### Video Transcript

Given that π¦ is equal to seven cot five π₯ plus three cosec six π₯ to the negative one, find dπ¦ by dπ₯.

In this example, we have a reciprocal function. We could write this as a fraction and apply the quotient rule. Alternatively, we could use the chain rule. Letβs look at how we might use the chain rule. This says that if π¦ is a function in π’ and π’ itself is a function in π₯, then the derivative of π¦ with respect to π₯ is dπ¦ by dπ’ times dπ’ by dπ₯. We will let π’ be equal to seven cot of five π₯ plus three cosec six π₯. This means that π¦ is equal to π’ to the negative one. The derivative of π¦ with respect to π’ is fairly straightforward. Itβs negative one times π’ to the negative two. Then we quote the derivative for cot ππ₯ as being negative π cosec squared ππ₯ and the derivative of cosec ππ₯ as being negative π cosec ππ₯ cot ππ₯. So we see dπ’ by dπ₯ is equal to negative 35 cosec squared five π₯ minus 18 cosec six π₯ cot six π₯.

The derivative of π¦ with respect to π₯ is the product of these two. And we recall that we can write negative π’ to the negative two as one over negative π’ squared. We then divide through by negative one and replace π’ with seven cot five π₯ plus three cosec six π₯. And weβve obtained dπ¦ by dπ₯ to be equal to 35 cosec squared five π₯ plus 18 cosec six π₯ cot six π₯ over seven cot five π₯ plus three cosec six π₯ all squared.