1 respuesta
Respuesta de Valero Angel Serrano Mercadal
1
Haremos so de la regla de la cadena.
$$\begin{align}&\frac{\partial z}{\partial x}=f\left( \frac{x+y}{x-y} \right)·\frac{x-y-(x+y)}{(x-y)^2}=\\ &\\ &f\left( \frac{x+y}{x-y} \right)·\frac{-2y}{(x-y)^2}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\frac{\partial z}{\partial y}=f\left( \frac{x+y}{x-y} \right)·\frac{x-y+x+y}{(x-y)^2}=\\ &\\ &f\left( \frac{x+y}{x-y} \right)·\frac{2x}{(x-y)^2}\\ &\\ &\\ &Luego\\ &\\ &x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=\\ &f\left( \frac{x+y}{x-y} \right)·\frac{-2xy}{(x-y)^2}+f\left( \frac{x+y}{x-y} \right)·\frac{2xy}{(x-y)^2}=0\\ &\\ &\\ &\\ &\\ &\end{align}$$Y eso es todo.
