# Video: Differentiating a Trigonometric Function Using the Chain Rule

If π(π₯) = sinβ΄ (π₯ β 2), find πβ²(π).

03:15

### 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 thatβs 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, letβs 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₯, weβre 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₯.

Letβs 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, letβs 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, letβs 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, letβs 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.