# Question Video: Differentiating Parametric Functions Mathematics • Higher Education

Given that π₯ = cos π‘ and π¦ = sin 2π‘, find dπ¦/dπ₯.

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### Video Transcript

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

In this question, weβve been given a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter; here thatβs π‘. And so, we recall that to find dπ¦ by dπ₯ β thatβs 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 weβre 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 thatβs 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. Itβs two cos two π‘ over negative sin π‘. And since weβre 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‘.