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becca18
 3 years ago
Best ResponseYou've already chosen the best response.0y'[x] == (3 E^y + Derivative[1, 0][f][x, y])/(3 E^y + 3 E^y x + 3 E^y y  Derivative[0, 1][f][x, y])

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.2\[f_x=3e^y\] \[f_y=(3x+3y)e^y+3e^y\]

The_G.O.A.T
 3 years ago
Best ResponseYou've already chosen the best response.0f'x = 3e^y(3x+3y) f'y = 3e^y(3x+3y)

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0why dn't you just multpply them out... i beleive that zarkon is correct

The_G.O.A.T
 3 years ago
Best ResponseYou've already chosen the best response.0outside inside rule e^y will stay e^y always, leave inside then multiply inside respect to x and y.

The_G.O.A.T
 3 years ago
Best ResponseYou've already chosen the best response.0zarkon is probably right though go with zarkon.

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0if you multiply it out Goat you get 3xe^y+3ye^y fx =3e^y + 0 fy=3xe^y+3e^y=3e^y(x+1)

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0could be errors but i just did that without paper

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.2use the product rule for fy

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0ahh yes both y's so yeah

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0but the first is correct

The_G.O.A.T
 3 years ago
Best ResponseYou've already chosen the best response.0yep i see thanks i am working on my understanding of this as well.

barry_chau
 3 years ago
Best ResponseYou've already chosen the best response.0so how would I find fxx and fyy

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.2just take the derivative again

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.2\[f_{xx}=0\] \[f_{yy}=(3y+3x+6)e^y\]
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