### Video Transcript

Find dπ¦ by dπ₯, given that π¦ is equal to the cube root of negative five π₯ cubed plus four π₯ minus eight squared.

The question wants us to find dπ¦ by dπ₯, which is the first derivative of π¦ with respect to π₯. And we can see that our function π¦ is the cube root of the square of a polynomial. In other words, in its current form, itβs the composition of three functions. So one way we could evaluate this derivative is to use the chain rule twice. However, since our innermost cubic function only has three terms, we could also distribute the square over the parentheses. Then we would just need to apply the chain rule once. And this would also work, although thereβs actually an even easier method.

We can simplify this expression by using our laws of exponents. First, we know the cube root of π is equal to π to the power of one-third. Next, π to the power of π all raised to the power of π is just equal to π to the power of π times π. Applying both of these laws of exponents to our equation, we get that π¦ is equal to negative five π₯ cubed plus four π₯ minus eight all raised to the power of two over three.

And now we can see that π¦ is the composition of two functions. So we can differentiate this by using the chain rule. Weβll start by setting π’ to be our inner function. Thatβs negative five π₯ cubed plus four π₯ minus eight. So this gives us π¦ is equal to π’ to the power of two over three.

Letβs now recall the chain rule. The chain rule tells us if we have π¦ is a function of π’ and π’ is a function of π₯. Then we can find the derivative of π¦ with respect to π₯ by first finding the derivative of π¦ with respect to π’ and multiplying this by the derivative of π’ with respect to π₯. So by applying the chain rule, we have dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.

Letβs first find an expression for dπ¦ by dπ’. We know that π¦ is equal to π’ to the power of two over three. So dπ¦ by dπ’ is just the derivative of π’ to the power of two over three with respect to π’. Next, letβs find an expression for dπ’ by dπ₯. We have that π’ is equal to negative five π₯ cubed plus four π₯ minus eight. So the first derivative of π’ with respect to π₯ is equal to the derivative of this expression with respect to π₯.

And we can evaluate both of these derivatives by using the power rule for differentiation. For constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one. We multiply by the exponent of π₯ and reduce this exponent by one. Applying this to our first derivative, we multiply by the exponent of π’ β thatβs two-thirds β and reduce this exponent by one. This gives us two-thirds π’ to the power of negative one-third.

And weβll simplify this expression by remembering π’ to the power of negative one-third is the same as one divided by the cube root of π’. So we now want to evaluate the derivative of our second expression. Thatβs the derivative of negative five π₯ cubed plus four π₯ minus eight with respect to π₯. And we do this term by term by using the power rule for differentiation. We get negative 15π₯ squared plus four.

Remember, weβre finding an expression for the derivative of π¦ with respect to π₯. So we want to give our answer in terms of π₯. So weβll use our substitution π’ is equal to negative five π₯ cubed plus four π₯ minus eight. And by doing this, weβve shown that if π¦ is equal to the cube root of negative five π₯ cubed plus four π₯ minus eight squared. Then dπ¦ by dπ₯ is equal to two times negative 15π₯ squared plus four all divided by three times the cube root of negative five π₯ cubed plus four π₯ minus eight.