# Critical points where a partial derivate is 0

I have a function for which I have calculated:

$\dfrac{d}{dx}f(x,y)=0$

and

$\dfrac{d}{dy}f(x,y)=2y+\cos(y)$

How can I proceed to calculate the critical points?

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Looks like your function only depends on $y$! –  Mercy Sep 27 '12 at 20:55

## 1 Answer

Just as with equations in one variable, determine when the partial derivatives become 0. Here, one is already zero so no information there...

But the other one is $2y+\cos(y)=0$. It looks like there isn't a closed form for the solution, so you'd need an approximation...

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If I solve this with the calculator I get -0.45. What will the critical points be? (?,-0.45) I don't understand how to get the x value(s). –  Farmor Sep 27 '12 at 20:32
Since there are no restrictions on $x$, $(x,-0.45)$ is a critical point for all $x$! That is, The entire horizontal line consists of critical points. –  rschwieb Sep 27 '12 at 20:40
You got it right... You have no idea what it is, and actually there is no restriction there... It could be anything. –  N. S. Sep 27 '12 at 20:41
Was your original $f(x,y)=y^2+\sin(y)$? If it was, you can see that in 3-d, it's the same shape at every $x$ cross-section. –  rschwieb Sep 27 '12 at 20:41
Thanks very much. Yes, that was exactly the original function. –  Farmor Sep 27 '12 at 20:46