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kingdan1550

  • 4 years ago

How do I partially differentiate f = 2x/(x^2 +y^2) wrt x and y? Is it by the quotient rule? Any help is very much appreciated.

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  1. experimentX
    • 4 years ago
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    by treating one as constant

  2. anonymous
    • 4 years ago
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    quotient rule needed for \(f_x\) but not for \(f_y\) since in that case you view \(2x\) as a constant

  3. kingdan1550
    • 4 years ago
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    Thank you

  4. anonymous
    • 4 years ago
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    who nellie, you are differentiating not integrating.

  5. anonymous
    • 4 years ago
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    \[f_x=\frac{(x^2+y^2)2-x^2(2x)}{(x^2+y^2)^2}\] then some algebra to clean it up

  6. niki
    • 4 years ago
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    thnxxxxxx

  7. anonymous
    • 4 years ago
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    \(f_y\) is easier because you treat the numerator as a constant you get \[f_y=-\frac{2y}{(x^2+y^2)^2}\]

  8. kingdan1550
    • 4 years ago
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    I ended up getting 2/(x^2 + y^2) - 4x^2/(x^2 +y^2)^2 for the partial derivative wrt x and -4yx/(x^2 +y^2) for y. Any idea where I might be going wrong? Thanks again

  9. anonymous
    • 4 years ago
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    maybe i made a mistake. lets do it for \(f_y\)

  10. anonymous
    • 4 years ago
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    oh right, doh you are rigth about \(f_y\)

  11. anonymous
    • 4 years ago
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    well, except the denominator should be squared

  12. kingdan1550
    • 4 years ago
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    Why is that?

  13. anonymous
    • 4 years ago
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    quotient rule

  14. anonymous
    • 4 years ago
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    or in this case just \[\frac{d}{dy}\frac{1}{f(y)}=-\frac{f'(y)}{f^2(y)}\]

  15. anonymous
    • 4 years ago
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    first one it looks like you are using the product rule, so i might be the same as my answer

  16. anonymous
    • 4 years ago
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    oh yes, you are right. i missed a 2 from the 2x up top you are correct

  17. kingdan1550
    • 4 years ago
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    Ah, fantastic. Thanks a million.

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