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I read this definition in a book of multivariable calculus:

$f(x,y)$ is differentiable at $(x_0,y_0)$ if it can be expressed as the form $$f(x_0+\Delta x, y_0+\Delta y)=f(x_0,y_0)+A\Delta x+B\Delta y+\alpha \Delta x+\beta \Delta y$$ where $A,B$ are constants, $\alpha, \beta\rightarrow 0$ when $\Delta x,\Delta y\rightarrow 0$.

$A,B$ can easily proved is respectively $f'_x(x_0,y_0)$ and $f'_y(x_0,y_0)$ ($f'_x(x_0,y_0)$ and $f'_y(x_0,y_0)$ are the partial derivatives of $f$)

I don't understand why they can come up with this definition. In one-variable calculus, the definition of derivative is easily understood by the graph

Derivative of one-variable function

When we deal with two-variable function, it is much more difficult to imagine the image. Let me take a function which is differentiable at $(0,0)$: $f(0,0)=0$ and $f(x,y)=\frac{x^2y^4}{x^4+y^4}$ if $(x,y)\neq (0,0)$. Here is its graph:

enter image description here

If a function is differentiable at a point in 2-dimensional space, it has a tangent line at that point. How would it be if $f$ is in 3-dimensional space? Is it true that if $f$ is differentable at $(0,0)$ then it has a tangent plane at $(0,0)$? If yes, what is the equation of that plane?

To summarize, I have two questions:

  1. Why do $\alpha,\beta$ need to tend to $0$ by the above definition?

  2. What is the equation of the tangent (plane) at a point of a 2-variable function?

Thanks so much for any help. I have just learnt multivariable calculus recently, so some of my questions may not make sense. Hope they do not annoy you.

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I have always disliked the definition of differentiable given in introductory multivariable calculus texts. Wikipedia has a much nicer definition which I will try to spell out.

The derivative is not as easily visualized in higher dimensions. However, the idea is the same. The tangent line at a point $x$ is the line that best approximates the function at $x$. This idea of linear (or really, affine) approximation carries over to higher dimensions.

You're familiar with the usual definition of a 1-variable derivative: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$. It might not be clear at first how to generalize this to multivariable functions, but hopefully it will be after we rearrange the above equation: $$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \iff 0 = \lim_{h \to 0} \frac{f(x + h) - f(x) - f'(x)h}{h} \, . $$ Note that for a fixed $x$, the function $L(h) = f'(x) h$ is just a line through the origin with slope $f'(x)$, which is an example of a linear map in the sense of linear algebra. In general, the derivative of a function $f : \mathbb{R}^m \to \mathbb{R}^n$ at a point ${x} \in \mathbb{R}^m$ is defined to be a linear map $Df_{x} : \mathbb{R}^m \to \mathbb{R}^n$ such that $$ \lim_{h \to 0} \frac{f(x + h) - f(x) - Df_x(h)}{\|h\|} = 0 $$ where $\|h\|$ is the length of the vector $h \in \mathbb{R}^m$. One can show that such a linear map is unique if it exists. One can also show that this linear map $Df_x$ can be represented as left multiplication by the Jacobian matrix $$ [Df_x] = \begin{pmatrix} \left. \frac{\partial F_1}{\partial x_1} \right|_x & \cdots & \left. \frac{\partial F_1}{\partial x_m} \right|_x\\ \vdots & & \vdots\\ \left. \frac{\partial F_n}{\partial x_1} \right|_x & \cdots & \left. \frac{\partial F_n}{\partial x_m} \right|_x \end{pmatrix} $$ where the $F_i : \mathbb{R}^m \to \mathbb{R}$ are the component functions of $F$, i.e., $F(x) = (F_1(x), \ldots, F_n(x))$.

Okay, after all those abstract definitions, let's consider your particular example. For a function $f : \mathbb{R}^2 \to \mathbb{R}$, the derivative can indeed be visualized as the tangent plane to the graph of $f$. Writing $z = f(x,y)$ or $0 = f(x,y) - z$, then points on the graph of $f$ are of the form $(x,y,z) = (x,y,f(x,y))$. In this case, $Df_{(x,y)}$ is simply the gradient $\nabla f|_{(x,y)} = \left(\left.\frac{\partial f}{\partial x}\right|_{(x,y)}, \left.\frac{\partial f}{\partial y}\right|_{(x,y)}\right)$. Letting $g(x,y,z) = f(x,y) - z$, then $\nabla g = ([Df_{(x,y)}],-1) = \left(\left.\frac{\partial f}{\partial x}\right|_{(x,y)}, \left.\frac{\partial f}{\partial y}\right|_{(x,y)}, -1\right)$. This defines a vector that is orthogonal to the graph of $f$ and is the normal vector to the tangent plane at the point $(x,y)$. Thus, from this very abstract definition of a derivative given above, we recover the intuitive idea that the tangent plane should represent the derivative.

For instance, suppose we have the function $f(x,y) = x^2 + y^2$ and we'd like to find its derivative and tangent plane at the point $(2,-3)$. We compute the gradient $\nabla f = (2x, 2y)$, so $\left. \nabla f \right|_{(2,-3)} = (4, -6)$. Note that $f(2,-3) = 13$. Letting $g(x,y,z) = f(x,y) - z$ as above, then $\left.\nabla g\right|_{(2,-3,13)} = (4, -6, -1)$ This is the normal vector of the tangent plane of the graph of $f$ at $(2,-3)$, which we compute as $$ 4x - 6y - z = (4, -6, -1) \cdot (x,y,z) = (4, -6,-1) \cdot (2, -3,13) = 8 + 18 - 13 = 13 $$ so the tangent plane is given by $z = 4x - 6y -13$.

For more background, I recommend Apostol's Mathematical Analysis.

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    $\begingroup$ Many thanks for your extremely useful answer! $\endgroup$ – Tien Kha Pham Mar 13 '15 at 6:09

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