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

1. becca18

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

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

3. The_G.O.A.T

f'x = 3e^y(3x+3y) f'y = 3e^y(3x+3y)

4. Outkast3r09

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

5. Zarkon

i am ;)

6. The_G.O.A.T

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

zarkon is probably right though go with zarkon.

8. Outkast3r09

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

could be errors but i just did that without paper

10. Zarkon

use the product rule for fy

11. Outkast3r09

ahh yes both y's so yeah

12. Outkast3r09

but the first is correct

13. The_G.O.A.T

yep i see thanks i am working on my understanding of this as well.

14. barry_chau

so how would I find fxx and fyy

15. Zarkon

just take the derivative again

16. Zarkon

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

17. barry_chau

and fxy = 3e^y?

18. Zarkon

yes

19. barry_chau

thank you!