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artix_17
Group Title
Partial derivative, pls help !!
f(w,z)= w(w^2+z^2)^1
Find f'w
 one year ago
 one year ago
artix_17 Group Title
Partial derivative, pls help !! f(w,z)= w(w^2+z^2)^1 Find f'w
 one year ago
 one year ago

This Question is Closed

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
What is the biggest problem for you?
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
I can't seem to get the answer
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Do you know how to calculate the ordinary derivative, not the partial?
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
Yes, for this question I use product rule, but just can't get the answer
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Write here what you've got.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
You can draw it using the "Draw" button.
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
f'w = w X (1)(w^2+z^2)^2 X 2w + (w^2+z^2)^1 (1) = 2w^3 (w^2+z^2) + (w^2+z^2)^1
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
= 2w^3 (w^2+z^2)^2 + (w^2+z^2)^1
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
You have to check the last answer very carefully. You are doing this right, but you are making mistakes. Check them.
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
Which part is the mistake?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
f'w = w X (1)(w^2+z^2)^2 X 2w + (w^2+z^2)^1  This is OK, but this = 2w^3 (w^2+z^2) + (w^2+z^2)^1 is not. Try to find the common denominator in the first one and subtract the fractions.
 one year ago

artix_17 Group TitleBest ResponseYou've already chosen the best response.0
oh okay got it, thanks
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
You answered your question by yourself, I did not help you.
 one year ago
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