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

Let 𝑔(π‘₯) = (arcsin (π‘₯⁴))Β³. Find 𝑔′(π‘₯).

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Video Transcript

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

𝑔 of π‘₯ is expressed as a function of a function of a function. It’s a composite function. We’re therefore going to need to use the chain rule to find the derivative 𝑔 prime of π‘₯. The chain rule says that if 𝑦 is a function of 𝑒 and 𝑒 is a function of π‘₯, then d𝑦 by dπ‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯.

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

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

So we’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‘₯. We’ll say that 𝑦 is equal to arcsin of 𝑒 and 𝑒 is equal to π‘₯ to the power of four.

To use the chain rule, we’re going to need to find the derivative of each of these. The derivative of 𝑒 with respect to π‘₯ is four π‘₯ cubed. We’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 d𝑦 by d𝑒 is one over the square root of one minus 𝑒 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 𝑒 squared times four π‘₯ cubed. Remember that we’re trying to differentiate this with respect to π‘₯. So we’re going to use the fact that we let 𝑒 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 𝑔 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 𝑔 prime of π‘₯, the derivative of our function 𝑔 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.

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