Answer
(a) $f'(x) = \frac{2-2x^2}{\sqrt{2-x^2}}$
(b) We can see a sketch of the graphs below.
Work Step by Step
(a) $f(x) = x~\sqrt{2-x^2}$
We can find $f'(x)$:
$f'(x) = \sqrt{2-x^2}+\frac{1}{2}(2-x^2)^{-1/2}(-2x)(x)$
$f'(x) = \frac{2-x^2}{\sqrt{2-x^2}}-\frac{x^2}{\sqrt{2-x^2}}$
$f'(x) = \frac{2-2x^2}{\sqrt{2-x^2}}$
(b) We can see a sketch of the graphs below.
$f'(x)$ 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.