### Video Transcript

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

So what we actually have here is a pair of parametric equations. So to differentiate this to find ππ¦ ππ₯, what we can actually do is use the chain rule which will allow us to differentiate them individually. And then, put it together. So to actually use the chain rule in this question, what we can say is that ππ¦ ππ₯ is gonna be equal to ππ¦ ππ‘ multiplied by ππ‘ ππ₯. So therefore, as I said, what it does is it allows us to actually differentiate each of our parametric equations separately. And then bring them together to find ππ¦ ππ₯.

So what Iβm gonna do is start by differentiating each of our equations. So Iβm gonna start with π¦ equals four π‘ squared plus five π‘. So Iβm gonna differentiate this to find ππ¦ ππ‘ which is gonna give us eight π‘ plus five. And just as a quick reminder of how we actually did that, Iβm looking at the first term. So what weβve got is the coefficient multiplied by the exponent, so a four by two, which gives us eight π‘. And then, what weβve done is actually reduce the exponent by one, so two minus one. So it just gives us π‘ to the power of one or π‘. Okay, so great. Weβve now differentiated π¦ equals four π‘ squared plus five π‘. Now, letβs look at the next equation.

Well, the next equation is π₯ is equal to four π‘ squared plus one. So therefore, if we differentiate this, weβre gonna find ππ₯ ππ‘. And this is gonna be equal to eight π‘. And we donβt have anything else to add on. Because if we differentiate one, we just get zero.

Okay, so we now got ππ¦ ππ‘ and ππ₯ ππ‘. However, if we look at the chain rule, what it says is that ππ¦ ππ₯ is gonna be equal to ππ¦ ππ‘ multiplied by ππ‘ ππ₯. So we can actually say, βAh, this is gonna be the reciprocal of what we found cause we found ππ₯ ππ‘β. But, if we want to actually find what ππ‘ ππ₯ is, what we can actually do is just find the reciprocal of ππ₯ ππ‘. So therefore, we can say that ππ‘ ππ₯ is equal to one over eight π‘. Okay, fantastic. Weβve got everything we need.

So what we can now do is actually find ππ¦ ππ₯ by applying our chain rule. So therefore, we can say that ππ¦ ππ₯ is equal to eight π‘ plus five multiplied by one over eight π‘. And thatβs because itβs ππ¦ ππ‘ multiplied by ππ‘ ππ₯. So we can say finally that given that π₯ equals four π‘ squared plus one and π¦ equals four π‘ squared plus five π‘, then ππ¦ ππ₯ is gonna be equal to eight π‘ plus five over eight π‘.