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$f(x,y)= \ln(x^2 + y^2 +1)$

The partial derivatives of $f(x,y)$ being:

$f_x(x,y) = \frac{2x}{x^2+y^2+1}$ and $f_y(x,y)=\frac{2y}{x^2+y^2+1}$

Setting each partial derivative equal to zero, adding the two linear equations two, and solving for $y$, I get:

$y = - x$

In this problem Finding the $x$ and $y$ values such that the partial derivatives are zero simultaneously, there was only one ordered pair. In the problem before us, I solved it the same way I did in the problem given in the link Does the solution above imply that there are an infinite amount of $(x,y)$ pairs, such that there partial derivatives are zero? I ask, because this particular problem is an even number, and the answer key provides no answers for such questions.

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  • $\begingroup$ No, you get $x = y = 0$. $\endgroup$ – vonbrand Mar 10 '13 at 16:04
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Hint: Since the denominators of these partial derivatives are never zero, then the partial derivatives are equal to zero precisely when their numerators are.

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  • $\begingroup$ Yes, that is what I did. Then I had simultaneous equations, $2y = 0$ and $2x = 0$. When I added the two equations together, I got $2x +2y = 0 \implies y = -x$. $\endgroup$ – Mack Mar 10 '13 at 14:42
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    $\begingroup$ As in the other problem, you've added your equations together unnecessarily. You should only add two equations together if it eliminates variables. In this case, your variables are all but solved for. $2y=0$ implies $y=0$ and $2x=0$ implies $x=0$. Thus, $(0,0)$ is the solution point. $\endgroup$ – Cameron Buie Mar 10 '13 at 14:45
  • $\begingroup$ Oh, I see. Thank you, Cameron! $\endgroup$ – Mack Mar 10 '13 at 14:47
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    $\begingroup$ P.S.: If you do add equations together to eliminate variables, don't forget to back-substitute at the end, so you can solve for all of the variables! $\endgroup$ – Cameron Buie Mar 10 '13 at 14:49

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