Answer
\[\frac{{dy}}{{dx}} = \frac{{ - 2{x^3}\cos x - {x^4}\sin x - x\sin x + \cos x}}{{\,{{\left( {1 + {x^3}} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{x\cos x}}{{1 + {x^3}}} \hfill \\
\hfill \\
by\,\,the\,\,quotient\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\left( {1 + {x^3}} \right)\left[ {x\cos x} \right]' - \left( {x\cos x} \right)\left[ {1 + {x^3}} \right]'}}{{{{\left( {1 + {x^3}} \right)}^2}}} \hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\,\left( {1 + {x^3}} \right)\,\left( { - x\sin x + \cos x} \right) - x\cos x\,\left( {3{x^2}} \right)}}{{\,{{\left( {1 + {x^3}} \right)}^2}}} \hfill \\
\hfill \\
multiply\,\,and\,\,simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{ - x\sin x + \cos x - {x^4}\sin x + {x^3}\cos x - 3{x^3}\cos x}}{{\,{{\left( {1 + {x^3}} \right)}^2}}} \hfill \\
\hfill \\
reduce\,\,line\,\,terms \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{ - 2{x^3}\cos x - {x^4}\sin x - x\sin x + \cos x}}{{\,{{\left( {1 + {x^3}} \right)}^2}}} \hfill \\
\end{gathered} \]
