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I can't really understand how can the partial derivatives of a function exist at a point where the function is not continuous.
For example, say we have a function f(x,y) that has domain all of R^2 except (0,0). Then in the definition of the partial derivatives, how can we even compute the limit if f(0,0) does not exist?
Thank you.

NOTE: The particular example I am bothered by can be found in See Colley's "Vector Calculus" at p.123, example 7. The example used there is for the function $\dfrac{x^2y^2}{x^4+y^4}$

This is the full solution from the book

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    $\begingroup$ That's because it is continuous restricted to the coordinate directions, but it's not continuous in all directions. $\endgroup$
    – Crostul
    Aug 20 '16 at 16:50
  • $\begingroup$ Make a picture. $\endgroup$ Aug 20 '16 at 22:51
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The existence of partial derivatives for $f:\Bbb R^n\to \Bbb R^p$ at a point $a=(a_1,\dots,a_n)$ where $f$ is defined (see edit and comments below) corresponds to the differentiability of the (single-variable) functions $$f_i : x\mapsto f(a_1,\dots,a_{i-1},a_i+x,a_{i+1},\dots,a_n)$$ for $1\leq i\leq n$. In other words, $f$ is differentiable (hence continuous) when restricted to the lines parallel to the $n$ coordinate axis of $\Bbb R^n$ passing through $a$. But this doesn't suffice to ensure continuity of $f$ because this continuity means that $f$ is continuous when restricted to all the directions around $a$ (not only the directions of the coordinate axis). For example, see my answer in Differentiability of Multivariable Functions

Edit: A function which is undefined at a point $a$ can't have partial derivatives at this point (simply because the functions $f_i$ of my answer above are undefined and hence, as in single-variable calculus, they have no derivative at this point because the quotient $\dfrac{f_i(h)-f_i(0)}{h}$ doesn't make sense, since $f_i(0)=f(a)$ doesn't exist!).

2nd Edit: In the example given, there is no problem: the function is well-defined at $(0,0)$ by $f(0,0)=0$ (see 2nd line after the brace). By definition, we have $f(0,0)=0$ (this has nothing to do with the definition of $f$ at the other points, she may have chosen another arbitrary value for $f(0,0)$ ). Hence the computation of the limits of $\dfrac{f_i(h)-f_i(0)}{h}$ (for $i=1$ or $2$) makes sense and it gives 0 because, as it's explained in your book, $\forall x\in\Bbb R, f(x,0)=0$ and $\forall y\in\Bbb R, f(0,y)=0$.

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  • $\begingroup$ But, if say the point (0,0) is not defined for a function f: R^2--> R , meaning f is not continuous at (0,0), then doesn't that mean that any partial derivative at (0,0) should not exist? I have found that there are cases that they do exist at points like that, but I don't quite understand why $\endgroup$ Aug 20 '16 at 17:00
  • $\begingroup$ A function which is undefined at a point can't have partial derivatives at this point (simply because the functions $f_i$ of my answer above are undefined and hence, as in single-variable calculus, they have no derivative at this point). $\endgroup$
    – paf
    Aug 20 '16 at 17:09
  • $\begingroup$ But I found an example of such a function where the partials at the origin do exist. See Colley p.123, example 7. $\endgroup$ Aug 20 '16 at 17:11
  • $\begingroup$ Could you edit your question by adding a description of Colley's example? I don't have this book with me. $\endgroup$
    – paf
    Aug 20 '16 at 17:13
  • $\begingroup$ I edited. Sorry for not using LaTex $\endgroup$ Aug 20 '16 at 17:16

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