how to use $f(x,y)$ to find $df(x,y)$ If I have $f(x,y)= 36-x^2-4y^2$ and I need to figure out $df(x,y)$, is it just $dx/dy$?
Also how to find the best affine approximation to $f$ at $(x,y)=(2,1)$
 A: Find the best affine approximation at $(x,y)=(2,1)$ given
$$f(x,y)= 36-x^2-4y^2$$
We calculate
$$\nabla f(x, y) = (-2x, -8y) \implies \nabla f(2, 1) = (-4, -8)$$
At $(x, y) = (2,1)$, we have
$$f(2, 1) = 36 -(2)^2 -4(1)^2 = 28$$
The best affine approximation is given by
$$A(x, y) = \nabla f(\textbf{c}) \cdot (\textbf{x − c}) + f(\textbf{c}) = (-4,-8) \cdot(x-2,y-1)+ 28 = -4x - 8y + 44$$
A: The affine approximation to $f(x,y)$ at $(x_0,y_0)$, also called the linearization of $f$ at $(x_0,y_0)$ is the equation of the tangent plane to the surface $z=f(x,y)$ at the point $(x_0,y_0)$.
Accordingly, it is a function $z=L(x,y)$ with the properties that


*

*$L(x_0,y_0)=f(x_0,y_0)$

*$L_x(x_0,y_0)=f_x(x_0,y_0)$

*$L_y(x_0,y_0)=f_y(x_0,y_0)$
The function satisfying these requirements is
$$ L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$
If you apply this to your problem you should obtain the result
$$ L(x,y)=44-4x-8y $$
A: $$f(x,y)= 36-x^2-4y^2$$
$$df(x,y)=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = -2xdx - 8ydy$$
And affine approximation:
$$f(x_0 + dx ,y_0 + dy )=f(x_0, y_0) + df(x_0, y_0) = f(x_0, y_0) -2x_0dx - 8y_0dy$$
Let $dx=x-x_0, \, dy = y-y_0$
$$f(x,y)=f(x_0, y_0) -2x_0(x - x_0) - 8y_0(y - y_0)$$
where $x_0=2, \, y_0=1$
