#### Answer

$x+\frac{y}{4}$

#### Work Step by Step

Given that f is a differentiable function with $f(x,y)=\sqrt {xy}$ at $(1,4)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=\sqrt {xy}$
$f_{x}(a,b)=\frac{\sqrt y}{2\sqrt x}$
$f_{y}(a,b)=\frac{\sqrt x}{2\sqrt y}$
At $(1,4)$
$f(1,4)=\sqrt {1\times 4}=2$
$f_{x}(1,4)=\frac{\sqrt 4}{2\sqrt 1}=1$
$f_{y}(1,4)=\frac{\sqrt 1}{2\sqrt 4}=\frac{1}{4}$
The linearization $L(x,y)$ of function at $(1,4)$ is
$L(x,y)= f(1,4)+f_{x}(1,4)(x-1)+f_{y}(1,4)(y-4)$
$=2+1(x-1)+\frac{1}{4}(y-4)$
$=2+x-1+\frac{y}{4}-1$
$=x+\frac{y}{4}$
Hence, the linearization $L(x,y)$ of the function at $(1,4)$ is $ x+\frac{y}{4}$