A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
partial derivative f(w,z) = w/(w^2+z^2)
anonymous
 3 years ago
partial derivative f(w,z) = w/(w^2+z^2)

This Question is Closed

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

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

anonymous
 3 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
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. Though it can be simplified by gathering like terms in the numerator.

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

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yah looks good for fz c:
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.