# Video: Finding the First Derivative of a Function Defined by Parametric Equations

Given that π₯ = 3π‘Β³ + 1 and π¦ = 5π‘Β² β π‘, find ππ¦/ππ₯.

03:19

### Video Transcript

Given that π₯ equals three π‘ cubed plus one and π¦ equals five π‘ squared minus π‘, find ππ¦ ππ₯.

Well, here we actually have a pair of parametric equations. And actually to find ππ¦ ππ₯, we donβt have to write them in a Cartesian format. We can actually solve it using the chain rule. And the chain rule actually states that ππ¦ ππ₯ is equal to ππ¦ and in this case ππ‘, because π‘ is our other variable multiplied by ππ‘ ππ₯. And this is great because it enables us to actually differentiate each of our parametric equations separately and then combine them to find ππ¦ ππ₯.

So letβs get on and do that. So first of all, Iβm gonna start with π¦ equals five π‘ squared minus π‘. So to find ππ¦ ππ‘, Iβm gonna differentiate that. And my first term is just 10π‘. And we get that because itβs two multiplied by five because thatβs our exponent multiplied by a coefficient, which gives us 10. And thatβs π‘ to the power of two minus one. And thatβs because actually we subtract one from the exponent when we actually differentiate.

And then the second term differentiates two minus one. So great! Weβve got 10π‘ minus one. So now we can actually move on and find the second part. But the second part as you can see, as weβre trying to find ππ‘ ππ₯, weβre gonna have to do that in a slightly different way. And the way we do that is we actually find ππ₯ ππ‘ and then find the reciprocal of that because that will give us ππ‘ ππ₯.

Okay, so letβs get on and find that now. So we have π₯ is equal to three π‘ cubed plus one. So therefore weβre gonna get ππ₯ ππ‘ is equal to 9π‘ squared. Again and we got that because we have three multiplied by three cause itβs the exponent multiplied by the coefficient. And thatβs π‘ to the power of three subtract one. And thatβs that because we actually subtract one when weβre differentiating.

So weβve got 9π‘ squared. And positive one just differentiates to zero. So great! Weβve got 9π‘ squared for ππ₯ ππ‘. So now we can say that, therefore, ππ‘ ππ₯ is equal to one over 9π‘ squared. Thatβs because as we said before, we find a reciprocal of 9π‘ squared.

So great! Weβve got ππ¦ ππ‘ and ππ‘ ππ₯. Letβs use the chain rule to find ππ¦ ππ₯. So therefore using the chain rule, we can say that ππ¦ ππ₯ is equal to 10π‘ minus one multiplied by one over 9π‘ squared. So therefore we can say that ππ¦ ππ₯ is equal to 10π‘ minus one over 9π‘ squared, given that π₯ is equal to three π‘ cubed plus one and π¦ is equal to five π‘ squared minus π‘.

And now weβve got the final answer. I just wanna do a quick recap to remind us of what weβve done. So first of all, we can see that actually we have parametric equations. So weβre gonna use the chain rule to find ππ¦ ππ₯. And then we actually differentiate both of our parametric equations. And that gives us in this case ππ¦ ππ‘ and then ππ₯ ππ‘. But remembering that we actually had to find the reciprocal of ππ₯ ππ‘ to find ππ‘ ππ₯, then finally we apply the chain rule to get our final answer.