Video: Using the Chain Rule to Evaluate the Derivative of a Trigonometric Function

If 𝑦 = cosΒ³ (5 βˆ’ 4π‘₯), find d𝑦/dπ‘₯.

02:10

Video Transcript

If 𝑦 equals cos cubed of five minus four π‘₯, find d𝑦 by dπ‘₯.

Let’s begin by rewriting our function slightly. We’re going to write it as cos of five minus four π‘₯ all cubed. And when we write it like this, we notice we have a composite function. That’s a function of another function. And so to find d𝑦 by dπ‘₯, that’s the derivative of 𝑦 with respect to π‘₯, we need to use the chain rule. This says that if 𝑦 is some function of 𝑒 and 𝑒 itself is some function of π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. We let 𝑒 be equal to the inner part of our composite function. So 𝑒 is equal to cos of five minus four π‘₯. And that means that 𝑦 is equal to 𝑒 cubed. We can see from the chain rule that we need to evaluate d𝑦 by d𝑒 and d𝑒 by dπ‘₯.

Well, d𝑦 by d𝑒 is quite straightforward. We multiply that entire term by the exponent and then reduce that exponent by one. So d𝑦 by d𝑒 is three 𝑒 squared. But what about the derivative of cos of five minus four π‘₯? Well, we’re going to use the chain rule to evaluate this too. We’ll let 𝑣 be equal to five minus four π‘₯, such that 𝑒 is equal to cos of 𝑣. We can then reword our chain rule slightly. And we say that the derivative of 𝑒 with respect to π‘₯ here will be equal to the derivative of 𝑒 with respect to 𝑣 times the derivative of 𝑣 with respect to π‘₯. Then we see that d𝑣 by dπ‘₯ is negative four. And of course, the derivative of cos of 𝑣 with respect to 𝑣 is negative sin 𝑣. d𝑒 by dπ‘₯ is then the product of these. It’s negative four multiplied by negative sin 𝑣. That’s four sin 𝑣.

And since 𝑣 is equal to five minus four π‘₯ and we want d𝑒 by dπ‘₯ to be in terms of π‘₯, we can say that this is equal to four sin of five minus four π‘₯. Now that we know d𝑒 by dπ‘₯ and d𝑦 by d𝑒, we can work out d𝑦 by dπ‘₯ by finding their product. That’s three 𝑒 squared times four sin of five minus four π‘₯. We can multiply the three and the four to give us a coefficient of 12. And then we replace 𝑒 with our earlier substitution with cos of five minus four π‘₯. And so d𝑦 by dπ‘₯ is 12 times cos of five minus four π‘₯ all squared times sin of five minus four π‘₯. And then we know that cos of five minus four π‘₯ squared can be written as cos squared of five minus four π‘₯. Writing our terms in increasing powers, and we see that d𝑦 by dπ‘₯ is 12 sin of five minus four π‘₯ times cos squared of five minus four π‘₯.

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