Answer
$g'\left( x\right) = \frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}\cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \sin \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right) \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\sin \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)} \right] \cr
& {\text{Use the chain Rule }}\frac{d}{{dx}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dx}} \cr
& g'\left( x \right) = \cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)\underbrace {\frac{d}{{dx}}\left[ {\frac{{{e^x}}}{{1 + {e^x}}}} \right]}_{{\text{Use quotient rule}}} \cr
& g'\left( x \right) = \cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)\left[ {\frac{{\left( {1 + {e^x}} \right)\frac{d}{{dx}}\left[ {{e^x}} \right] - {e^x}\frac{d}{{dx}}\left[ {1 + {e^x}} \right]}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& g'\left( x \right) = \cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)\left[ {\frac{{\left( {1 + {e^x}} \right){e^x} - {e^x}\left( {{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right] \cr
& g'\left( x \right) = \cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)\left[ {\frac{{{e^x} + {e^{2x}} - {e^{2x}}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right] \cr
& g'\left( x \right) = \cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right)\left[ {\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right] \cr
& g'\left( x \right) = \frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}\cos \left( {\frac{{{e^x}}}{{1 + {e^x}}}} \right) \cr} $$
