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