# Video: Finding the First Derivative of a Function Defined Implicitly Using Implicit Differentiation and the Product Rule

Given that 2π₯Β³ + 5π¦Β³ = 7π₯π¦, determine ππ¦/ππ₯.

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### Video Transcript

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

The first thing we can see here is that actually our function is defined implicitly. So therefore, to determine ππ¦ ππ₯ 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, weβre actually gonna deal with each term separately. So first of all, weβre gonna deal with our first term which is two π₯ cubed. So if we differentiate this with respect to π₯, weβre 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 weβre gonna have to deal with our second term differently. And thatβs because if we wanna find the derivative, with respect to π₯, of five π¦ cubed, itβs gonna be equal to the derivative with respect to π¦ multiplied by ππ¦ ππ₯. So therefore, our second term is gonna be 15π¦ squared because thatβs the derivative, with respect to π¦, of five π¦ cubed multiplied by ππ¦ ππ₯. So thatβs fantastic. Weβve 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 weβre 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 ππ¦ ππ₯ is gonna be equal to π’ ππ£ ππ₯ plus π£ ππ’ ππ₯. Okay, great. So letβs use this to actually differentiate seven π₯π¦.

So first of all, letβs decide whatβs gonna be π’ and π£. So Iβm gonna take π’ to be seven π₯ and π£ to be π¦. So then next, I wanna find ππ’ ππ₯. Well, ππ’ ππ₯ is just going to be seven. And thatβs what we get if we differentiate seven π₯. And then, we need to find ππ£ ππ₯. Well, again, using the same rule we did earlier with the chain rule for our second term, we can say that ππ£ ππ₯ is gonna be equal to the derivative of π¦ β which is our π£ with respect to π¦, so π ππ¦ of π¦ β and then multiplied by ππ¦ ππ₯, which is just gonna be equal to ππ¦ ππ₯. Because if we differentiate π¦, we just get one. So one multiplied by ππ¦ ππ₯ is just ππ¦ ππ₯.

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

So the first stage is to actually get our terms with ππ¦ ππ₯ on the same side of our equation. So we have six π₯ squared minus seven π¦ is equal to seven π₯ ππ¦ ππ₯ minus 15π¦ squared ππ¦ ππ₯. 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 weβve taken ππ¦ ππ₯ out as a factor, so ππ¦ ππ₯ 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 ππ¦ ππ₯.

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