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Given that two π₯ cubed plus five π¦ cubed equals seven π₯π¦, determine ππ¦ ππ₯.
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The first thing we can see here is that actually our function is defined implicitly.
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So therefore, to determine ππ¦ ππ₯ and find our derivative, what we want to actually do is differentiate this implicitly.
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And the first stage of differentiating this function using implicit differentiation, is to actually differentiate it with respect to π₯.
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So in order to do this, weβre actually gonna deal with each term separately.
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So first of all, weβre gonna deal with our first term which is two π₯ cubed.
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So if we differentiate this with respect to π₯, weβre gonna get six π₯ squared.
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And we get that because we actually multiply the coefficient by the exponents, so two by three which gives a six.
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And then, we reduce the exponent by one, so six π₯ squared.
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Then weβre gonna have to deal with our second term differently.
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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₯.
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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₯.
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So thatβs fantastic.
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Weβve got our first two terms.
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And now, this is will equal to, the derivative of seven π₯π¦ with respect to π₯.
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And to enable us to do this, what weβre gonna use is the product rule.
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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₯.
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Okay, great.
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So letβs use this to actually differentiate seven π₯π¦.
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So first of all, letβs decide whatβs gonna be π’ and π£.
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So Iβm gonna take π’ to be seven π₯ and π£ to be π¦.
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So then next, I wanna find ππ’ ππ₯.
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Well, ππ’ ππ₯ is just going to be seven.
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And thatβs what we get if we differentiate seven π₯.
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And then, we need to find ππ£ ππ₯.
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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₯.
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Because if we differentiate π¦, we just get one.
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So one multiplied by ππ¦ ππ₯ is just ππ¦ ππ₯.
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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₯.
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So first of all, weβre gonna get seven π₯ multiplied by ππ¦ ππ₯.
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And thatβs because seven π₯ is our π’ and ππ¦ ππ₯ is our ππ£ ππ₯.
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And this is plus seven π¦ because π¦ is our π£ and seven is our ππ’ ππ₯.
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So great, weβve now found the derivative of seven π₯π¦ with respect to π₯.
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Okay, so weβre now on to the next phase for our implicit differentiation.
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And we want to do is actually rearrange to make ππ¦ ππ₯ the subject.
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So the first stage is to actually get our terms with ππ¦ ππ₯ on the same side of our equation.
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So we have six π₯ squared minus seven π¦ is equal to seven π₯ ππ¦ ππ₯ minus 15π¦ squared ππ¦ ππ₯.
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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.
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So then, we can actually divide through by seven π₯ minus 15π¦ squared.
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So then, we get six π₯ squared minus seven π¦ over seven π₯ minus 15π¦ squared is equal to ππ¦ ππ₯.
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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.