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# Use the Chain Rule and the Product Rule to give an alternative proof of the Quotient Rule.[ $Hint:$ Write $f(x)/ g(x) = f(x)[g(x)]^{-1}.]$

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

All right, So we're going to find an alternative of the quotient rule our way to prove the quotient rule by taking the derivative of a product and using the chain rule. And so what we're aiming for is the derivative of a quotient. And so what we're going to do is take the derivative of this product instead. So according to the product rule, we have the first f of x times, the derivative of the second. Now, when we go to take the derivative of the second, we have to use the chain rule. So we're going to bring down the negative one, and then we're going to raise G of X to the negative second power, and then we're going to multiply by the derivative of G of X, which is G prime of X. Okay, so so far, we have the first times the derivative of the second. Now we're going to have plus the second, which was G of X, the negative one times the derivative of the first. And that would be the derivative of F. Okay, now our goal is just to manipulate this and change how it looks until it looks like the quotient rule. So let's notice that we have some negative exponents so we can turn some things into fractions. So the whole first term here is going to be the opposite of F of X times G prime of X over G of X squared. And that's looking pretty good because that looks a lot like our quotient rule, at least part of it. And then this whole second term here because we have a negative exponents, we can write that as f of X over g of X. Now, we have different denominators in these fractions. And if we want a common denominator, we should multiply the second fraction by G of X over G of X. Okay, that's going to give us our denominator of g of x quantity squared, which is what we want for the quotient rule and for the numerator. What I'm going to do, IHS Well, I'm gonna go back and write my prime sign on that F that f prime of X there it lost its prime Simon's prime sign in translation. Okay, now for the numerator. So I'm just going to write this part first, so that would be G of X Times f prime of X and then I'm going to write this part second. So I have minus f of x times g prime of X and this is the quotient rule.

Oregon State University

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