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

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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

Mathematics
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you would substitute 1 in for x and 2 in for y
(1^3+2^4)
HI!!

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is it \[\huge f_{xy}-f_{yx}\]?
partial derivatives?
no that can't be right, both second partials are zero must be something else
lol yes it is. sorry
both partials are zero everywhere
\[f(x,y) = (x^3+y^4) \]\] \[f_x(x,y)=3x^2\] so \[f_{xy}(x,y)=0\]
lol now i'm confused..
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
ooh okay sorry about that. thanks:D
maybe i am not understanding the question correctly
is the answer 8?
its a multiple choice question. i'm not really sure how to approach the question. could it be that the question is incorrect?
can you post a screen shot or sommat?
okay just give me a second
my pc is a little bit slow this evening, please bear with me
are you able to make that out?
lol you forgot the exponent!!
\[f(x,y) = (x^3+y^4)^5\]
LOL oh right i did, hehe SO SORRY!!
thats correct:D
How would i approach this question?
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
is that part clear?
oops gotta run, you will get more help i am sure
yeah sort of..
okay cool thanks:D
what exactly are you learning? mixed partials should be the same ordinarily
partial derivatives
its a new section of work so im really not too familiar with it
ah! so you need to go through the process of doing the derivative in that case, go for it.
okay cool, thanks dude:D

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