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.


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.

  • 1
    $\begingroup$ Many thanks for your extremely useful answer! $\endgroup$ – Tien Kha Pham Mar 13 '15 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.