## Calculus with Applications (10th Edition)

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