if f : R2 -> R is differentiable and F(u,v) = f(u+v, u-v), show that (these are the curly d's) (dF/du)(dF/dv) = (df/dx)^2 - (df/dy)^2, where the functions on the right-hand side are evaluated at .
This is the exact question as written in the book. I think it's got a lot to do with the chain rule of multivariable calc, but I'm stuck... And very, very tired/exhausted. Any help would be greatly appreciated.
This is the exact question as written in the book. I think it's got a lot to do with the chain rule of multivariable calc, but I'm stuck... And very, very tired/exhausted. Any help would be greatly appreciated.
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(∂F/∂u) (∂F/∂v) = (∂f/∂x ∂x/∂u + ∂f/∂y ∂y/∂u) (∂f/∂x ∂x/∂v + ∂f/∂y ∂y/∂v)
. . . . . . . . . . . = (∂f/∂x ∂/∂u (u+v) + ∂f/∂y ∂/∂u (u-v)) (∂f/∂x ∂/∂v (u+v) + ∂f/∂y ∂/∂v (u-v))
. . . . . . . . . . . = (∂f/∂x (1) + ∂f/∂y (1)) (∂f/∂x (1) + ∂f/∂y (-1))
. . . . . . . . . . . = (∂f/∂x + ∂f/∂y) (∂f/∂x - ∂f/∂y)
. . . . . . . . . . . = (∂f/∂x)² - (∂f/∂y)²
Ματπmφm
. . . . . . . . . . . = (∂f/∂x ∂/∂u (u+v) + ∂f/∂y ∂/∂u (u-v)) (∂f/∂x ∂/∂v (u+v) + ∂f/∂y ∂/∂v (u-v))
. . . . . . . . . . . = (∂f/∂x (1) + ∂f/∂y (1)) (∂f/∂x (1) + ∂f/∂y (-1))
. . . . . . . . . . . = (∂f/∂x + ∂f/∂y) (∂f/∂x - ∂f/∂y)
. . . . . . . . . . . = (∂f/∂x)² - (∂f/∂y)²
Ματπmφm