hey guys, some help please..
if f(x,y) = (x^3+y^4) then f(xy)-f(yx) at the point (1,2) is.. how would i solve this question. Thanks in advance:D

- anonymous

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- ayee_ciera

you would substitute 1 in for x and 2 in for y

- ayee_ciera

(1^3+2^4)

- misty1212

HI!!

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## More answers

- misty1212

is it \[\huge f_{xy}-f_{yx}\]?

- misty1212

partial derivatives?

- misty1212

no that can't be right, both second partials are zero
must be something else

- anonymous

lol yes it is. sorry

- misty1212

both partials are zero everywhere

- misty1212

\[f(x,y) = (x^3+y^4) \]\]
\[f_x(x,y)=3x^2\] so \[f_{xy}(x,y)=0\]

- anonymous

lol now i'm confused..

- misty1212

the partial wrt x, treat y as a constant so it is just \(3x^2\)
then if you take the partial wrt y, there is no y in it, so it is just zero

- anonymous

ooh okay sorry about that. thanks:D

- misty1212

maybe i am not understanding the question correctly

- IrishBoy123

is the answer 8?

- anonymous

its a multiple choice question. i'm not really sure how to approach the question. could it be that the question is incorrect?

- misty1212

can you post a screen shot or sommat?

- anonymous

okay just give me a second

- anonymous

my pc is a little bit slow this evening, please bear with me

- anonymous

##### 1 Attachment

- anonymous

are you able to make that out?

- misty1212

lol you forgot the exponent!!

- misty1212

\[f(x,y) = (x^3+y^4)^5\]

- anonymous

LOL oh right i did, hehe SO SORRY!!

- anonymous

thats correct:D

- anonymous

How would i approach this question?

- misty1212

i think it is still has to be zero, but at now we can compute it
first
\[f_x=5(x^3+y^3)^4\times 3x^2\] via the chain rule

- misty1212

is that part clear?

- misty1212

oops gotta run, you will get more help i am sure

- anonymous

yeah sort of..

- anonymous

okay cool thanks:D

- IrishBoy123

what exactly are you learning?
mixed partials should be the same ordinarily

- anonymous

partial derivatives

- anonymous

its a new section of work so im really not too familiar with it

- IrishBoy123

ah!
so you need to go through the process of doing the derivative
in that case, go for it.

- anonymous

okay cool, thanks dude:D

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