Question

`\begin{gathered} \rm{Let \: u=(x-y)/(2) \: and \: v=(x+y)/(2)} \: it \: changes \: f (u, v) \\ \rm into \: f (x, y).Use \: an \: appropriate \: form \\ \rm{of \: the \: chain \: rule \: to \: express \: the \: partial} \\ \rm{derivatives \: (\partial F)/(\partial x) and \:(\partial F)/(\partial y) \: in \: terms \: of } \\ \rm{ the \: partial \: derivatives \: (\partial f)/(\partial u) \: and \: (\partial f)/(\partial v) .}\end{gathered}` ​

Answer 1

If u = (x - y)/2 and v = (x + y)/2, then right away we have partial derivatives

∂u/∂x = 1/2

∂u/∂y = -1/2

∂v/∂x = 1/2

∂v/∂y = 1/2

By the chain rule, for a function F(x, y) = F(x(u, v), y(u, v)), we have

∂F/∂x = ∂F/∂u • ∂u/∂x + ∂F/∂v • ∂v/∂x

Then for the "given" function f(x, y), we have

∂f/∂x = 1/2 ∂f/∂u + 1/2 ∂f/∂v

and

∂f/∂y = -1/2 ∂f/∂u + 1/2 ∂f/∂v