Need help with this.
Suppose that $G(u, v)$ is a differentiable function of two variables and that $g(x, y) = G(x/y , y/x)$. Show that $xg_x(x, y) + yg_y(x, y) = 0$.
Where $g_x(x,y)$ and $g_y(x,y)$ are the partials.
2026-04-17 18:06:14.1776449174
Show that $xg_x(x, y) + yg_y(x, y) = 0$.
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Applying the chain rule $$\partial_x g(x,y) = G_x(x/y,y/x)\partial_x (x/y) + G_y(x/y,y/x)\partial_x(y/x) $$ $$ = \frac{1}{y} G_x(x/y,y/x) - \frac{y}{x^2}G_y(x/y,y/x)$$ and similarly $$ \partial_y g(x,y) = G_x(x/y,y/x)\partial_y(x/y) + G_y(x/y,y/x)\partial_y(y/x) $$ $$ = \frac{-x}{y^2}G_x(x/y,y/x) + \frac{1}{x}G_x(x/y,y/x). $$ Multiply the first equation by $x$ and the second by $y$, then add them to get your result.