### Video Transcript

Given that π₯ equals cos π‘ and π¦ equals sin two π‘, find dπ¦ by dπ₯.

In this question, weβve been given a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter; here thatβs π‘. And so, we recall that to find dπ¦ by dπ₯ β thatβs the derivative of π¦ with respect to π₯ β as long as dπ₯ by dπ‘ is not equal to zero, we can divide dπ¦ by dπ‘ by dπ₯ by dπ‘. And so, it should be fairly clear to us that weβre going to need to differentiate π₯ with respect to π‘ and the equation for π¦ with respect to π‘.

And we might recall this cycle that helps us to differentiate sin and cos. The derivative of sin π₯ is cos π₯. The derivative of cos π₯ with respect to π₯ is negative sin π₯. Differentiating again, and we get negative cos π₯. And differentiating once more, and we get sin π₯. This means that the derivative of cos π‘ with respect to π‘ is negative sin π‘. Then, we could quote the general result for the derivative of sin of some constant of π‘. But we could also use the chain rule.

By letting π’ be equal to two π‘, we find that dπ¦ by dπ‘ is equal to the derivative of π’ with respect to π‘ β thatβs the derivative of two π‘ with respect to π‘ β times the derivative of sin π’ with respect to π’. Well, the derivative of two π‘ with respect to π‘ is two whereas the derivative of sin π’ with respect to π’ is cos π’. But since we let π’ be equal to two π‘, we find that dπ¦ by dπ‘ is two cos of two π‘. Then, dπ¦ by dπ₯ is the quotient of these. Itβs two cos two π‘ over negative sin π‘. And since weβre dividing a positive function by a negative function, we know that our final result will be a negative function. So, dπ¦ by dπ₯ is negative two cos two π‘ over sin π‘.