Question

If f(x,y) = x(raised to 3) + y (raised to 3)
---------------------------
x-y
; where x=/= y,
and f(x,y) = 0; when x=y.
Then find partial derivatives with respect to x (i.e. df/dx )and with respect to y.. (i.e. df/dy)

I'm stumped at this question. Can someone help me with this?

Answer 1

first you must find out whether it is differentiable at x=y

Answer 2

no you take partial derivatives

Answer 3

basically it means when respect to x, all y's are constants

Answer 4

and vice versa

Answer 5

i have a feeling limitless is doing this he's been typing for quite a while

Answer 6

@Outkast3r09, are you certain this isn't just easy partial differentiation?

I think one has to apply the product rule and that's the hardest part. Besides that, you separate constants and apply a substitution.

Answer 7

So u would first take a partial derivative with respect to x keeping y constant and then with repect to y keeping x constant. and yes I hope Limitless is solving. :) - Thats wat i feel too.. :3

Answer 8

Nope, I was solving. I saw that, based on my understanding of partials, it was fairly easy if you knew the right path.

Answer 9

well , f(x,y)=0 might not be differentiable i'm not sure off the top of my head as this was right out of my head at the end of calc 3

Answer 10

If \(f(x,y)=0,\) and it is not differentiable at \(x=y\), wouldn't you simply state that in your solution?

Answer 11

yes however you have to check to make sure

Answer 12

this is one of thos problems where you have topick specific paths to find if a limit exists

Answer 13

@Outkast3r09, I am not as familiar with the conditions of differentiation. Couldn't you simply say, "At \(x=y\), the function is undefined. Ergo, it is not differentiable there."

Answer 14

^Specific paths?????

Answer 15

i've found a problem much like this let me show you

Answer 16

kkk plz. :)

Answer 17

it must be continous to differentiate

Answer 18

once you've shown that it is not differentiable all you have to do is show that when x doesn't equal y your answer will be the partials

Answer 19

Wait how do u get -3x^(2)/2x^(2) in the question u gave??

Answer 20

So, take left and right hand limits. Presto.

Answer 21

they approached it from y=x so instead of as x,y)-> (0,0) it was as (x,x)-> 0

@dpaInc some assistance please

Answer 22

the denominator has an x, you have to use product rule and you might have to use an y=f(x) sub

Answer 23

\(\partial\) (\partial) is used for partial derivatives, not \(d.\)

Answer 24

^Yyeah i know. Dunno how to insert with keyboard. I assume u added through the equation button??

Answer 25

anyways, f(x,y)=0 is differential i believe as going from any line will get you 0

Answer 26

and since it simply says to take the derivatives, it's obvious that the first part of piecewise is also continuous however you' have to check it to make sure

Answer 27

if you didn't ask

Answer 28

Can i get something straight???? (Yes I'm terrible @derivatives/integration etc).
If u solve delta f/delta x = x^3+y^3/x-y wouldn't u do it using the U/v methood????

Answer 29

you could or you can do it by product rule

\[\frac{\delta f}{dx}=[(x^3+y^3)(x-y)^{-1}]'_x\]

Answer 30

\[\begin{align}
\frac{\partial f}{\partial x}&=\frac{\partial}{\partial x}\frac{x^3}{x-y}+\frac{\partial}{\partial x}\frac{y^3}{x-y}\\
&=\underbrace{\frac{\partial}{\partial x}\frac{x^3}{x-y}}_{\text{product rule}}+y^3\underbrace{\frac{\partial}{\partial x}\frac{1}{x-y}}_{u \text{ sub}}
\end{align}\]
Can you work from here?

Answer 31

hence, = [(derivative of u)v-(derivative of v)u]/v^2???
In this case => =[(3x^2)(x-y)] - [(1)(x^3+y^3)[/ (x-y)^2

Answer 32

\[f(x,y) = {x^3 + y^3 \over x - y} = (x^3 + y^3 )(x-y)^{-1}\]
\[{\partial f \over \partial x} = (x^3 + y^3) (-x+y)^{-2} + (x-y)^{-1} (3x)\]
\[{\partial f \over \partial y} = (x^3 + y^3) (x-y)^{-2} + (x-y)^{-1} (3y)\]

Answer 33

I can't remember, for partials, do you need to make a sub to get rid of y's or was that something else

Answer 34

with partials all other variables are constant

Answer 35

I'm really sorry, I dunno how to use the euqation thing. :/ So it may be very hard for u to understand wat I'm typing. :(

Answer 36

i mean't at the product i don't think it's needed , anyways

eyust answer is corect you just need to make a common denominator

Answer 37

I remember something in calculus 3 you had to change everything... i think that was when you had a parameter and you had to switch it all out of x and y's to s and t's

Answer 38

http://www.wolframalpha.com/input/?i=partial+derivative+with+respect+to+x+of+%28x^3%2By^3%29%2F%28x-y%29

http://www.wolframalpha.com/input/?i=partial+derivative+with+respect+to+y+of+%28x^3%2By^3%29%2F%28x-y%29

Answer 39

I'm pretty sure its not the switching both x and y. to s and t. nor to r and :theta:

Answer 40

o no not this question i'm just saying that later on you'll do something like that and it's insanely annoying.

anyways did you make a common denominator?

Answer 41

@eyust707 thanx for the wolfram links. Seems really helpful. Also Outkast I really appreciate ur help. :) I think I'll be ab;le to easily figure out from the link information. :D thank u again guys. (ALso @Limitless)

Answer 42

You're welcome. :D

Answer 43

:) I guess I close this thread now. Thanx to u folks. :)