Answer
$$\frac{{dy}}{{dx}} = \cot x$$
Work Step by Step
$$\eqalign{
& y = \ln \left| {5\sin x} \right| \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left( {\ln \left| {5\sin x} \right|} \right) \cr
& {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right).{\text{ consider }}u = 5\sin x \cr
& \frac{{dy}}{{dx}} = \left( {\frac{1}{{5\sin x}}} \right){D_x}\left( {5\sin x} \right) \cr
& {\text{solve the derivative }} \cr
& \frac{{dy}}{{dx}} = \left( {\frac{1}{{5\sin x}}} \right)\left( {5\cos x} \right) \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = \frac{{\cos x}}{{\sin x}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \cot x \cr} $$
