Given,
$\LARGE f(x)=\frac{e^{x}}{1+e^{x}}$
Differentiating with respect to x,
$\large f'(x)=\frac{(1+e^{x}).\frac{\mathrm{d} }{\mathrm{d} x}e^{x}-e^{x}\frac{\mathrm{d} }{\mathrm{d} x}(1+e^{x})}{(1+e^{x})^{2}}$
$\large f'(x)=\frac{1+e^{2x}-e^{2x}}{(1+e^{x})^{2}}$
$\large f'(x)=\frac{1}{(1+e^{x})^{2}}$
$\large f'(x)$ is always greater than 0 , as denominator $\large (1+e^{x})^{2}$ >0 and numerator is 1 so ,
$\large f'(x)>0$ in $\large \left [ -\infty,+\infty \right ]$.
It indicates that if we draw a tangent in the graph for $\large f(x)=\frac{e^{x}}{1+e^{x}}$at any point in $\large \left [ -\infty,+\infty \right ]$ the slop of the tangent for the graph is always positive which indicates that the graph is monotonically increasing.
The graph will look like this ,
so correct answer is option (a).