# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 22

Let π(π₯) = (arcsin (π₯β΄))Β³. Find πβ²(π₯).

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### Video Transcript

Let π of π₯ be equal to arcsin of π₯ to the power of four cubed. Find π prime of π₯.

π of π₯ is expressed as a function of a function of a function. Itβs a composite function. Weβre therefore going to need to use the chain rule to find the derivative π prime of π₯. The chain rule says that if π¦ is a function of π’ and π’ is a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.

A special case of the chain rule is the general power rule. And this says that if π’ is a function of π₯, then the derivative of π’ to the power of π can be written as π times π’ to the power of π minus one multiplied by the derivative of π’ with respect to π₯.

Weβre actually going to apply both of these rules during this question. We can use the general power rule to begin finding π prime of π₯. Since our function in π₯ is arcsin of π₯ to the power of four and then thatβs being cubed, we can say that π prime of π₯ must be equal to three times that function arcsin of π₯ to the power of four, and then thatβs squared. And we multiply that by the derivative of arcsin of π₯ to the power of four with respect to π₯.

So weβre going to need to use the chain rule to actually evaluate the derivative of arcsin of π₯ to the power of four with respect to π₯. Weβll say that π¦ is equal to arcsin of π’ and π’ is equal to π₯ to the power of four.

To use the chain rule, weβre going to need to find the derivative of each of these. The derivative of π’ with respect to π₯ is four π₯ cubed. Weβll also use the fact that the derivative of arcsin of π₯ with respect to π₯ is one over the square root of one minus π₯ squared. This means that dπ¦ by dπ’ is one over the square root of one minus π’ squared.

We substitute this back into the formula for the chain rule. And we see that the derivative of arcsin of π₯ to the power of four with respect to π₯ is one over the square root of one minus π’ squared times four π₯ cubed. Remember that weβre trying to differentiate this with respect to π₯. So weβre going to use the fact that we let π’ be equal to π₯ to the power of four. And when we substitute this back into the expression for the derivative, we get four π₯ cubed over the square root of one minus π₯ to the power of four all squared.

Now, in fact, π₯ to the power of four squared is π₯ to the power of eight. And we can replace this in our original equation for π prime of π₯. And we get three arcsin of π₯ to the power of four squared times four π₯ cubed over the square root of one minus π₯ to the power of eight.

Simplifying just a little, and we find that π prime of π₯, the derivative of our function π with respect to π₯, is 12 arcsin of π₯ to the power of four squared over the square root of one minus π₯ to the power of eight.