Answer
$$f'\left( x \right) = {x^2}\cos x + 2x\sin x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2}\sin x,{\text{ }}\left[ {0,2\pi } \right] \cr
& \left( {\text{a}} \right){\text{ Using a computer algebra system }}\left( {{\text{Geogrebra}}} \right){\text{ we obtain }} \cr
& {\text{the derivative of the function}}{\text{.}} \cr
& f'\left( x \right) = {x^2}\left( {\cos x} \right) + \sin x\left( {2x} \right) \cr
& f'\left( x \right) = {x^2}\cos x + 2x\sin x \cr
& \cr
& \left( {\text{b}} \right){\text{ Sketch the graph of }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ }}\left( {{\text{See graph below}}} \right) \cr
& \left( {\text{c}} \right){\text{From the graph we obtain the critical numbers}} \cr
& x = 2.29{\text{ and }}x = 5.09 \cr
& \cr
& \left( {\text{d}} \right){\text{ From the graph, we can see that the function is:}} \cr
& {\text{Increasing on: }}\left( {0,2.29} \right){\text{ and }}\left( {5.09,2\pi } \right) \cr
& {\text{Decreasing on: }}\left( {2.29,5.09} \right) \cr} $$
