Answer
$\frac{y-1}{x+1} \approx x+y-1$
Work Step by Step
The linear approximation is provided by the formula:
$L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$
Given $f(x,y)=\frac{y-1}{x+1}$ we find that:
$z_0=f(0,0)=\frac{0-1}{0+1}=-1$
$f_x=-\frac{y-1}{(x+1)^2}$
$f_y=\frac{1}{x+1}$
Evaluating the functions at the point $(0,0)$
$f_x(0,0)=1$
$f_y(0,0)=1$
Using the linear approximation formula, the linear approximation of $f$ at $(x_0,y_0)=(0,0)$ becomes:
$L(x,y)=1(x-0)+1(y-0)+(-1)=x+y-1$
