anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ayee_ciera
  • ayee_ciera
you would substitute 1 in for x and 2 in for y
ayee_ciera
  • ayee_ciera
(1^3+2^4)
misty1212
  • misty1212
HI!!

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misty1212
  • misty1212
is it \[\huge f_{xy}-f_{yx}\]?
misty1212
  • misty1212
partial derivatives?
misty1212
  • misty1212
no that can't be right, both second partials are zero must be something else
anonymous
  • anonymous
lol yes it is. sorry
misty1212
  • misty1212
both partials are zero everywhere
misty1212
  • misty1212
\[f(x,y) = (x^3+y^4) \]\] \[f_x(x,y)=3x^2\] so \[f_{xy}(x,y)=0\]
anonymous
  • anonymous
lol now i'm confused..
misty1212
  • 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
  • anonymous
ooh okay sorry about that. thanks:D
misty1212
  • misty1212
maybe i am not understanding the question correctly
IrishBoy123
  • IrishBoy123
is the answer 8?
anonymous
  • 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
  • misty1212
can you post a screen shot or sommat?
anonymous
  • anonymous
okay just give me a second
anonymous
  • anonymous
my pc is a little bit slow this evening, please bear with me
anonymous
  • anonymous
anonymous
  • anonymous
are you able to make that out?
misty1212
  • misty1212
lol you forgot the exponent!!
misty1212
  • misty1212
\[f(x,y) = (x^3+y^4)^5\]
anonymous
  • anonymous
LOL oh right i did, hehe SO SORRY!!
anonymous
  • anonymous
thats correct:D
anonymous
  • anonymous
How would i approach this question?
misty1212
  • 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
  • misty1212
is that part clear?
misty1212
  • misty1212
oops gotta run, you will get more help i am sure
anonymous
  • anonymous
yeah sort of..
anonymous
  • anonymous
okay cool thanks:D
IrishBoy123
  • IrishBoy123
what exactly are you learning? mixed partials should be the same ordinarily
anonymous
  • anonymous
partial derivatives
anonymous
  • anonymous
its a new section of work so im really not too familiar with it
IrishBoy123
  • IrishBoy123
ah! so you need to go through the process of doing the derivative in that case, go for it.
anonymous
  • anonymous
okay cool, thanks dude:D

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