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.

Question

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.

anonymous · Answer

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?

experimentx · Answer

that is just partial derivative, the function is extended function ...use the inequality to find the derivative.

experimentx · Answer

use g(x) = 2xy/(x+y)

anonymous · Answer

Why would that work, given that at (0,0) the inequality does not hold necessarily?

experimentx · Answer

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.

anonymous · Answer

Is it always the case that if we apply a 'gluing job' to all discontinuities, the function will become continuous?

experimentx · Answer

yeah ... since this is removable discontinuity.

anonymous · Answer

By the way, here is the limit at x=2 the same coming from both sides/|dw:1351706587261:dw|