Question

How do I take the x-partial derivative of -5e^(-x^2 -y^2)(-2x^2 +2xy+1) fx=
These is what I have tried:
fx= [-5e^(-x^2 -y^2)(-4x +2y)] +[(-2x^2 +2xy+1)-5e^(-x^2 -y^2)*-2x]
-fx= [-5e^(-x^2 -y^2)(-4x +2y)] +[(-2x^2 +2xy+1)10xe^(-x^2 -y^2)]
-fx= [-5(-4x +2y)e^(-x^2 -y^2)] +[10x(-2x^2 +2xy+1)e^(-x^2 -y^2)]
-fx= -5(-4x +2y)e^(-x^2 -y^2) +10x(-2x^2 +2xy+1)e^(-x^2 -y^2) Stuck

Answer 1

Treat y​ as a constant

Take the constant - 5 out:

- 5 d / dx [ e^ ( - x^2 - y^2 ) ( - 2 x^2 + 2 x y +1) ]

Apply the Product Rule:

( f · g )' = f ' · g + f · g'

where

f = e^ ( - x^2 - y^2 ) , g = - 2 x^2 + 2 x y +1

f ' = - 2 e^ ( - x^2 - y^2 ) · x , g' = - 4 x + 2 y


- 5 d / dx [ e^ ( - x^2 - y^2 ) ( - 2 x^2 + 2 x y +1) ] =

- 5 [ - 2 e^ ( - x^2 - y^2 ) · x · ( - 2 x^2 + 2 x y +1 ) + e^ ( - x^2 - y^2 ) · ( - 4 x + 2 y ) ] =

- 5 e^ ( - x^2 - y^2 ) · [ - 2 · x · ( - 2 x^2 + 2 x y +1 ) - 4 x + 2 y ] =

- 5 e^ ( - x^2 - y^2 ) · ( 4 x^3 - 4 x^2 y - 2 x - 4 x + 2 y ) =

- 5 e^ ( - x^2 - y^2 ) · ( 4 x^3 - 4 x^2 y - 6 x + 2 y )

Answer 2

and you can factor out a final factor of 2, if you wish:
-10 e^(-x^2 - y^2) (2x^3 - 2x^2 y - 3x + y)

Answer 3

uhh, There is a typo with the function. It suppose to be 5e^(-x^2 -y^2)(-2x^2 +2xy+1). Still use Bosian same method to solve and end with oobleck answer right?

Answer 4

well duh - if the only change is to turn -5 to 5, then the final answer just changes sign.