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Find the partial derivatives with respect to (a) $x$ and (b) $y$.$$f(x, y)=\frac{x^{2}+y^{2}}{x^{2}-y^{2}}$$

(a) $\frac{-4 x y^{2}}{\left(x^{2}-y^{2}\right)^{2}}$(b) $\frac{4 x^{2} y}{\left(x^{2}-y^{2}\right)^{2}}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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05:31

Find the partial derivativ…

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02:43

05:17

01:49

Find the first partial der…

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06:48

05:27

00:55

03:33

02:44

00:59

Find all the second partia…

So if we want to find the partial derivative of this with respect to both X and why remember, what we're going to do is whichever partially take, we just assume all other variables are going to be constant. And we would treat it just like we would in a one dimensional or a one variable calculus sense. So for the first part, they wants to find the partial with respect. X. So we do, del by Dell X of this someone the left we can rewrite. This is F sub X, which means f with partial, the partial derivative of F with respect tax and then over here to take the derivative that will need to use quotient rule. So I always have to write this out. Otherwise I do it wrong. So the d X of hi over low So this is equal toe low d hi minus high de lo all over the square of what's below. So I'm just gonna write this up. First will be X squared minus y squared times del by Dell X oh x squared plus y squared minus so high x squared, plus y squared times del by Dell X of X squared minus y squared and then all over X squared minus y squared, squared. Now again, since we're taking the partial with respect to acts here, both of these wise here. When we take these partials, we treat as if they were constants. So this first one is just going to be two X plus zero because a constant square is still going to be a constant. And then over here will be two X minus zero. Yeah, on then. Since we have to accept every week ago at a factor, those two x is out. So it be two x time. So X squared minus y squared. And then we distribute this negative here. So be negative. X squared minus y squared and then all over X squared minus y squared. Spread on. Now these X words cancel out with each other, and then that would be negative To Weiss groups, this is going to be negative. Four x y squared all over X squared minus y squared, squared. And so this is our partial with respect to X, um, let me come back here and drag this down. So now to get our partial with respect, toe Why? So we do, Del by Dell. Why eso We'd still have to use quotient room. So over here we have f sub y or the partial derivative of effort. Perspective. Why? And then let me just write this out So low d hi, Minus hi. De lo and then all over the square of what's below? Yeah. Now, since we're doing the partial with respect to why these two axis here we treat this Constance. So this is going to be zero, um, plus two. Why? And then over here, this will be zero minus two. Why? Yeah, and at this point, we can factor out these two wise here. But it also knows this negative, and this negative will cancel out. So go ahead and factor out the two y. Then we'll be left with X squared minus y squared, plus x squared plus y squared all over, X squared minus. Why squared? And then all of this squared there are wise words. Cancel out. And then we would just end up with two x squared. So I'll give us two acts. 24 x squared y all over. Expert minus y squared squared. Okay. And so then This is our partial with respect to why so again, just to kind of, um, refresh what we did. Whenever we're taking these parcels, we just think all of the other variables are constants, and we just take the derivative just like we would in a single variable case.

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