How to find in which points both partial derivatives are equal to zero?

2,207

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.

Share:
2,207
javaniewbie
Author by

javaniewbie

Want to know all ins and outs of programming, one step at a time.

Updated on August 01, 2022

Comments

  • javaniewbie
    javaniewbie over 1 year

    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!

    • Daniel Fischer
      Daniel Fischer almost 10 years
      The conditions you get from each partial derivative are alternatives, from the first, you get $y = 0$ or $x = 1$.
  • javaniewbie
    javaniewbie almost 10 years
    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?
  • mathlove
    mathlove almost 10 years
    @javaniewbie: I added a bit. I hope this helps.
  • javaniewbie
    javaniewbie almost 10 years
    mathlove, you rock! I got it! Thank you very much!
  • mathlove
    mathlove almost 10 years
    You are welcome!