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kingdan1550
 3 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.
kingdan1550
 3 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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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0by treating one as constant

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1quotient rule needed for \(f_x\) but not for \(f_y\) since in that case you view \(2x\) as a constant

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1who nellie, you are differentiating not integrating.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1\[f_x=\frac{(x^2+y^2)2x^2(2x)}{(x^2+y^2)^2}\] then some algebra to clean it up

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1\(f_y\) is easier because you treat the numerator as a constant you get \[f_y=\frac{2y}{(x^2+y^2)^2}\]

kingdan1550
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1maybe i made a mistake. lets do it for \(f_y\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1oh right, doh you are rigth about \(f_y\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1well, except the denominator should be squared

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1or in this case just \[\frac{d}{dy}\frac{1}{f(y)}=\frac{f'(y)}{f^2(y)}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1first one it looks like you are using the product rule, so i might be the same as my answer

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1oh yes, you are right. i missed a 2 from the 2x up top you are correct

kingdan1550
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, fantastic. Thanks a million.
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