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Here's my confusion: My teacher, as well as some online sources such as Khan Academy, seem to assume that the derivative expression being undefined at a point implies that the function being differentiated is not differentiable at that point.

Consider, for example, $g(x)=x^{1/3}$. The second derivative is $g''(x)=-2/9*x^{-5/3}$. In this Khan Academy video, the speaker concludes that the second derivative doesn't exist at $x=0$ because if you plug zero into the $g''$ expression you end up dividing by zero. But why does that conclusion follow? How can we be sure the expression is defined wherever the function is twice differentiable?

Along similar lines, when doing implicit differentiation in class, we were taught that a function $y$ fails to be differentiable when the expression we get for $dy/dx$ is undefined. For example, if $x^2+2xy+2y^2=1$, then we found $\frac{dy}{dx}=\frac{(-2x-2y)}{(2x+4y)}$. We were told that to find where $y$ fails to be differentiable, we should set $2x+4y = 0$, because that is the denominator of our derivative expression and we can't divide by zero. But again, as asked above, why can we be certain that having the derivative expression undefined implies that $y$ isn't differentiable?

Finally, I'll note that there's at least one example where I've noticed disconnect between where the expression is defined and where the derivative exists. That example is $f(x)=\ln(x)$. That is obviously not differentiable for $x<0$, yet the derivative expression, $f'(x)=1/x$, is defined for $x<0$ (it's only undefined if $x=0$). How do we reconcile that with what I've written in the preceding paragraphs?

I hope my question is clear. I can clarify if necessary - I realize it's a bit complicated and lengthy.

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    $\begingroup$ Regarding your second paragraph, the reasoning is incorrect. For example, the function defined by $f(x) = x^2\sin(1/x)$ and $f(0)=0$ is differentiable at $x=0,$ but the "differentiated expression" obtained by using the usual short-cut rules for differentiation is not defined at $x=0.$ The short-cut rules all assume differentiability for their application, so it is incorrect reasoning (in the sense of assuming the converse of a statement follows from the statement) to infer lack of differentiability from using the short-cut rules. $\endgroup$ Jan 14, 2020 at 17:31
  • $\begingroup$ @DaveL.Renfro Good point, but that seems a little different from the scenario in my second paragraph. Applying the short-cut rules occurs in both your example and mine, but yours is a piece-wise function where applying the rules to one piece clearly isn't sufficient. That doesn't entirely answer my question, I think. $\endgroup$
    – Will
    Jan 14, 2020 at 20:06
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    $\begingroup$ Regarding your comment, to be fully rigorous all appropriate (finite) differentiability assumptions need to be established prior to the use of the various short-cut rules. Note that the statement of the power rule here includes the assumption that $f$ is (finitely) differentiable. By careful (i.e. rigorous) proof, one can show that when the result of the power rule (and most other such short-cut rules applied to most any function explicitly defined by the use of formulas involving "precalculus type expressions") (continued) $\endgroup$ Jan 15, 2020 at 8:49
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    $\begingroup$ leads to an undefined value, then the function is not (finitely) differentiable at that value, and thus for elementary calculus purposes this kind of stuff is usually ignored (or included in carefully written "stealth language" for actual theorem statements), since the primary focus on this stuff in elementary calculus is on techniques such as mechanically applying the short-cut rules and quickly spotting where expressions are undefined. $\endgroup$ Jan 15, 2020 at 8:50

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Remember that derivative of $f$ at point $a$ is defined as

$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$

So in your example of $f(x) = \ln x$, the derivative of $f$ at point $a$ is

$$f'(a) = \lim_{x \to a} \frac{\ln x - \ln a}{x-a}$$.

Now, if you wanted to compute this limit for $a < 0$, would it make sense?

The answer is no, because in order to have a limit at a point $a$, we need that $f$ is defined in some open interval containing $a$.

So in this sense, $f$ is not defined for $x < 0$

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  • $\begingroup$ Thanks. But my question wasn't primarily about that example - I think I understand it (and did even before asking the current question). Rather, my question is, if in that case the (natural) domain of the derivative formula found from the shortcut rules differs from the actual set of differentiable points, then how can we be sure they don't differ in other cases, like the first two I noted? $\endgroup$
    – Will
    Jan 14, 2020 at 20:18
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You have a function, $f(x)=\ln x$, defined on $(0,\infty)$. And on $(0,\infty)$, its derivative is $1/x$.

But now you introduce another function, $g(x)=1/x$, defined on $\Bbb R-\{0\}$. It agrees with the derivative of $g$ on the interval $(0,\infty)$; but there is no reason to expect that this should magically make $f$ differentiable on $(-\infty,0)$. I could just as well define a function $h(x)$ which equals $1/x$ on $(0,\infty)$, and equals $e^{x^3}$ on $(-\infty, 0)$. Would you see this as evidence that the derivative of $f$ on $(-\infty,0)$ exists?

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  • $\begingroup$ Thanks. But my question wasn't primarily about that example - I think I understand it (and did even before asking the current question). Rather, my question is, if in that case the (natural) domain of the derivative formula found from the shortcut rules differs from the actual set of differentiable points, then how can we be sure they don't differ in other cases, like the first two I noted? $\endgroup$
    – Will
    Jan 14, 2020 at 20:11
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    $\begingroup$ @Will: I'm not sure I understand. Obviously the domain of the derivative can't be greater than the domain of the original function. But equally obviously, it can be smaller. So what is your problem here? (Also, note that your second example is not actually a function at all, because it is two-valued in most of its domain.) $\endgroup$
    – TonyK
    Jan 14, 2020 at 20:28
  • $\begingroup$ I get why you're confused by my question. How about this ... forget about the $ln(x)$ example and just look at the $dy/dx$ question in my original post. $\endgroup$
    – Will
    Jan 14, 2020 at 20:43
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    $\begingroup$ @Will: OK: you ask "How can we be sure the expression is defined wherever the function is twice differentiable?" The simple answer is: we can't. You must approach such problems on a case-by-case basis. (The two examples that you give are easy to resolve, because the tangent to the curve is vertical at all the points in question.) $\endgroup$
    – TonyK
    Jan 14, 2020 at 21:19
  • $\begingroup$ Can you explain more? Why does it help that the tangents are vertical, rather than just completely non-existent? Something about limits? $\endgroup$
    – Will
    Jan 14, 2020 at 23:43

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