Answer
$e^xcos(xy) \approx x+1$
Work Step by Step
The linear approximation is provided by the formula:
$L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$
Given $f(x,y)=e^xcos(xy)$ we find that:
$z_0=f(0,0)=e^0cos(0*0)=1$
$f_x=e^xcos(xy)-ye^xsin(x)$
$f_y=-xe^xsin(xy)$
Evaluating the functions at the point $(0,0)$
$f_x(0,0)=1$
$f_y(0,0)=0$
Using the linear approximation formula, the linear approximation of $f$ at $(x_0,y_0)=(0,0)$ becomes:
$L(x,y)=1(x-0)+0(y-0)+1=x+1$
