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 2 years ago
If \[g(x)=\left\{\begin{matrix}
\frac{2xy}{x^2+y^2} \text{ if }(x,y) \ne (0,0)\\
0\text{ if }(x,y) = (0,0)
\end{matrix}\right. \]
Show that \[ g_y(0,0) \] and \[ g_x(0,0) \]exist.
 2 years ago
If \[g(x)=\left\{\begin{matrix} \frac{2xy}{x^2+y^2} \text{ if }(x,y) \ne (0,0)\\ 0\text{ if }(x,y) = (0,0) \end{matrix}\right. \] Show that \[ g_y(0,0) \] and \[ g_x(0,0) \]exist.

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henpen
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not sure how to approach this how do you find \[ g_y(x,y) \] in the first place if \[ g(x,y) \] is 2 functions glued together?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0that is just partial derivative, the function is extended function ...use the inequality to find the derivative.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0use g(x) = 2xy/(x+y)

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Why would that work, given that at (0,0) the inequality does not hold necessarily?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0the function is continuous since f(0,0) = 0 and both limits are zero. show that the limit of partial derivatives f_x and f_y exits.

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Is it always the case that if we apply a 'gluing job' to all discontinuities, the function will become continuous?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0yeah ... since this is removable discontinuity.

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0By the way, here is the limit at x=2 the same coming from both sides/dw:1351706587261:dw
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