It is both kinda surprising and disappointing that the easiest way to compute the derivative of \(arc\ sin\) is to solve the integral of \(\sqrt{1-x^2}\) geometrically. 
 
Just by looking on a quarter of a circle with angle in it you know that:
$$\int{\sqrt{1-x^2}}dx = \frac{arc\ sinx}{2} + \frac{x\sqrt{1-x^2}}{2}$$

By deriving everything we get:
$$\sqrt{1-x^2} = \frac{(arc\ sinx)'}{2} + \frac{(x\sqrt{1-x^2})'}{2}$$

And now it's easy to get \((arc\ sinx)'\) out of the equation and solve it. 

It is \(\frac{1}{\sqrt{1-x^2}}\) if anyone asked. 

Then you can compute \((arc\ cosx)'\):

$$(arc\ cosx)' = (arc\ sin(1-x))' =$$
$$= (1-x)'\frac{1}{\sqrt{1-(1-x)^2}} = ...$$

Similarily and even easier \((arc\ tanx)'\) because it's just a derivative of a fraction of functions.