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Find the partial derivative of f(x,y)=(3x+3y)e^y

Mathematics
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y'[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])
\[f_x=3e^y\] \[f_y=(3x+3y)e^y+3e^y\]
f'x = 3e^y(3x+3y) f'y = 3e^y(3x+3y)

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Other answers:

why dn't you just multpply them out... i beleive that zarkon is correct
i am ;)
outside inside rule e^y will stay e^y always, leave inside then multiply inside respect to x and y.
zarkon is probably right though go with zarkon.
if you multiply it out Goat you get 3xe^y+3ye^y fx =3e^y + 0 fy=3xe^y+3e^y=3e^y(x+1)
could be errors but i just did that without paper
use the product rule for fy
ahh yes both y's so yeah
but the first is correct
yep i see thanks i am working on my understanding of this as well.
so how would I find fxx and fyy
just take the derivative again
\[f_{xx}=0\] \[f_{yy}=(3y+3x+6)e^y\]
and fxy = 3e^y?
yes
thank you!

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