#### Answer

Equation of tangent: $x-2y+1=0$.

#### Work Step by Step

Given, $y=\frac{2x}{x+1}$. Differentiating $y$ with respect to $x$, at $(1,1)$, to find the slope of the tangent at the given point. Using quotient's rule,
$$\frac{dy}{dx}=\frac{2(x+1) - 2x}{(x+1)^2}$$
$$\frac{dy}{dx}\bigg|_{x=1}=\frac{2(1+1) - 2\times1}{(1+1)^2}=\frac{1}{2}.$$
Using point-slope form of a line, the equation of tangent through $(1,1)$ is,
$$y-1=\frac{1}{2}(x-1)$$
$$2y-2=x-1\implies x-2y+1=0.$$