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I was reading this answer regarding the span of a tangent plane here.

The answer says the graph of $f$ is also the graph of the map $F(x,y) = (x,y,f(x,y))$. The tangent plane is spanned by $(1,0,f_x), (0,1,f_y)$.

However, I don't understand why. If I were to find the tangent plane at a point $c = (x_0,y_0)$, I would do the following

$$\nabla F(c) ( (x,y) -c) = f_x(x - x_0) + f_y(y - y_0) - (f(x,y) - f(c)) = 0$$

The tangent plane would be set $T$ that satisfies the equation above.

Two questions:

(1) Is $f(x,y) = f(c)$?

(2) Suppose that I have a set $T$ that satisfies the equation above. Then each element of $T$ can be written as $[a_1 (1,0,f_x), \ a_2 (0,1,f_y)]$. If I plug that into the equation, I get $$f_x(a_1 - x_0) + f_y (a_2 - y_0) - (f(x,y) - f(c)) = 0$$

But $f(x,y) = a_1 f_x + a_2 f_y$, then I have

$$f_x(a_1 - x_0) + f_y (a_2 - y_0) - (a_1 f_x + a_2 f_y - f(c)) = 0$$ which simplifies to

$$ f(c) - (f_x x_0 + f_y y_0) = 0$$

How do you show that this equality holds?

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I’m not sure how you’re calculating your tangent plane above. The general approach is to observe that the tangent plane should contain every tangent line: given any direction $(dx, dy)$, restrict $F$ to the line passing though $(x_0, y_0)$ in that direction:

$$G(t) = F(x_0 + tdx, y_0+ tdy).$$

This is now a one-dimensional function whose derivative gives the slope of a tangent line to $G$; therefore the vector $(dx, dy, G’(0))$ should lie on the tangent plane of $F$.

To find a basis for the tangent plane, pick any linearly independent pair of directions $(dx, dy)$. $(1,0)$ and $(1,0)$ are particularly convenient.

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  • $\begingroup$ Why do you have $G'(0)$ in $(dx,dy,G'(0))$? $\endgroup$ – user1691278 Feb 18 '18 at 23:11

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