For a dynamical system governed by the equation $f(x) = \mathrm{d}y/\mathrm{d}x = 2(1-x^2)^{1/2}$. Find stable and unstable fixed points.
The fixed points for the above equation are $+1$ and $-1$. I took the derivative of given equation and set it equal to zero. If $f'(x)<0$, it's a stable point and if $f'(x)>0$, it's an unstable point. The derivative is $$\frac{-2x}{(1-x^2)^{1/2}}$$ Now if I plug $x=1$ or $x=-1$, it's not defined. How do I find the stability of fixed points in this case?
If what you mean is that the system is $dx/dt=f(x)$, then just draw the phase portrait on the interval $[-1,1]$, using that $dx/dt > 0$ for $-1<x<1$.