## Calculus 8th Edition

$2x+y-1$
Given that f is a differentiable function with $f(x,y)=x^{2}e^{y}$ at $(1,0)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=x^{2}e^{y}$ $f_{x}(a,b)=2xe^{y}$ $f_{y}(a,b)=x^{2}e^{y}$ At $(1,0)$ $f(1,0)=1^{2}e^{0}=1$ $f_{x}(1,0)=2(1)e^{0}=2$ $f_{y}(1,0)=1^{2}e^{0}=1$ The linearization $L(x,y)$ of function at $(1,0)$ is $L(x,y)= f(1,0)+f_{x}(1,0)(x-1)+f_{y}(1,0)(y-0)$ $=1+2(x-1)+1(y-0)$ $=1+2x-2+y$ $=2x+y-1$ Hence, the linearization $L(x,y)$ of the function at $(1,4)$ is $2x+y-1$