Itβs 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 thatβs much more straightforward. Itβs three times π to the power of three π₯. And thatβs just a result that we can quote, the derivative of π to the power of ππ₯ with respect to π₯ is π times π to the power of ππ₯ for real constants π. Now, of course, weβre looking to find an expression for dπ¦ by dπ₯. So, weβre 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¦.