Many posts here have answered questions regarding the partial derivatives of multivariable functions that are continuous at some point.

Consider the equation $(xy)^\frac{1}{3}$ which is continuous at $(0,0)$. When we take the partial derivatives, we see that they yield an undefined result, so we need to resort to the limit definition. I suppose this makes sense since the limit definition will yield the partial derivative without possibly being at the point (it does not even require continuity for that matter).

For the above example, the limit definitions yield $0$ for both.

$\dfrac {\partial f}{\partial x} = \frac{y^\frac13}{3x^\frac{2}{3}}$ and $\dfrac {\partial f}{\partial y} = \frac{x^\frac13}{3y^\frac{2}{3}}$

Even though the limit definition solves the problem, I still find it difficult to see that taking the partial derivative through differentiation and evaluating is not exactly equivalent to finding the partial derivative at a point by the limit definition. Surely the limit definition provides the framework for taking the partial derivatives the "normal" way. Why does this "discrepancy" occur?

  • 1
    $\begingroup$ The exponents in the denominator should be $2/3$, not $-2/3$. $\endgroup$
    – angryavian
    Jun 9, 2018 at 20:29
  • $\begingroup$ The issue here is that $x \mapsto \sqrt[3]{x}$ is not differentiable at $x=0$. If you take the partials at the origin, then the function is zero, so the composite is differentiable (along the axes). There is no inconsistency. $\endgroup$
    – copper.hat
    Jun 9, 2018 at 20:35
  • $\begingroup$ Yes I'm sorry. I guess I overdid it with the fraction and negative exponent. Will edit. $\endgroup$
    – Ian L
    Jun 9, 2018 at 20:46

1 Answer 1


The problem in this particular example is that you have used a differentiation rule incorrectly. To get $$\frac{\partial f}{\partial x}=\frac{y^{1/3}}{3x^{2/3}},$$ I'm guessing you used the following rule: if $g(x)=ch(x)$ for some constant $c$ and for all $x$, then $g'(x)=ch'(x)$. (Here $y$ is fixed, $c=y^{1/3}$, $g(x)=f(x,y)$, and $h(x)=x^{1/3}$.) However, this rule is not literally correct. The correct rule is that if $h'(a)$ exists and $g(x)=ch(x)$ for all $x$ (in a neighborhood of $a$), then $g'(a)$ exists and $g'(a)=ch'(a)$. For $a=0$ and $h(x)=x^{1/3}$, $h'(a)$ does not exist, so you cannot apply the rule and conclude that your formula for $\frac{\partial f}{\partial x}$ is correct at $x=0$.

(Note that if $c\neq 0$, then the naive rule $g'(x)=ch'(x)$ is valid in the sense that for each $x$ either both derivatives exist and the equation is true or neither derivative exists. That is because if $g'(a)$ exists for any particular $a$, you can write $h(x)=c^{-1}g(x)$ and then use the rule with $c^{-1}$ in place of $c$ to conclude that $h'(a)$ exists and $h'(a)=c^{-1}g'(a)$. But when $c=0$, it is possible that $h'(a)$ is not defined but $g'(a)$ is, and that is exactly what is happening in this example, since $c=y^{1/3}$ is $0$ when $y=0$.)

  • $\begingroup$ So I suppose a single variable analog would be $(0\cdot-\frac{1}{x})$. Even though the derivative of $h$ is undefined at $0$, $g$ still exists. $\endgroup$
    – Ian L
    Jun 9, 2018 at 20:53
  • $\begingroup$ Well, that example is a little problematic because the function $g$ would not even be defined at $0$ (though it can be extended to be differentiable at $0$). A better example would be $g(x)=0\cdot x^{1/3}$. $\endgroup$ Jun 9, 2018 at 20:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .