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barry_chau

  • 4 years ago

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

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  1. becca18
    • 4 years ago
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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])

  2. Zarkon
    • 4 years ago
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    \[f_x=3e^y\] \[f_y=(3x+3y)e^y+3e^y\]

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

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

  5. Zarkon
    • 4 years ago
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    i am ;)

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

  7. The_G.O.A.T
    • 4 years ago
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    zarkon is probably right though go with zarkon.

  8. Outkast3r09
    • 4 years ago
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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)

  9. Outkast3r09
    • 4 years ago
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    could be errors but i just did that without paper

  10. Zarkon
    • 4 years ago
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    use the product rule for fy

  11. Outkast3r09
    • 4 years ago
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    ahh yes both y's so yeah

  12. Outkast3r09
    • 4 years ago
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    but the first is correct

  13. The_G.O.A.T
    • 4 years ago
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    yep i see thanks i am working on my understanding of this as well.

  14. barry_chau
    • 4 years ago
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    so how would I find fxx and fyy

  15. Zarkon
    • 4 years ago
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    just take the derivative again

  16. Zarkon
    • 4 years ago
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    \[f_{xx}=0\] \[f_{yy}=(3y+3x+6)e^y\]

  17. barry_chau
    • 4 years ago
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    and fxy = 3e^y?

  18. Zarkon
    • 4 years ago
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    yes

  19. barry_chau
    • 4 years ago
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    thank you!

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