Video: Differentiating a Trigonometric Function Using the Chain Rule

If 𝑓(π‘₯) = sin⁴ (π‘₯ βˆ’ 2), find 𝑓′(πœ‹).

03:15

Video Transcript

If 𝑓 of π‘₯ equals sin to the fourth power of π‘₯ minus two, find 𝑓 prime of πœ‹.

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 that’s all raised to the fourth power. We want to find the first derivative of this function and then evaluate it at πœ‹.

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

Let’s begin by finding d𝑦 by d𝑒. If 𝑦 equals 𝑒 to the fourth power, we can differentiate this using the power rule. This gives us four 𝑒 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 d𝑒 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 d𝑒 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, let’s now use the chain rule to find d𝑦 by dπ‘₯. We have that d𝑦 by dπ‘₯ equals d𝑦 by d𝑒, which is four 𝑒 to the third power, multiplied by cos of π‘₯ minus two, which is d𝑒 by dπ‘₯. But remember here that earlier we let 𝑒 equal sin of π‘₯ minus two. So, we can actually replace 𝑒 to give us d𝑦 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 𝑓 of π‘₯. So, let’s use the notation that we were given in the question. Now that we found 𝑓 prime of π‘₯, we can find 𝑓 prime of πœ‹. To do this, all we need to do is make the substitution π‘₯ equals πœ‹. This gives us our final answer of 𝑓 prime of πœ‹ equals four sin cubed of πœ‹ minus two cos of πœ‹ minus two.

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