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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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henpenBest ResponseYou've already chosen the best response.0
I'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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
that is just partial derivative, the function is extended function ...use the inequality to find the derivative.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
use g(x) = 2xy/(x+y)
 one year ago

henpenBest ResponseYou've already chosen the best response.0
Why would that work, given that at (0,0) the inequality does not hold necessarily?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
the 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.
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
yeah ... since this is removable discontinuity.
 one year ago

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