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anonymous
 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.
anonymous
 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.

This Question is Closed

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0quotient rule needed for \(f_x\) but not for \(f_y\) since in that case you view \(2x\) as a constant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0who nellie, you are differentiating not integrating.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f_x=\frac{(x^2+y^2)2x^2(2x)}{(x^2+y^2)^2}\] then some algebra to clean it up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(f_y\) is easier because you treat the numerator as a constant you get \[f_y=\frac{2y}{(x^2+y^2)^2}\]

anonymous
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe i made a mistake. lets do it for \(f_y\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh right, doh you are rigth about \(f_y\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, except the denominator should be squared

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or in this case just \[\frac{d}{dy}\frac{1}{f(y)}=\frac{f'(y)}{f^2(y)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first one it looks like you are using the product rule, so i might be the same as my answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yes, you are right. i missed a 2 from the 2x up top you are correct

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