Answer
$$\frac{{\cot \left( {2x - 1} \right)}}{x} - \left( {2\ln \left| x \right|} \right){\csc ^2}\left( {2x - 1} \right)$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {\left[ {\ln \left| x \right|} \right]\left[ {\cot \left( {2x - 1} \right)} \right]} \right) \cr
& {\text{By the product rule}} \cr
& {\text{ = }}\left[ {\cot \left( {2x - 1} \right)} \right]\frac{d}{{dx}}\left[ {\ln \left| x \right|} \right] + \left[ {\ln \left| x \right|} \right]\frac{d}{{dx}}\left[ {\cot \left( {2x - 1} \right)} \right] \cr
& {\text{Computing derivatives}} \cr
& {\text{ = }}\left[ {\cot \left( {2x - 1} \right)} \right]\left( {\frac{1}{x}} \right) + \left[ {\ln \left| x \right|} \right]\left[ { - {{\csc }^2}\left( {2x - 1} \right)} \right]\frac{d}{{dx}}\left[ {2x - 1} \right] \cr
& {\text{ = }}\left[ {\cot \left( {2x - 1} \right)} \right]\left( {\frac{1}{x}} \right) + \left[ {\ln \left| x \right|} \right]\left[ { - {{\csc }^2}\left( {2x - 1} \right)} \right]\left( 2 \right) \cr
& {\text{Simplifying}} \cr
& {\text{ = }}\frac{{\cot \left( {2x - 1} \right)}}{x} - \left( {2\ln \left| x \right|} \right){\csc ^2}\left( {2x - 1} \right) \cr} $$
