Derivative of $\sqrt{\sin (x^2)}$ I have problems calculating derivative of $f(x)=\sqrt{\sin (x^2)}$.
I know that $f'(\sqrt{2k \pi + \pi})= - \infty$ and $f'(\sqrt{2k \pi})= + \infty$ because $f$ has derivative only if $ \sqrt{2k \pi} \leq |x| \leq \sqrt{2k \pi + \pi}$.
The answer says that for all other values of $x$,  $f'(0-)=-1$ and $f'(0+)=1$.
Why is that? All I get is $f'(x)= \dfrac{x \cos x^2}{\sqrt{\sin (x^2)}} $.

I don't know if you did it this way, so I figured that I would at least display it.
\begin{align}
   y &= \sqrt{\sin x^2}\\
   y^2 &= \sin x^2\\
   2yy' &= 2x \cos x^2\\
   y' &= \frac{x \cos x^2}{\sqrt{\sin x^2}}
\end{align}
