Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 57

a.) Suppose that $f(x) = x \sqrt{2 -x ^2}$, find $f'(x)$


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (x \sqrt{2 -x ^2})
\\
\\
f'(x) =& \left[ (x) \frac{d}{dx} \sqrt{2 -x ^2} \right] + \left[ \sqrt{2 -x ^2} \frac{d}{dx} (x) \right]
\\
\\
f'(x) =& \left[ (x) \frac{d}{dx} (2 - x^2)^{\frac{1}{2}} \right] + \left[ (2 - x^2)^{\frac{1}{2}} \frac{d}{dx} (x) \right]
\\
\\
f'(x) =& \left[ (x) \left( \frac{1}{2} \right) (2 - x^2)^{\frac{-1}{2}} \frac{d}{dx} (2 - x^2)\right] + \left[ (2 - x^2)^{\frac{1}{2}} (1) \right]
\\
\\
f'(x) =& \left[ (x) \left( \frac{1}{2} \right) (2 - x^2)^{\frac{-1}{2}} (0-2x)\right] + (2 - x^2)^{\frac{1}{2}}
\\
\\
f'(x) =& \left( \frac{x}{\cancel{2}} \right) (2 - x^2)^{\frac{-1}{2}} ( \cancel{2} x) + (2 - x^2)^{\frac{1}{2}}
\\
\\
f'(x) =& -x^2 (2 - x^2)^{\frac{-1}{2}} + (2 - x^2)^{\frac{1}{2}}
\\
\\
f'(x) =& \frac{-x^2}{(2 - x^2)^{\frac{1}{2}}} + (2 - x^2)^{\frac{1}{2}}
\\
\\
f'(x) =& \frac{-x^2 + 2 -x^2}{(2 - x^2)^{\frac{1}{2}}}
\\
\\
f'(x) =& \frac{2 - 2x^2}{(2 - x^2)^{\frac{1}{2}}} \text{ or } \frac{2 - 2x^2}{\sqrt{2 - x^2}}
\\
\\


\end{aligned}
\end{equation}
$


b.) Graph $f$ and $f'$