0
$\begingroup$

I am very aware that a partial derivative could be not continuous at a certain point. However, I don't understand when can a partial derivative doesn't exist. Because on the book, they always say "If the partial derivative exists ...." or sometimes "If the partial derivative exists and is continuous at (x,y)...." So I assume, there must be a situation where it doesn't even exist

Also, unlike single variable calculus, where we know that some functions such as $|x|$ is not differentiable and thus we have no derivative. But I learnt that (Hopefully I am correct) that partial derivative exists even when the function is not differentiable

However, as for me, it is just about find a derivative of a function, and I can just do that by following derivation rules so I can't think of a situation where I can't find a partial derivative of a function (I feel there should always be an expression regardless if that derivative is continuous at (x,y) or not.

I ask this question because when I was checking past calculus exams, there are always questions where they ask if the partial derivatives exists, while the function given are always in forms such as: $$\displaystyle f(x,y) = \frac{x^2y^2}{x^4+y^4} \text{ that involves fractions}$$

Can someone give me an example and explain a bit for me?

NB: I am just learning multivariable calculus, so from what I have learned so far I can't think of any situation like this. Not sure if this happens in more advanced math

$\endgroup$
5
  • 1
    $\begingroup$ Using your example, consider $f(x,y)=|x+y|$ at $(0,0)$. The sentence "partial derivative exists even when the function is not differentiable" should mean "partial derivative CAN exists even when the function is not differentiable", as differentiable means differentiable from all direction. $\endgroup$ Oct 24 '20 at 20:17
  • $\begingroup$ @David Cheng But is a function involving absolute value, the only possible situation where the partial derivative doesn't exist? $\endgroup$
    – Yan Zhuang
    Oct 24 '20 at 20:21
  • 1
    $\begingroup$ There are even lot's of example in single variable calc. $\endgroup$ Oct 24 '20 at 20:22
  • $\begingroup$ @YanZhuang For the function $f$ you mention, first you need to define that is $f(0,0)$. It seems to me that whatever you choose, the function will not be even continuous, as it has no limit in the origin. $\endgroup$ Oct 24 '20 at 20:22
  • $\begingroup$ @PeterFranek Oh okay, thanks for the reminder. I am not yet too used to the rigorousness of math. $\endgroup$
    – Yan Zhuang
    Oct 24 '20 at 20:24
2
$\begingroup$

Take, for instance,$$\begin{array}{rccc}f\colon&\Bbb R^2&\longrightarrow&\Bbb R\\&(x,y)&\mapsto&|x|.\end{array}$$Then $\frac{\partial f}{\partial x}(0,0)$ doesn't exist, since the limit$$\lim_{h\to0}\frac{f(h,0)}h\left(=\lim_{h\to0}\frac{|h|}h\right)$$doesn't exist.

$\endgroup$
3
  • $\begingroup$ But this is the only case when it happens (I mean when the function involves absolute value)? Are there other situations? $\endgroup$
    – Yan Zhuang
    Oct 24 '20 at 20:20
  • 1
    $\begingroup$ Sure. If, say $f(x,y)=x\sin\left(\frac1x\right)$ if $x\ne0$ and $f(0,y)=0$, then $\frac{\partial f}{\partial x}(0,0)$ doesn't exist. $\endgroup$ Oct 24 '20 at 20:22
  • 2
    $\begingroup$ Yes, for example discontinuous functions. And there are even nowhere differentiable functions. $\endgroup$
    – LL 3.14
    Oct 24 '20 at 20:22
2
$\begingroup$

Consider te equation of a cone $z=a\sqrt{x^2+y^2}$. This is not derivable at $(x,y)=(0,0)$ where there is the vertex of the cone.

$\endgroup$

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.