### Video Transcript

If π of π₯ equals sin to the fourth power of π₯ minus two, find π prime of π.

We could equivalently write sin to the fourth power of π₯ minus two as sin of π₯ minus two all raised to the fourth power. Writing it in this way helps us to see that we have here a function within a function. We have sin of π₯ minus two as our inner function. And thatβs all raised to the fourth power. We want to find the first derivative of this function and then evaluate it at π.

So, to differentiate this, letβs use the chain rule. The chain rule says that if π¦ equals π of π’ and π’ equals π of π₯, then dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯. So, for this question, if we let π¦ equal sin of π₯ minus two all raised to the fourth power and π’ equal to sin of π₯ minus two, then actually we can rewrite this expression for π¦ in terms of π’. This is π’ to the fourth power. So, in order to find dπ¦ by dπ₯, weβre going to need to find dπ¦ by dπ’ and dπ’ by dπ₯. These are the derivative of π¦ with respect to π’ and the derivative of π’ with respect to π₯.

Letβs begin by finding dπ¦ by dπ’. If π¦ equals π’ to the fourth power, we can differentiate this using the power rule. This gives us four π’ to the third power. And we get this by multiplying by the power and then subtracting one from the power. So, now, letβs also find dπ’ by dπ₯. To do this, we recall the useful rule that the derivative with respect to π₯ of sin of π₯ equals cos of π₯. And this gives us that dπ’ by dπ₯ equals cos of π₯ minus two.

Note that we can think of this as a kind of another application of the chain rule. If we had a different expression other than π₯ minus two in this bracket, we would have to multiply by the derivative of this bracket. But in this case, the derivative of π₯ minus two is just one. So, this is actually one multiplied by cos of π₯ minus two, which is just cos of π₯ minus two.

So, letβs now use the chain rule to find dπ¦ by dπ₯. We have that dπ¦ by dπ₯ equals dπ¦ by dπ’, which is four π’ to the third power, multiplied by cos of π₯ minus two, which is dπ’ by dπ₯. But remember here that earlier we let π’ equal sin of π₯ minus two. So, we can actually replace π’ to give us dπ¦ by dπ₯ equals four multiplied by sin of π₯ minus two cubed multiplied by cos of π₯ minus two. And we can equivalently write this as four sin cubed of π₯ minus two cos of π₯ minus two.

Remember here that our original function was given as π of π₯. So, letβs use the notation that we were given in the question. Now that we found π prime of π₯, we can find π prime of π. To do this, all we need to do is make the substitution π₯ equals π. This gives us our final answer of π prime of π equals four sin cubed of π minus two cos of π minus two.