Answer
$6x+4y-23$
Work Step by Step
Given that f is a differentiable function with $f(x,y)=1+xln(xy-5)$ at $(2,3)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=1+xln(xy-5)$
$f_{x}(a,b)=ln(xy-5)+\frac{xy}{(xy-5)}$
$f_{y}(a,b)=\frac{x^{2}}{(xy-5)}$
At $(2,3)$
$f_{x}(2,3)=ln((2).(3)-5)+\frac{(2).(3)}{(2.3-5)}=ln1+6=6$
$f_{y}(a,b)=\frac{2^{2}}{(6-5)}=4$
The linearization $L(x,y)$ of function at $(2,3)$ is
$L(x,y)= f(2,3)+f_{x}(2,3)(x-2)+f_{y}(2,3)(y-3)$
$= f(2,3)+6(x-2)+4(y-3)$
$=1+6(x-2)+4(y-3)$
$=1+6x-12+4y-12$
$=6x+4y-23$
Hence, the linearization $L(x,y)$ of the function at $(2,3)$ is $ 6x+4y-23$