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anonymous

  • one year ago

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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  1. ayee_ciera
    • one year ago
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    you would substitute 1 in for x and 2 in for y

  2. ayee_ciera
    • one year ago
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    (1^3+2^4)

  3. misty1212
    • one year ago
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    HI!!

  4. misty1212
    • one year ago
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    is it \[\huge f_{xy}-f_{yx}\]?

  5. misty1212
    • one year ago
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    partial derivatives?

  6. misty1212
    • one year ago
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    no that can't be right, both second partials are zero must be something else

  7. anonymous
    • one year ago
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    lol yes it is. sorry

  8. misty1212
    • one year ago
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    both partials are zero everywhere

  9. misty1212
    • one year ago
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    \[f(x,y) = (x^3+y^4) \]\] \[f_x(x,y)=3x^2\] so \[f_{xy}(x,y)=0\]

  10. anonymous
    • one year ago
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    lol now i'm confused..

  11. misty1212
    • one year ago
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    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

  12. anonymous
    • one year ago
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    ooh okay sorry about that. thanks:D

  13. misty1212
    • one year ago
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    maybe i am not understanding the question correctly

  14. IrishBoy123
    • one year ago
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    is the answer 8?

  15. anonymous
    • one year ago
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    its a multiple choice question. i'm not really sure how to approach the question. could it be that the question is incorrect?

  16. misty1212
    • one year ago
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    can you post a screen shot or sommat?

  17. anonymous
    • one year ago
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    okay just give me a second

  18. anonymous
    • one year ago
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    my pc is a little bit slow this evening, please bear with me

  19. anonymous
    • one year ago
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  20. anonymous
    • one year ago
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    are you able to make that out?

  21. misty1212
    • one year ago
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    lol you forgot the exponent!!

  22. misty1212
    • one year ago
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    \[f(x,y) = (x^3+y^4)^5\]

  23. anonymous
    • one year ago
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    LOL oh right i did, hehe SO SORRY!!

  24. anonymous
    • one year ago
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    thats correct:D

  25. anonymous
    • one year ago
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    How would i approach this question?

  26. misty1212
    • one year ago
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    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

  27. misty1212
    • one year ago
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    is that part clear?

  28. misty1212
    • one year ago
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    oops gotta run, you will get more help i am sure

  29. anonymous
    • one year ago
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    yeah sort of..

  30. anonymous
    • one year ago
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    okay cool thanks:D

  31. IrishBoy123
    • one year ago
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    what exactly are you learning? mixed partials should be the same ordinarily

  32. anonymous
    • one year ago
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    partial derivatives

  33. anonymous
    • one year ago
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    its a new section of work so im really not too familiar with it

  34. IrishBoy123
    • one year ago
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    ah! so you need to go through the process of doing the derivative in that case, go for it.

  35. anonymous
    • one year ago
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    okay cool, thanks dude:D

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