Answer
$$\frac{{dy}}{{dx}} = \frac{{2\cos x}}{{{{\left( {\sin x + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\sin x - 1}}{{\sin x + 1}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{\sin x - 1}}{{\sin x + 1}}} \right] \cr
& {\text{use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {\sin x + 1} \right) \cdot {D_x}\left( {\sin x - 1} \right) - \left( {\sin x - 1} \right) \cdot {D_x}\left( {\sin x + 1} \right)}}{{{{\left( {\sin x + 1} \right)}^2}}} \cr
& {\text{solving derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {\sin x + 1} \right)\cos x - \left( {\sin x - 1} \right)\cos x}}{{{{\left( {\sin x + 1} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sin x\cos x + \cos x - \sin x\cos x + \cos x}}{{{{\left( {\sin x + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{2\cos x}}{{{{\left( {\sin x + 1} \right)}^2}}} \cr} $$
