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



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becca18
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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])

Zarkon
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\[f_x=3e^y\]
\[f_y=(3x+3y)e^y+3e^y\]

The_G.O.A.T
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f'x = 3e^y(3x+3y)
f'y = 3e^y(3x+3y)

Outkast3r09
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why dn't you just multpply them out... i beleive that zarkon is correct

Zarkon
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i am ;)

The_G.O.A.T
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outside inside rule e^y will stay e^y always, leave inside then multiply inside respect to x and y.

The_G.O.A.T
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zarkon is probably right though go with zarkon.

Outkast3r09
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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)

Outkast3r09
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could be errors but i just did that without paper

Zarkon
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use the product rule for fy

Outkast3r09
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ahh yes both y's so yeah

Outkast3r09
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but the first is correct

The_G.O.A.T
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yep i see thanks i am working on my understanding of this as well.

barry_chau
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so how would I find fxx and fyy

Zarkon
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just take the derivative again

Zarkon
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\[f_{xx}=0\]
\[f_{yy}=(3y+3x+6)e^y\]

barry_chau
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and fxy = 3e^y?

Zarkon
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yes

barry_chau
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thank you!