If a function is undefined at a point, would its derivative be undefined at that point as well? If $f(x)$ is defined everywhere except at $x=x_0$, would $f'(x_0)$ be undefined at $x=x_0$ as well?
One example is: $$f(x)=\ln(x)\rightarrow f'(x)=\frac{1}{x}$$
In this particular case, both $f(x)$ and $f'(x)$ are undefined at $x=0$. I wonder if this always holds true.
Thank you.
 A: This depends on some conventions, but the typical answer is yes, because if a function is not defined somewhere, it cannot have a slope there! In other words, we certainly don't have a slope where there is no function value!
We need a value $f(x_0)$ to plug into the limit definition of the derivative, after all.
A: The usual definition, $f'(x)=\lim_{h \to 0}\frac {f(x+h)-f(x)}h$ clearly requires that we be able to evaluate $f(x)$ in a neighborhood of $x$.  A challenging case is where there is a removable discontinuity.  For example, take $f(x)=\frac {x(x-1)}x$.  It cannot be evaluated at $x=0$, but otherwise is the same as $g(x)=x-1$.  For a function which is differentiable at $x$, another expression is $f'(x)=\lim_{h \to 0}\frac {f(x+h)-f(x-h)}{2h}$.  This one avoids the problem of evaluating $f(x)$ at the point in question, so will give the sensible answer of $1$ for $f'(0)$.  Should $f(x)$ have a derivative at $0$?  If you think so, you can propose the second as a replacement definition.  The fact that you can give a sensible answer to another type of problem is a point in favor.  That is how definitions get changed-people find the new one more useful than the old one.
A: In order to have a first derivative at a point, a function must be continuous at that point. 
Differentiation requires continuity. A function that is not defined at some point cannot be continuous at that point because it does not exist at that point.
If the first derivative $f(x)$ is defined then all functions that have that derivative are defined as
$$F(x) = \int f(x) dx$$
which is defining a family of functions that differ only by a constant. The Riemann integral requires that a function is defined at the region of integration because at any interval we have to be able to create a mesh with arbitrarily small sizes and the evaluation of the integral cannot depend on the choice of a mesh.
Technically, we could ignore or circumvent a point where a function is not defined, but then we cannot speak about the derivative of that function at that point.
As an option, sometimes, we still attach a value to the function at that point so that it becomes defined. Typically, we choose such value so that the function becomes continuous as well.
Example:
$$F(x)=\frac{\sin(x)}{x}$$
is not defined at $0$ as the expression that we get by simply replacing $x=0$, $0/0$ is undefined. Yet we can additionally specify $F(0)=1$ getting a perfectly continuous and differentiable function at $0$.
There is a definition of symmetric derivative that can exist even when normal derivative does not exist
$$\lim_{h \to 0}\frac{f(x+h) - f(x-h)}{2h}$$
This is for example defined for $|x|$ where the function does not have a normal derivative. Symmetric derivative evaluates to the same value as ordinary derivative in case both exists.
Apart from these two amends, the answer is strictly no, a function cannot be undefined at a point and have a derivative at that point.
Notice that the first derivative at $x_0$ is defined precisely as
$$\lim_{h \to 0}\frac{f(x_0+h) - f(x_0)}{h}$$
so if anything makes this expression not evaluating, the first derivative does not exist. And in your case $f(x_0)$ is not defined, thus first derivative at $x_0$ is not defined as well.
