Answer
$\frac{x}{3}+y$
Work Step by Step
Given that f is a differentiable function with $f(x,y)=y+sin(\frac{x}{y})$ at $(0,3)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=y+sin(\frac{x}{y})$
$f_{x}(a,b)=\frac{1}{y}+cos(\frac{x}{y})$
$f_{y}(a,b)=1-\frac{x}{y^{2}}cos(\frac{x}{y})$
At $(0,3)$
$f(0,3)=3+sin(\frac{0}{3})=3$
$f_{x}(0,3)=\frac{1}{3}+cos(\frac{0}{3})=\frac{1}{3}$
$f_{y}(0,3)=1-\frac{0}{3^{2}}cos(\frac{0}{3})=1$
The linearization $L(x,y)$ of the function at $(0,3)$ is
$L(x,y)= f(0,3)+f_{x}(0,3)(x-0)+f_{y}(0,3)(y-3)$
$=3+\frac{1}{3}(x-0)+1(y-3)$
$=3+\frac{x}{3}+y-3$
$=\frac{x}{3}+y$
Hence, the linearization $L(x,y)$ of the function at $(1,3)$ is $\frac{x}{3}+y$
