### Video Transcript

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

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

So letβs get on and do that. So first of all, Iβm gonna start with π¦ equals five π‘ squared minus π‘. So to find ππ¦ ππ‘, Iβm gonna differentiate that. And my first term is just 10π‘. And we get that because itβs two multiplied by five because thatβs our exponent multiplied by a coefficient, which gives us 10. And thatβs π‘ to the power of two minus one. And thatβs because actually we subtract one from the exponent when we actually differentiate.

And then the second term differentiates two minus one. So great! Weβve 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 weβre trying to find ππ‘ ππ₯, weβre gonna have to do that in a slightly different way. And the way we do that is we actually find ππ₯ ππ‘ and then find the reciprocal of that because that will give us ππ‘ ππ₯.

Okay, so letβs get on and find that now. So we have π₯ is equal to three π‘ cubed plus one. So therefore weβre gonna get ππ₯ ππ‘ is equal to 9π‘ squared. Again and we got that because we have three multiplied by three cause itβs the exponent multiplied by the coefficient. And thatβs π‘ to the power of three subtract one. And thatβs that because we actually subtract one when weβre differentiating.

So weβve got 9π‘ squared. And positive one just differentiates to zero. So great! Weβve got 9π‘ squared for ππ₯ ππ‘. So now we can say that, therefore, ππ‘ ππ₯ is equal to one over 9π‘ squared. Thatβs because as we said before, we find a reciprocal of 9π‘ squared.

So great! Weβve got ππ¦ ππ‘ and ππ‘ ππ₯. Letβs use the chain rule to find ππ¦ ππ₯. So therefore using the chain rule, we can say that ππ¦ ππ₯ is equal to 10π‘ minus one multiplied by one over 9π‘ squared. So therefore we can say that ππ¦ ππ₯ 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 weβve got the final answer. I just wanna do a quick recap to remind us of what weβve done. So first of all, we can see that actually we have parametric equations. So weβre gonna use the chain rule to find ππ¦ ππ₯. And then we actually differentiate both of our parametric equations. And that gives us in this case ππ¦ ππ‘ and then ππ₯ ππ‘. But remembering that we actually had to find the reciprocal of ππ₯ ππ‘ to find ππ‘ ππ₯, then finally we apply the chain rule to get our final answer.