Inverse Jacobian matrix of spherical coordinates

I found inverse transformation from spherical coordinates to cartesian coordinates (on $x>0$, $y>0$ and $z>0$). I have $$r = w_1(x,y,z) = \sqrt{x^2+y^2+z^2}$$ $$\theta = w_2(x,y,z) = \arccos\biggl(\frac{z}{\sqrt{x^2+y^2+z^2}}\biggr)$$ $$\phi = w_3(x,y,z) = \arcsin\biggl(\frac{y}{\sqrt{x^2+y^2}}\biggr)$$

Now I have to find the Jacobian matrix of this application. What is the better way? By direct derivation, I have a problem with $\arccos$ and $\arcsin$ functions. However, inverting the Jacobian matrix of the inverse transformation (cartesian to spherical) is a long task...

Any help?

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It's not that tedious. $\frac{\partial w_2(x,y,z)}{\partial z} = -\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2}$, for example. –  copper.hat Mar 4 '13 at 18:14
@copper.hat Right... I will try to do it explicitly! –  R. M. Mar 4 '13 at 18:31
Another approach would be to use the inverse function theorem, which states that (under appropriate conditions) $D(f^{-1})(f(x)) = (Df(x))^{-1}$. So, compute the Jacobian of the transformation $(r,\theta, \phi) \to (x,y,z)$ and invert (with appropriate argument, of course). –  copper.hat Mar 4 '13 at 18:33
No; it $\arccos ' (x) = -\frac{1}{\sqrt{1-x^2}}$. –  copper.hat Mar 4 '13 at 19:30