### Video Transcript

Given that two π₯ cubed plus five π¦ cubed equals seven π₯π¦, determine ππ¦ ππ₯.

The first thing we can see here is that actually our function is defined implicitly. So therefore, to determine ππ¦ ππ₯ and find our derivative, what we want to actually do is differentiate this implicitly. And the first stage of differentiating this function using implicit differentiation, is to actually differentiate it with respect to π₯. So in order to do this, weβre actually gonna deal with each term separately. So first of all, weβre gonna deal with our first term which is two π₯ cubed. So if we differentiate this with respect to π₯, weβre gonna get six π₯ squared. And we get that because we actually multiply the coefficient by the exponents, so two by three which gives a six. And then, we reduce the exponent by one, so six π₯ squared.

Then weβre gonna have to deal with our second term differently. And thatβs because if we wanna find the derivative, with respect to π₯, of five π¦ cubed, itβs gonna be equal to the derivative with respect to π¦ multiplied by ππ¦ ππ₯. So therefore, our second term is gonna be 15π¦ squared because thatβs the derivative, with respect to π¦, of five π¦ cubed multiplied by ππ¦ ππ₯. So thatβs fantastic. Weβve got our first two terms.

And now, this is will equal to, the derivative of seven π₯π¦ with respect to π₯. And to enable us to do this, what weβre gonna use is the product rule. And the product rule tells us that if we have a function in the form π¦ equals π’π£, then ππ¦ ππ₯ is gonna be equal to π’ ππ£ ππ₯ plus π£ ππ’ ππ₯. Okay, great. So letβs use this to actually differentiate seven π₯π¦.

So first of all, letβs decide whatβs gonna be π’ and π£. So Iβm gonna take π’ to be seven π₯ and π£ to be π¦. So then next, I wanna find ππ’ ππ₯. Well, ππ’ ππ₯ is just going to be seven. And thatβs what we get if we differentiate seven π₯. And then, we need to find ππ£ ππ₯. Well, again, using the same rule we did earlier with the chain rule for our second term, we can say that ππ£ ππ₯ is gonna be equal to the derivative of π¦ β which is our π£ with respect to π¦, so π ππ¦ of π¦ β and then multiplied by ππ¦ ππ₯, which is just gonna be equal to ππ¦ ππ₯. Because if we differentiate π¦, we just get one. So one multiplied by ππ¦ ππ₯ is just ππ¦ ππ₯.

Okay, so now that we have π’, π£, ππ’ ππ₯, and ππ£ ππ₯, we can actually use the product rule to find the derivative of seven π₯π¦ with respect to π₯. So first of all, weβre gonna get seven π₯ multiplied by ππ¦ ππ₯. And thatβs because seven π₯ is our π’ and ππ¦ ππ₯ is our ππ£ ππ₯. And this is plus seven π¦ because π¦ is our π£ and seven is our ππ’ ππ₯. So great, weβve now found the derivative of seven π₯π¦ with respect to π₯. Okay, so weβre now on to the next phase for our implicit differentiation. And we want to do is actually rearrange to make ππ¦ ππ₯ the subject.

So the first stage is to actually get our terms with ππ¦ ππ₯ on the same side of our equation. So we have six π₯ squared minus seven π¦ is equal to seven π₯ ππ¦ ππ₯ minus 15π¦ squared ππ¦ ππ₯. So now, we can actually factor the right-hand side of our equation which will give us six π₯ squared minus seven π¦ is equal to, now weβve taken ππ¦ ππ₯ out as a factor, so ππ¦ ππ₯ multiplied by seven π₯ minus 15π¦ squared. So then, we can actually divide through by seven π₯ minus 15π¦ squared. So then, we get six π₯ squared minus seven π¦ over seven π₯ minus 15π¦ squared is equal to ππ¦ ππ₯.

So therefore, we can say that given that two π₯ cubed plus five π¦ cubed equals seven π₯π¦, ππ¦ ππ₯ is equal to six π₯ squared minus seven π¦ over seven π₯ minus 15π¦ squared.