Function being differentiable vs. derivative expression being undefined 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. 
 A: 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 $x < 0$ 
A: 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?
