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Find the derivative of the function.

$ F(t) = \frac {t^2}{\sqrt {t^3 + 1}} $

$\frac{2 t\left(t^{3}+1\right)^{1 / 2}-\frac{3}{2} t^{4}\left(t^{3}+1\right)^{-1 / 2}}{t^{3}+1}$

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Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

here we have f of tea and we have a quotient. So to find the derivative, we're going to end up using the quotient rule. But inside the quotient, we have a composite function. So we will also be using the chain rule. And before I get going on that, I'm going to rewrite the function changing the square root signed to a 1/2 power. So I have t squared over the quantity t cubed plus one to the 1/2 power. Okay, so now for the derivative using the quotient rule, we have the bottom T Q plus one to the 1/2 times the derivative of the top to T minus the top T squared times the derivative of the bottom. Bring down the 1/2 and raise t cubed plus one to the negative 1/2 and then multiplied by the derivative of the inside three t squared. Now this is all going to be over the denominator squared. And when you square something to the 1/2 power is just going to be to the first power. Now there's a lot of things we could potentially do to simplify this, and our goal is going to be to get it to the point where it looks like the answer that was in the book. But just keep in mind that different instructors are going to have different expectations about simplifying, so you may be expected to do more or less than this. Okay, so the first thing I'm going to do is just right the first term over again. But I'm going to put the two T in the front so we have to tee times t cubed, plus one to the 1/2. And now, with second term, I'm going to notice that there's a 1/2 and there's a three so we could call that three house, and we also have a T squared and the T squared so minus will call that T to the fourth. So we have minus three halves Times T to the fourth Power Times T Q plus one to the negative, 1/2 over T cubed plus one. Now this is how they wrote the answer in the back of the books. For us, I can tell I probably would take it another step or two in terms of simplifying, so I'll show you some optional simplifying steps, all right, since we don't typically leave a negative exponents in there, one thing we could do is we could multiply the top and the bottom by T Q plus one to the 1/2 power. So when you distribute that, what's going to happen is you're gonna multiply it by this first part. And when you multiply T Q plus one to the 1/2 by T Q plus one to the 1/2 you get T Q plus one so that first part is going to be to tee times T Q plus one and then when you multiply it by the second part, the T T the T Q plus one to the negative 1/2 multiplied by the T Q plus one to the positive 1/2 is just going to multiply to give you T Q plus one of the zero and that is one. So then the second term is just minus three halves T to the fourth power, and then what we have on the bottom, we're gonna add the powers so one plus 1/2 is 1.5. So we have t cubed plus one to the three halfs power. Okay, now we can distribute our to tea. I'll take this over here, and we have to t to the fourth plus two t minus three halves t to the fourth all over t cubed plus one to the three house. So now we have some, like, terms. We have to t to the fourth. And we have negative three hats t to the fourth. When we subtract those, we just have 1/2 to the fourth plus two t overtaking plus one to the three house. Now, we don't typically leave a fraction inside of fractions, So the way to compensate for that is to multiply the top and bottom by two. Now, if we thought about that earlier, we could have multiplied the top and bottom by two. At the same time, we were multiplying the top and bottom by cheese squared, plus one to the 1/2. But that's OK. So then, for a final simplified answer, we would have t to the fourth plus 40 over two times t cubed plus one to the three house

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