I am supposed to determine the angle between the left and the right tangent to the graph g, in the point $$\left [ 1, \frac{\sqrt{3}\pi }{6} \right ]$$where $$g(x)=\frac{1}{\sqrt{3}}\arcsin \frac{2x}{1+x^{2}}$$ I tried to do first derivation, which is equal to $$ \frac{2\sqrt{3}}{3+3x^{2}}$$ and limits: $$\lim_{x\rightarrow 1^{+}} \frac{2\sqrt{3}}{3+3x^{2}}=\frac{\sqrt{3}}{3}$$ and $$\lim_{x\rightarrow 1^{-}} \frac{2\sqrt{3}}{3+3x^{2}}=-\frac{\sqrt{3}}{3}$$ but I do not know, what to do next.
Can anyone help me?
Thanks
As has been mentioned in the comments, the correct derivative is
$$g'(x) = \frac{-2(x-1)(x+1)}{\sqrt3 (x^2+1)\cdot |x-1| \cdot |x+1|}$$
However, your limits are correct. Considering that the derivative $g'(x)$ is equal to the slope of the tangent at the point $(x, g(x))$, the angle between the tangent and the $x$ axis is $\alpha=\arctan(g'(x))$.
Hence, if $\alpha_1$ and $\alpha_2$ are the angles of the left tangent and the right tangent respectively, then the angle between them is $$\alpha_1-\alpha_2 = \arctan\left(\frac{\sqrt3}{3}\right) - \arctan\left(-\frac{\sqrt3}{3}\right) = \frac{\pi}{3}$$