Answer
$$\left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{{{e^x}}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {2x + 1} \right) \cr
& f'\left( x \right) = \frac{2}{{2x + 1}} \cr
& {\text{Calculating the inverse of the function}} \cr
& y = \ln \left( {2x + 1} \right) \cr
& x = \ln \left( {2y + 1} \right) \cr
& {e^x} = {e^{\ln \left( {2y + 1} \right)}} \cr
& {e^x} = 2y + 1 \cr
& {e^x} - 1 = 2y \cr
& y = \frac{{{e^x} - 1}}{2} \cr
& {f^{ - 1}}\left( x \right) = \frac{{{e^x} - 1}}{2} \cr
& {\text{By the formula 2: }}\left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{\frac{2}{{2\left( {\frac{{{e^x} - 1}}{2}} \right) + 1}}}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{1}{{\frac{2}{{{e^x} - 1 + 1}}}} = \frac{1}{{\frac{2}{{{e^x}}}}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{{{e^x}}}{2} \cr
& \cr
& {\text{Differentiating }}\left( {{f^{ - 1}}} \right)\left( x \right){\text{ directly}} \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{e^x} - 1}}{2}} \right] \cr
& \left( {{f^{ - 1}}} \right)'\left( x \right) = \frac{{{e^x}}}{2} \cr} $$
