## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = - \frac{{2\cos x}}{{{{\sin }^3}x}} + \frac{{2\cos x}}{{{{\sin }^2}x}}$$
\eqalign{ & y = \frac{1}{{{{\sin }^2}x}} - \frac{2}{{\sin x}} \cr & {\text{write as}} \cr & y = {\left( {\sin x} \right)^{ - 2}} - 2{\left( {\sin x} \right)^{ - 1}} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {\sin x} \right)}^{ - 2}}} \right] - 2\frac{d}{{dx}}\left[ {{{\left( {\sin x} \right)}^{ - 1}}} \right] \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dx}} = - 2{\left( {\sin x} \right)^{ - 3}}\frac{d}{{dx}}\left[ {\sin x} \right] - 2\left( { - 1} \right){\left( {\sin x} \right)^{ - 2}}\frac{d}{{dx}}\left[ {\sin x} \right] \cr & {\text{find derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = - 2{\left( {\sin x} \right)^{ - 3}}\left( {\cos x} \right) - 2\left( { - 1} \right){\left( {\sin x} \right)^{ - 2}}\left( {\cos x} \right) \cr & \frac{{dy}}{{dx}} = - \frac{{2\cos x}}{{{{\sin }^3}x}} + \frac{{2\cos x}}{{{{\sin }^2}x}} \cr}