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

Given that 2π‘₯Β³ + 5𝑦³ = 7π‘₯𝑦, determine 𝑑𝑦/𝑑π‘₯.

04:14

Video Transcript

Given that two π‘₯ cubed plus five 𝑦 cubed equals seven π‘₯𝑦, determine 𝑑𝑦 𝑑π‘₯.

The first thing we can see here is that actually our function is defined implicitly. So therefore, to determine 𝑑𝑦 𝑑π‘₯ 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, we’re actually gonna deal with each term separately. So first of all, we’re gonna deal with our first term which is two π‘₯ cubed. So if we differentiate this with respect to π‘₯, we’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 we’re gonna have to deal with our second term differently. And that’s because if we wanna find the derivative, with respect to π‘₯, of five 𝑦 cubed, it’s gonna be equal to the derivative with respect to 𝑦 multiplied by 𝑑𝑦 𝑑π‘₯. So therefore, our second term is gonna be 15𝑦 squared because that’s the derivative, with respect to 𝑦, of five 𝑦 cubed multiplied by 𝑑𝑦 𝑑π‘₯. So that’s fantastic. We’ve got our first two terms.

And now, this is will equal to, the derivative of seven π‘₯𝑦 with respect to π‘₯. And to enable us to do this, what we’re gonna use is the product rule. And the product rule tells us that if we have a function in the form 𝑦 equals 𝑒𝑣, then 𝑑𝑦 𝑑π‘₯ is gonna be equal to 𝑒 𝑑𝑣 𝑑π‘₯ plus 𝑣 𝑑𝑒 𝑑π‘₯. Okay, great. So let’s use this to actually differentiate seven π‘₯𝑦.

So first of all, let’s decide what’s gonna be 𝑒 and 𝑣. So I’m gonna take 𝑒 to be seven π‘₯ and 𝑣 to be 𝑦. So then next, I wanna find 𝑑𝑒 𝑑π‘₯. Well, 𝑑𝑒 𝑑π‘₯ is just going to be seven. And that’s what we get if we differentiate seven π‘₯. And then, we need to find 𝑑𝑣 𝑑π‘₯. Well, again, using the same rule we did earlier with the chain rule for our second term, we can say that 𝑑𝑣 𝑑π‘₯ is gonna be equal to the derivative of 𝑦 β€” which is our 𝑣 with respect to 𝑦, so 𝑑 𝑑𝑦 of 𝑦 β€” and then multiplied by 𝑑𝑦 𝑑π‘₯, which is just gonna be equal to 𝑑𝑦 𝑑π‘₯. Because if we differentiate 𝑦, we just get one. So one multiplied by 𝑑𝑦 𝑑π‘₯ is just 𝑑𝑦 𝑑π‘₯.

Okay, so now that we have 𝑒, 𝑣, 𝑑𝑒 𝑑π‘₯, and 𝑑𝑣 𝑑π‘₯, we can actually use the product rule to find the derivative of seven π‘₯𝑦 with respect to π‘₯. So first of all, we’re gonna get seven π‘₯ multiplied by 𝑑𝑦 𝑑π‘₯. And that’s because seven π‘₯ is our 𝑒 and 𝑑𝑦 𝑑π‘₯ is our 𝑑𝑣 𝑑π‘₯. And this is plus seven 𝑦 because 𝑦 is our 𝑣 and seven is our 𝑑𝑒 𝑑π‘₯. So great, we’ve now found the derivative of seven π‘₯𝑦 with respect to π‘₯. Okay, so we’re now on to the next phase for our implicit differentiation. And we want to do is actually rearrange to make 𝑑𝑦 𝑑π‘₯ the subject.

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

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

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