## Calculus: Early Transcendentals (2nd Edition)

$y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}}$
$\begin{gathered} y = \frac{{\left( {{x^2} - 1} \right)\sin x}}{{\sin x + 1}} \hfill \\ \hfill \\ Using\,\,quotient\,\,rule \hfill \\ \hfill \\ y' = \frac{{\,\,{{\left[ {\,\left( {{x^2} - 1} \right)\sin x} \right]}^\prime } \cdot \,\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x \cdot \,{{\left( {\sin x + 1} \right)}^\prime }}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ y' = \frac{{\,\,\left[ {2x\sin x + \,\left( {{x^2} - 1} \right)\cos x} \right]\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x\,\left( {\sin x + 1} \right) - \,\left( {{x^2} - 1} \right)\sin x\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ y' = \frac{{2x\sin x\,\left( {\sin x + 1} \right) + \,\left( {{x^2} - 1} \right)\cos x}}{{\,{{\left( {\sin x + 1} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered}$