### Video Transcript

If π¦ equals cos cubed of five minus four π₯, find dπ¦ by dπ₯.

Letβs begin by rewriting our function slightly. Weβre 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. Thatβs a function of another function. And so to find dπ¦ by dπ₯, thatβs the derivative of π¦ with respect to π₯, we need to use the chain rule. This says that if π¦ is some function of π’ and π’ itself is some function of π₯, then the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. We let π’ be equal to the inner part of our composite function. So π’ is equal to cos of five minus four π₯. And that means that π¦ is equal to π’ cubed. We can see from the chain rule that we need to evaluate dπ¦ by dπ’ and dπ’ by dπ₯.

Well, dπ¦ by dπ’ is quite straightforward. We multiply that entire term by the exponent and then reduce that exponent by one. So dπ¦ by dπ’ is three π’ squared. But what about the derivative of cos of five minus four π₯? Well, weβre going to use the chain rule to evaluate this too. Weβll let π£ be equal to five minus four π₯, such that π’ is equal to cos of π£. We can then reword our chain rule slightly. And we say that the derivative of π’ with respect to π₯ here will be equal to the derivative of π’ with respect to π£ times the derivative of π£ with respect to π₯. Then we see that dπ£ by dπ₯ is negative four. And of course, the derivative of cos of π£ with respect to π£ is negative sin π£. dπ’ by dπ₯ is then the product of these. Itβs negative four multiplied by negative sin π£. Thatβs four sin π£.

And since π£ is equal to five minus four π₯ and we want dπ’ 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 dπ’ by dπ₯ and dπ¦ by dπ’, we can work out dπ¦ by dπ₯ by finding their product. Thatβs three π’ 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 π’ with our earlier substitution with cos of five minus four π₯. And so dπ¦ 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 dπ¦ by dπ₯ is 12 sin of five minus four π₯ times cos squared of five minus four π₯.