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Find $ y' $ if $ y $ = In $ (x^2 + y^2). $

$y^{\prime}=\frac{2 x}{x^{2}+y^{2}-2 y}$

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Mati A.

November 6, 2019

where did the 1- go ?

Campbell University

Oregon State University

Harvey Mudd College

Boston College

So if we want to find why, Prime, um, we could try to take the derivative of this directly. But one thing I want to do is first apply E on each side and you'll maybe see why I want to do this in a moment. So this is going to give us e to the y is equal to X squared plus y squared. All right, so now what I'm going to do is go ahead and take the derivative here. So we have deep I d. X so on the left side will have to apply chain rule to this eso It would be e to the y times the derivative of why, which is just why prime And then over here, we'd end up with two x plus, uh, in the derivative of y squared will have to use chain rules. Well, so would be to y times the derivative of why? Which is why prime And now we can go ahead and get our white crimes on the same side. So I'm going to subtract this over so it be e to the Y times Why prime minus two y y prime is equal to x then we can factor out the white primes. We get white prime, Um, times e to the y minus two y is equal to two X. Then we could go ahead and divide this over. So we did it with why Prime is equal to two X over e to the Y, minus two y. And so now this, I believe, looks a little bit different from what they give us the answer in the back of the book. Um, so let's go ahead and rewrite this so we can get it toe look more like that solution. So, um, in the solution, they have this to extra number in the minus two. Why there? So all we really need to do is somehow get rid of this e to the Y. Well, if we come back up to here, notice how we have eat the Y is equal to x squared plus y squared. So we could come down here and just plug that in. So the why Prime is equal to two x over X squared plus y squared minus two by which is exactly what we were given. So I mean, both of these are valid solutions I mean, the only reason why I really rewrote it in this way is toe make it look more like the answer in the back of the book. Um, but you can write it is either one of these and these are still valid solutions. Um, if you were to have just taken the derivative directly. So like, if we didn't do this exponentially ation of e on each side and just did the derivative directly, then you would have got this second one a little bit more straightforward. Um, but doing it there was a lot more algebra involved. So that's why I prefer to do it this way. So, yeah, I encourage you try to do it the first way more directly, and then you would see that. Oh, yeah. Doing it that way is a lot more work. Yeah, I mean, Either way, though, you get the same answer

University of North Texas