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anonymous
 4 years ago
partial derivative f(w,z) = w/(w^2+z^2)
anonymous
 4 years ago
partial derivative f(w,z) = w/(w^2+z^2)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you thinking of using quotient rule or product rule for this one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how would you use this as a product rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large f(w,z) =\frac{w}{w^2+z^2} = w\left(w^2+z^2\right)^{1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Though I don't think it would be much better that way. Just thought I'de get your take on it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think the whole point of this problem is to remind myself of quotient rule lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:) If \(f = \dfrac{u}{v}\), then \(f' = \dfrac{u'v  uv'}{v^2}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeap so then \[1(w^2+z^2)w(2w)/ (w^2+z^2)^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. Though it can be simplified by gathering like terms in the numerator.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I figured you did :) But better to mention it than not.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for fz is 2zw/(w^2+z^2)^2?

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Yah looks good for fz c:
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