Answer
$y=\dfrac{1}{2} x$
Work Step by Step
We have: $y(x)=\dfrac{x}{1+e^x}; x=0$
Now, $y(0)=\dfrac{0}{1+e^0}=0$
We differentiate both sides with respect to $x$.
$y^{\prime}(x)=\dfrac{1+e^x-xe^x}{(1+e^x)^2}$
The equation of a tangent line at $(0,0)$ is:
$y-y_1=m(x-x_1)\\ y-0=\dfrac{1}{2} (x-0) \\ y=\dfrac{1}{2} x$
