Answer
$$\frac{{\cos x - 2{e^{ - x}} + 2x{e^{ - x}}}}{{\sin x - 2x{e^{ - x}}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln \left| {\sin x - 2x{e^{ - x}}} \right|} \right] \cr
& {\text{Use the basic rule of differentiation }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{{u'}}{u} \cr
& = \frac{{\left[ {\sin x - 2x{e^{ - x}}} \right]'}}{{\sin x - 2x{e^{ - x}}}} \cr
& = \frac{{\left[ {\sin x} \right]' - \left[ {2x{e^{ - x}}} \right]'}}{{\sin x - 2x{e^{ - x}}}} \cr
& {\text{By the product rule}} \cr
& = \frac{{\left[ {\sin x} \right]' - {e^{ - x}}\left[ {2x} \right]' - 2x\left[ {{e^{ - x}}} \right]'}}{{\sin x - 2x{e^{ - x}}}} \cr
& {\text{Computing derivatives}} \cr
& = \frac{{\cos x - {e^{ - x}}\left( 2 \right) - 2x\left( { - {e^{ - x}}} \right)}}{{\sin x - 2x{e^{ - x}}}} \cr
& {\text{Simplifying}} \cr
& = \frac{{\cos x - 2{e^{ - x}} + 2x{e^{ - x}}}}{{\sin x - 2x{e^{ - x}}}} \cr} $$
