$f(x,y) = e^x \sin(y)$
$f_x = e^x \sin(y)$
$f_y = e^x \cos(y)$
$L(x,y) = f(0,0) + f_x(0,0)x +f_y(0,0)y$
Solving:
$f(0,0) = e^0 \sin(0) = 0$
$f_x(0,0)x = e^0 \sin(0) \cdot x = 0$
$f_y(0,0)y = e^0 \cos(0) \cdot y = y$
So L(x,y) = y
yet here it says $1+y$, how come?
