Answer
$$f'\left( x \right) = \frac{{18 - 4{x^2}}}{{\sqrt {9 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2x\sqrt {9 - {x^2}} ,{\text{ }}\left[ { - 3,3} \right] \cr
& \left( {\text{a}} \right){\text{ Using a computer algebra system }}\left( {{\text{Geogrebra}}} \right){\text{ we obtain }} \cr
& {\text{the derivative of the function}}{\text{.}} \cr
& f'\left( x \right) = \frac{{ - 2{x^2} + 2\left( { - {x^2} + 9} \right)}}{{\sqrt { - {x^2} + 9} }} \cr
& f'\left( x \right) = \frac{{ - 2{x^2} - 2{x^2} + 18}}{{\sqrt {9 - {x^2}} }} \cr
& f'\left( x \right) = \frac{{18 - 4{x^2}}}{{\sqrt {9 - {x^2}} }} \cr
& \cr
& \left( {\text{b}} \right){\text{ Sketch the graph of }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ }}\left( {{\text{See graph below}}} \right) \cr
& \left( {\text{c}} \right){\text{The critical numbers are:}} \cr
& \frac{{18 - 4{x^2}}}{{\sqrt {9 - {x^2}} }} = 0 \cr
& 18 - 4{x^2} = 0 \to {x^2} = \frac{9}{2},{\text{ }}x = \pm \frac{3}{{2\sqrt 2 }} \cr
& \left( {\text{d}} \right){\text{ From the graph, we can see that the function is:}} \cr
& {\text{Increasing on: }}\left( { - \frac{3}{{2\sqrt 2 }},\frac{3}{{2\sqrt 2 }}} \right) \cr
& {\text{Decreasing on: }}\left( {0, - \frac{3}{{2\sqrt 2 }}} \right){\text{ and }}\left( {\frac{3}{{2\sqrt 2 }},3} \right) \cr} $$
