Question

f(x,y) = (xy + x^2)/(x^2 + y^2) if (x,y) is not (0,0)
0 if (x,y) = (0,0)

How do I find the partial derivatives?

Answer 1

when taking the derivative with respect to only one variable, you treat the others as constants
for example\[\frac{\partial }{\partial y}(xy^2)=x\frac{\partial }{\partial y}(y^2)=2xy\]\[\frac{\partial }{\partial x}(xy^2)=y^2\frac{\partial }{\partial x}(x)=y^2\]do you see the idea now?

Answer 2

I can do that because it's a differentiable function right?

Answer 3

Do what?
You can only differentiate a function if it is differentiable on that interval, but if you are talking about why we treat the other variable we are not taking the derivative with respect to as constant that is due to the fact that it is a *partial* derivative of f(x,y) with respect to either x or y.

Answer 4

I though maybe because it has that separate point for (x,y) = (0,0) I shouldn't differentiate partially without doing some checking of differentiablity or something like that first.

Answer 5

Let me ask you another question, I have this other exercise here:

Be h: R -> R a derivable function such that h'(1) = 4. Consider f(x,y) = h(x/y). Calculate df/dx (1,1) and df/dy(1,1).

Answer 6

The reason that the function has a special definition for (x,y)=(0,0) is that (0,0) is undefined in the original function, but since the limit as x->0 and y->0 for f(x,y)=0 redefining f(0,0)=0 makes the function continuous. That is the only reason they mention that; so you can ignore that technicality.

Answer 7

Phew.

Answer 8

And what about my other question?

Answer 9

for the other guy you have to use the chain rule, but let me check my notes to remember exactly how :P

Answer 10

ok

Answer 11

oh I get it now, no notes required
just chain rule as follows\[f_x(x,y)=h'(\frac xy)\frac{\partial}{\partial x}\left(\frac xy\right)=?\]

Answer 12

fx(1,1) = h'(1) * 1 = 4
fy(1,1) = h'(1) * (-1) = -4

Answer 13

?

Answer 14

yep

Answer 15

Thank you.

Answer 16

pretty easy, huh?
enjoy it while it lasts, calc II gets pretty tricky later :)
welcome!

Answer 17

calc III*

Answer 18

That's actually from list 10, we're on 15 now. I'm not doing them in order. For some reason I'm forgetting many things lately....