Video: Differentiating Parametric Functions

Given that π‘₯ = cos 𝑑 and 𝑦 = sin 2𝑑, find d𝑦/dπ‘₯.

01:47

Video Transcript

Given that π‘₯ equals cos 𝑑 and 𝑦 equals sin two 𝑑, find d𝑦 by dπ‘₯.

In this question, we’ve been given a pair of parametric equations. These are equations for π‘₯ and 𝑦 in terms of a third parameter; here that’s 𝑑. And so, we recall that to find d𝑦 by dπ‘₯ β€” that’s the derivative of 𝑦 with respect to π‘₯ β€” as long as dπ‘₯ by d𝑑 is not equal to zero, we can divide d𝑦 by d𝑑 by dπ‘₯ by d𝑑. And so, it should be fairly clear to us that we’re going to need to differentiate π‘₯ with respect to 𝑑 and the equation for 𝑦 with respect to 𝑑.

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 𝑑 with respect to 𝑑 is negative sin 𝑑. Then, we could quote the general result for the derivative of sin of some constant of 𝑑. But we could also use the chain rule.

By letting 𝑒 be equal to two 𝑑, we find that d𝑦 by d𝑑 is equal to the derivative of 𝑒 with respect to 𝑑 β€” that’s the derivative of two 𝑑 with respect to 𝑑 β€” times the derivative of sin 𝑒 with respect to 𝑒. Well, the derivative of two 𝑑 with respect to 𝑑 is two whereas the derivative of sin 𝑒 with respect to 𝑒 is cos 𝑒. But since we let 𝑒 be equal to two 𝑑, we find that d𝑦 by d𝑑 is two cos of two 𝑑. Then, d𝑦 by dπ‘₯ is the quotient of these. It’s two cos two 𝑑 over negative sin 𝑑. And since we’re dividing a positive function by a negative function, we know that our final result will be a negative function. So, d𝑦 by dπ‘₯ is negative two cos two 𝑑 over sin 𝑑.

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