The arcsine function:

The arcsin function, $\arcsin$, is the inverse of the Principle Sine Function, ${\rm Sin}$, defined by ${\rm Sin}:[-\pi/2,\pi/2]\rightarrow[-1,1]$with ${\rm Sin}(x) = \sin(x)$. Algebraically this means $\arcsin({\rm Sin}(a)) = a$ and ${\rm Sin}(\arcsin(a)) = a$ so long as domain requirements are respected. Since the tangent line to $y=\sin(x)$ is horizontal at $x=\pm\pi/2$ it is futile to look for the slope of the tangent line to $y=\arcsin(x)$ at $x = \pm 1$.

To find the slope of the tangent line to $y=\arcsin(x)$ at the point $(a,\arcsin(a))$ for $a\neq \pm 1$,we reason that if the slope of this line is m, then the slope of the tangent line to $y = {\rm Sin}(x)$ at $(\arcsin(a),a)$ should be 1/m, since when the graph of $y = {\rm Sin}(x)$ is reflected onto the graph of $y=\arcsin(x)$, the point $(\arcsin(a),a)$ is carried onto the point $(a,\arcsin(a))$. Well, since ${\rm Sin}'(x) = \cos(x)$, we know that the slope of the tangent line to $y = {\rm Sin}(x)$ at $(\arcsin(a),a)$ is given by $\cos(\arcsin(a))$. Surprisingly, this expression can be simplied. If you look at this picture of a right triangle,

you see that if $\theta = \arcsin(a)$ then the side opposite $\theta$ can have length a and the hypoteneuse can have length 1. This forces the adjacent side to have length $\sqrt{1-a^2}$. Now we see that the cosine of $\theta$ is $\sqrt{1-a^2}/1$, so $\cos(\arcsin(a)) = \sqrt{1 - a^2}$. Hence $1/m = \sqrt{1-a^2}$, that is $m = 1/\sqrt{1-a^2}$ is the slope of the tangent line to $y=\arcsin(x)$ at $(a,\arcsin(a))$. Hence we find that the rate of change of $\arcsin$ is given by $\arcsin'(a) = 1/\sqrt{1-a^2}$! Note that our formula make no sense at $a=\pm 1$, as expected.