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There are many questions on Math Stack Exchange asking if a function is differentiable if it has a removable discontinuity at $x=a$. But, I'm having trouble following the answers. I get the impression, from reading here on Stack Exchange and elsewhere that such a function is not differentiable but I don't understand why.

Consider the following equation: $f(x)=\frac{x^3}{x}$. The function is undefined at $x=0$ but it is clearly differentiable. First by simplifying:

$\frac{d}{dx}\frac{x^3}{x}=\frac{d}{dx}x^2=2x$

Or, using the quotient rule:

$\frac{d}{dx}\frac{x^3}{x}=\frac{3x^2\cdot x-x^3\cdot 1}{x^2}=\frac{3x^3-x^3}{x^2}=\frac{2x^3}{x^2}=2x$

And, finally, my calculator agrees.

Therefore, it is established that the function is differentiable and has a derivative at every x-value in its domain, including the troublesome $x=0$.

I conclude that a function is differentiable at $x=a$ if the discontinuity is removable.

The only thing I can think of that would make this untrue is the idea that I have changed the original function $f(x)$ by removing the discontinuity with some algebra and I am really differentiating a different function (call it $g(x)$) that is not quite the same at the one point of interest. That argument would be more persuasive if I had simply simplified first and then differentiated. There I was clearly differentiating a different equation. But, I did not simplify when applying the quotient rule and I obtained the same answer. So, the argument seems weak, at best.

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    $\begingroup$ This is not really the heart of your error, but you did simplify (in a way that is invalid for $x=0$) when using the quotient rule, in the last step. $\endgroup$ – Eric Wofsey Jul 21 '19 at 7:42
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    $\begingroup$ The function $x\mapsto \frac{x^3}x$ isn't discontinuous. It has a singularity, a point where it's undefined. That's different. $\endgroup$ – Arthur Jul 21 '19 at 7:58
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    $\begingroup$ "... but it is clearly differentiable": no, it can't be differentiable where it is undefined. $\endgroup$ – Yves Daoust Jul 21 '19 at 15:13
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The map$$\begin{array}{rccc}f\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\frac{x^2}x\end{array}$$is undefined at $0$, and therefore it is meaning less to ask whether or not it is differentiable there. It happens that we can extended it to one and only one continuous function $F\colon\mathbb R\longrightarrow\mathbb R$, which is defined by $F(x)=x$. And it happens that this function is differentiable at $0$.

However, if you take$$\begin{array}{rccc}g\colon&\mathbb R\setminus\{0\}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x>0\\-x&\text{ if }x<0,\end{cases}\end{array}$$then you can extend $g$ to one and only one continuous map $G\colon\mathbb R\longrightarrow\mathbb R$, which is $G(x)=\lvert x\rvert$, but the function $G$ is not differentiable at $0$.

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$$\frac{x^3}x$$ doesn't have a removable discontinuity. It is continuous and differentiable in $\mathbb R\setminus\{0\}$, where it equals $x^2$. It is undefined at $x=0$.


A discontinuous function is not differentiable at the discontinuity (removable or not).

It the discontinuity is removable, the function obtained after removal is continuous but can still fail to be differentiable.

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  • $\begingroup$ First, I assume that $\mathbb R\setminus\{0\}$ means "all real numbers except $0$". Also, I'm guessing that you are telling me that "undefined at a point" $\ne$ "discontinuous at that point". That's strange. I'll have to review my textbook again to figure that one out. If $f(x)$ is not differentiable at $x=0$, then how can I say with confidence that the slope (derivative) at $x=0$ is $0$ (which I believe it is)? Clearly I am missing something important here. $\endgroup$ – Randall Blake Jul 21 '19 at 16:02
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    $\begingroup$ @RandallBlake: your function is undefined at $x=0$, it can't have a derivative. Review your definition of the dérivative, your confidence is undue. $\endgroup$ – Yves Daoust Jul 21 '19 at 16:07
  • $\begingroup$ So, if I am asked if $f(x)$ is differentiable at $x=0$, the correct answer is "no" even though I can calculate a sensible derivative at $x=0$?. I suppose that it really does boil down to the difference between $f(x)$ and $g(x)$ that I mentioned in the original post. $f(x)$ has no "point of tangency" at $x=0$, but $g(x)$ does. Somehow that's a little unsatisfying. $\endgroup$ – Randall Blake Jul 21 '19 at 16:26
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    $\begingroup$ @RandallBlake: for the last time, $f(0)$ is undefined and there is nothing you can say about $f$ at $0$. It is impossible to compute the derivative. $\endgroup$ – Yves Daoust Jul 21 '19 at 16:35

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