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I need to find points in which both partial derivatives are equal to zero. Partial derivative with respect to $x$ is equal to: $(1-x)ye^{-x-y^2}$ With respect to $y$: $(1-2y^2)xe^{-x-y^2}$

When I set both of them equal to zero, I get from the first equation $y=0$, $x=1$ and from the second $x=0$ and $y=\pm 1/\sqrt 2$. But this seems that I am not on the right track.

Would be grateful for any tips. Thank you!

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  • $\begingroup$ The conditions you get from each partial derivative are alternatives, from the first, you get $y = 0$ or $x = 1$. $\endgroup$ – Daniel Fischer Jan 11 '14 at 11:14
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Since $$(1-x)ye^{-x-y^2}=0\Rightarrow x=1\ \text{or}\ y=0,$$ $$ (1-2y^2)xe^{-x-y^2}=0\Rightarrow x=0\ \text{or}\ y=1/\sqrt{2}\ \text{or}\ y=-1/\sqrt 2,$$ you'll have $$(1-x)ye^{-x-y^2}=0\ \text{and}\ (1-2y^2)xe^{-x-y^2}=0\Rightarrow (x,y)=(0,0),(1,\pm 1/\sqrt 2).$$

Edit : You have $2\times 3=6$ possibilities as the followings (Note that 'or' and 'and'):

$$x=1\ \text{and}\ x=0,$$ $$x=1\ \text{and}\ y=1/\sqrt 2,$$ $$x=1\ \text{and}\ y=-1/\sqrt 2,$$ $$y=0\ \text{and}\ x=0,$$ $$y=0\ \text{and}\ y=1/\sqrt 2,$$ $$y=0\ \text{and}\ y=-1/\sqrt 2.$$

You'll see some of these have to be excluded.

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  • $\begingroup$ Thank you very much for your answer! Could you be so kind to explain how you get from having roots of the equations to finding the points? $\endgroup$ – javaniewbie Jan 11 '14 at 11:31
  • $\begingroup$ @javaniewbie: I added a bit. I hope this helps. $\endgroup$ – mathlove Jan 11 '14 at 11:36
  • $\begingroup$ mathlove, you rock! I got it! Thank you very much! $\endgroup$ – javaniewbie Jan 11 '14 at 11:38
  • $\begingroup$ You are welcome! $\endgroup$ – mathlove Jan 11 '14 at 11:38

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