how to use $f(x,y)$ to find $df(x,y)$ [closed]

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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 $(x,y)=(2,1)$

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3 Answers

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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

$$A(x, y) = \nabla f(\textbf{c}) \cdot (\textbf{x − c}) + f(\textbf{c}) = (-4,-8) \cdot(x-2,y-1)+ 28 = -4x - 8y + 44$$

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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

1. $L(x_0,y_0)=f(x_0,y_0)$

2. $L_x(x_0,y_0)=f_x(x_0,y_0)$

3. $L_y(x_0,y_0)=f_y(x_0,y_0)$

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

$$ L(x,y)=44-4x-8y $$

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$$f(x,y)= 36-x^2-4y^2$$ $$df(x,y)=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = -2xdx - 8ydy$$

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 $x_0=2, \, y_0=1$

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