If 𝑓 of 𝑥 equals sin to the fourth power of 𝑥 minus two, find 𝑓 prime of 𝜋.
We could equivalently write sin to the fourth power of 𝑥 minus two as sin of 𝑥 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 𝑥 minus two as our inner function. And that’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, let’s use the chain rule. The chain rule says that if 𝑦 equals 𝑓 of 𝑢 and 𝑢 equals 𝑔 of 𝑥, then d𝑦 by d𝑥 equals d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. So, for this question, if we let 𝑦 equal sin of 𝑥 minus two all raised to the fourth power and 𝑢 equal to sin of 𝑥 minus two, then actually we can rewrite this expression for 𝑦 in terms of 𝑢. This is 𝑢 to the fourth power. So, in order to find d𝑦 by d𝑥, we’re going to need to find d𝑦 by d𝑢 and d𝑢 by d𝑥. These are the derivative of 𝑦 with respect to 𝑢 and the derivative of 𝑢 with respect to 𝑥.
Let’s begin by finding d𝑦 by d𝑢. If 𝑦 equals 𝑢 to the fourth power, we can differentiate this using the power rule. This gives us four 𝑢 to the third power. And we get this by multiplying by the power and then subtracting one from the power. So, now, let’s also find d𝑢 by d𝑥. To do this, we recall the useful rule that the derivative with respect to 𝑥 of sin of 𝑥 equals cos of 𝑥. And this gives us that d𝑢 by d𝑥 equals cos of 𝑥 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 𝑥 minus two in this bracket, we would have to multiply by the derivative of this bracket. But in this case, the derivative of 𝑥 minus two is just one. So, this is actually one multiplied by cos of 𝑥 minus two, which is just cos of 𝑥 minus two.
So, let’s now use the chain rule to find d𝑦 by d𝑥. We have that d𝑦 by d𝑥 equals d𝑦 by d𝑢, which is four 𝑢 to the third power, multiplied by cos of 𝑥 minus two, which is d𝑢 by d𝑥. But remember here that earlier we let 𝑢 equal sin of 𝑥 minus two. So, we can actually replace 𝑢 to give us d𝑦 by d𝑥 equals four multiplied by sin of 𝑥 minus two cubed multiplied by cos of 𝑥 minus two. And we can equivalently write this as four sin cubed of 𝑥 minus two cos of 𝑥 minus two.
Remember here that our original function was given as 𝑓 of 𝑥. So, let’s use the notation that we were given in the question. Now that we found 𝑓 prime of 𝑥, we can find 𝑓 prime of 𝜋. To do this, all we need to do is make the substitution 𝑥 equals 𝜋. This gives us our final answer of 𝑓 prime of 𝜋 equals four sin cubed of 𝜋 minus two cos of 𝜋 minus two.