Use the Chain Rule and some algebra to simplify:
d/dx(arcsin(sqrt{1-x^2}))
=1/sqrt{1-(sqrt{1-x^2})^2} * d/dx((1-x^2)^{1/2})
=1/sqrt{1-(1-x^2)} * 1/2 (1-x^2)^{-1/2} * (-2x)
=-x/(sqrt{x^2} sqrt{1-x^2})
=-x/sqrt{x^2-x^4}
This is true for all x\in (-1,0) cup (0,1) (for all x strictly between -1 and 1 except for x=0).
y=arcsin(sqrt{1-x^2}) is actually pretty interesting since you wouldn't expect ahead of time that it would fail to be differentiable at x=0. Here's what the graph of it (red) and its derivative (blue) look like:
![enter image source here]()
The discontinuity in the derivative at x=0 can also be seen by noting that sqrt{x^2}=|x| and therefore -x/sqrt{x^2}=-x/|x| equals -1 when x>0 and 1 when x<0. Therefore, f'(x)=-1/sqrt{1-x^2} when x>0 and f'(x)=1/sqrt{1-x^2} when x<0.
For the original function, f(x)=arcsin(sqrt{1-x^2}), evidently it's like taking the graph of arcsin(x), shifting it up by pi/2 units, and then reflecting the part of it for x>0 across the horizontal line y=pi/2.
In other words, arcsin(sqrt{1-x^2})=arcsin(x)+pi/2 when x<0 and arcsin(sqrt{1-x^2})=pi/2-arcsin(x) when x>0. Weird! Never would have guessed that!
The second equation (for x>0) makes sense if you draw a right triangle, label one angle arcsin(sqrt{1-x^2}), label the adjacent side sqrt{1-x^2} and the hypotenuse 1, then solve for the other side with the Pythagorean Theorem to get x. In that situation, arcsin(x) would be the complementary angle to the original one, illustrating why arcsin(sqrt{1-x^2})=pi/2-arcsin(x) when x>0.
Here's a graph of the original f(x) and f'(x), along with arcsin(x) to illustrate all this:
![enter image source here]()
