The definition of a partial derivative is the "derivative of a multi-variable function relative to a single variable when all other variables are held constant".

But isn't the regular derivative (for one-variable functions) just a trivial case of this, where there are no other variables to hold constant? Why do we need the separate notation for partial derivatives (that is, writing $\displaystyle \frac{\partial f}{\partial x}$ rather than just $\displaystyle \frac{df}{dx}$)?

  • 3
    $\begingroup$ Because when dealing with functions of several variables, the partial derivatives need to be distinguished from the total derivative. The total derivative need not exist when all partial derivatives exist. $\endgroup$ Jan 29, 2014 at 22:01
  • $\begingroup$ I like this question, and would be interested in seeing examples. $\endgroup$
    – DanielV
    Jan 29, 2014 at 22:06
  • $\begingroup$ Choose $f(x,y)=x$ for $|x|<y$ and $-x$ otherwise. $\endgroup$ Jan 29, 2014 at 22:29
  • 3
    $\begingroup$ @Daniel Fischer: What you say is certainly true, but I don't think it really answers the question. I believe the question is: "Why do we have to write $\frac{\partial f}{\partial x}$ instead of just $\frac{df}{dx}$?" And I think the answer to this is: we don't. I believe in fact that some famous mathematician once asked rhetorically "What other kind of derivative is there?" $\endgroup$ Jan 29, 2014 at 22:54

2 Answers 2


Because there exists a notion of "total derivative" that is the multivariate analogue of the 1-D derivative you're familiar. The total derivative of a function $f: \mathbb{R}^n \to \mathbb{R}^m$ is known as the Jacobian of f. You may have worked with this while doing change of variables with multiple integrals. See http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant. When we say $f$ is differentiable at a point $a$, we mean that its Jacobian exists at $a$. There is a theorem that says that if all of the partial derivatives of $f$ exist and are continuous at $a$ then $f$ is differentiable at $a$, but the converse isn't true. See Can "being differentiable" imply "having continuous partial derivatives"?.

Furthermore, the total derivative defines a linear map and is used as a linear approximation of $f$, via Taylor's Theorem (http://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables), as well as in the implicit function theorem (http://en.wikipedia.org/wiki/Implicit_function_theorem) among others.

Now, the notion of the total derivative of $f: \mathbb{R}^n \to \mathbb{R}^m$ can be extended to functions $f: \Omega \subset V \to W$ where $V$ and $W$ are normed vector spaces and $\Omega$ is open (probably further, but I haven't gotten there yet).

Definition: We say that $f: \Omega \subset V \to W$ is differentiable at a $\in \Omega$ if there exists a bounded linear operator $L_f[h]$ such that:

$f(a+h) = f(a) + L_f[h] + E[h]$

where $lim_{h \to 0} \frac{|E[h]|_W}{|h|_V} \to 0$, i.e. the error term vanishes as h approaches zero (think about the definition of the derivative in $\mathbb{R}$)

An aside: bounded linear operator here means that $|L_f(x) - L_f(y)|_W \leq C|x - y|_V$ for $x, y \in \Omega$ and a constant $C$.

  • $\begingroup$ (Note Some people use the term "Jacobian" for the determinant of the matrix of the total derivative.) $\endgroup$
    – Pedro
    Jan 29, 2014 at 22:36
  • $\begingroup$ I've seen it used that way myself as well. The determinant is what is used in change of variables in multiple integrals. Furthermore, the inverse function theorem requires that the determinant to be nonzero at $a$ in order for $f^{-1}$ to be differentiable at $a$ (think of the 1-D analogue). Recall that a linear transformation is invertible iff the determinant is nonzero. $\endgroup$
    – Lost
    Jan 29, 2014 at 22:40
  • $\begingroup$ I've generally called the Jacobian matrix the "derivative matrix", as that's more literally what it is. Don't know if that's correct usage, though. $\endgroup$
    – Joe Z.
    Jan 29, 2014 at 22:52

The total derivative/differential of Frechet (see the Lost answer) is the good generalization for several reasons. Two rather big:

(1) existence of partial derivatives does not imply continuity;

(2) the failure of the chain rule for partial derivatives.


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.