## Calculus: Early Transcendentals (2nd Edition)

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