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[tex]\begin{gathered} \rm{Let \: u=\frac{x-y}{2} \: and \: v=\frac{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 \: \frac{\partial F}{\partial x} and \:\frac{\partial F}{\partial y} \: in \: terms \: of } \\ \rm{ the \: partial \: derivatives \: \frac{\partial f}{\partial u} \: and \: \frac{\partial f}{\partial v} .}\end{gathered}[/tex] ​

Respuesta :

LammettHash

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

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