Answer
$f'(x) = \frac{-x}{\sqrt{1-x^2}}~arcsin~x+1$
We can see a sketch of the graphs below.
Work Step by Step
$f(x) = \sqrt{1-x^2}~arcsin~x$
We can find $f'(x)$:
$f'(x) = \sqrt{1-x^2}~arcsin~x$
$f'(x) = \frac{-x}{\sqrt{1-x^2}}~arcsin~x+\sqrt{1-x^2}\cdot \frac{1}{\sqrt{1-x^2}}$
$f'(x) = \frac{-x}{\sqrt{1-x^2}}~arcsin~x+1$
$f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.