Answer
$-x+\frac{y}{2}+\frac{3}{2}$
Work Step by Step
Given that f is a differentiable function with $f(x,y)=\frac{1+y}{1+x}$ at $(1,3)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=\frac{1+y}{1+x}$
$f_{x}(a,b)=\frac{-1-y}{(1+x)^{2}}$
$f_{y}(a,b)=\frac{1}{1+x}$
At $(1,3)$
$f(1,3)=\frac{1+3}{1+1}=2$
$f_{x}(1,3)=\frac{-1-3}{(1+1)^{2}}=-1$
$f_{y}(1,3)=\frac{1}{1+1}=\frac{1}{2}$
The linearization $L(x,y)$ of the function at $(1,3)$ is
$L(x,y)= f(1,3)+f_{x}(1,3)(x-1)+f_{y}(1,3)(y-3)$
$=2+(-1)(x-1)+\frac{1}{2}(y-3)$
$=2-x+1+\frac{y}{2}-\frac{3}{2}$
$=-x+\frac{y}{2}+\frac{3}{2}$
Hence, the linearization $L(x,y)$ of the function at $(1,3)$ is $-x+\frac{y}{2}+\frac{3}{2}$
