# 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?

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

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...
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
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