## Calculus 8th Edition

$e^xcos(xy) \approx x+1$
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: $L(x,y)=1(x-0)+0(y-0)+1=x+1$