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# Find the derivative of the function$F(x) = \frac {x^2 - 5x^3 + \sqrt{x}}{x^2}$in two ways: by using the Quotient Rule and by simplifying tirst. Show that >our an\uer\ are equivalent Which method do you prefer?

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Derivatives

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RR

Ryan R.

October 6, 2021

I think you copied the problem wrong.

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

All right, We've got a question here. Function of votes equal to X squared my five x, you plus the square root of X all divided by X work. Mhm. All right. And we are asked to find the derivative in two ways by using the quoting rule and by simplifying first. All right. Now, start off by using a quota rule, which is where we take the derivative of the numerator multiplied by the denominator. Basically, you take a square. Do you do X? The numerator multiplied by the denominator. Subtracting the derivative, the derivative of the A derivative of the denominator multiplied by the new All of that divided by the numerous. The denominator street. Okay, so the quote and rule states that if you take the derivative, the denominator multiplied. She's a derivative of the numerator, multiplied by the denominator, subtracting the derivative of the denominator multiplied by the numerator, all divided by denominated squared. You can get the first derivative of your function. You have X squared multiplied by two X minus 15 x squared, plus squirrels of extra riveted. It's the same thing. One over one. Over two, one over to root books. Uh, X squared minus five x cubed. Yeah. And then Drew. Experts sentence to acts all over X before have you got right? And then we'll get a two x cubed minus 30 ext four. It looks like there's a bit of a typo here. This is actually supposed to be extra for. So then this would be more X to the vote. All right, so then when we go ahead and simplify all of this, we will get two extra there minus five X fourth, minus three halves X would X all over extra. Fourth. We've simplified that will get two x five minus three. Uh huh. Excuse the negative five. That will be our final answer when we try to take the derivative using the quoting rule. And then if we use the who tried to take the derivative after a simplification, Basically, what we're doing is we're taking our equation x to the fourth minus five X cubed on his actual X over X squared. Getting you simplify, it will get X squared minus five x minus A. Excuse me was This should be a process. We have a plus a x to the negative. Excellent. Negative three half And then when you take the derivative of this, we'll have to X minus five was minus a. We have X, the negative five. All right, we can see that this was actually a lot simpler to do. There's gonna be situations where it's not easy to simplify, so you're going to use the court to rule and you'll you'll be able to get your answer. But in this specific situation, you can see that it's a lot easier if we had just simplified. All right, well, I hope that clarifies the question there. Thank you so much for watching.

The University of Texas at Arlington

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Differentiation

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