## Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = 2x\cot {x^2}$$
\eqalign{ & y = \ln \left| {\sin {x^2}} \right| \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = {D_x}\left( {\ln \left| {\sin {x^2}} \right|} \right) \cr & {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right).{\text{ consider }}u = \sin {x^2} \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{\sin {x^2}}}} \right){D_x}\left( {\sin {x^2}} \right) \cr & {\text{solve the derivative using }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right) \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{\sin {x^2}}}} \right)\left( {\cos {x^2}} \right) \cdot {D_x}\left( {{x^2}} \right) \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = \left( {\frac{1}{{\sin {x^2}}}} \right)\left( {\cos {x^2}} \right)\left( {2x} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = \frac{{2x\cos {x^2}}}{{\sin {x^2}}} \cr & or \cr & \frac{{dy}}{{dx}} = 2x\cot {x^2} \cr}