What does it mean for a function to be differentiable? What does it mean for the derivative of a function to exist at every point on the function's domain? It seems a very abstract thing to visualize. 
Can someone elaborate?
 Answers are much appreciated.
 A: I like to visualize it in terms of the definition and secant lines.  Remember that the definition of the derivative at a point $x_{0}$ is
$$f'(x_{0}) = \lim_{h\to 0}\frac{f(x_{0}+h) - f(x_{0})}{h}.$$
For any fixed value of $h$, the line $$y = \frac{f(x_{0}+h)-f(x_{0})}{h}\left(x - x_{0}\right) + f(x_{0})$$ is the secant line through the points $(x_{0},f(x_{0}))$ and $(x_{0}+h,f(x_{0}+h))$.  If the derivative of $f$ exists at $x_{0}$ then regardless of your choice of $h$, as you take it to $0$ the secant lines will all approach the same tangent line, namely $$y = f'(x_{0})(x-x_{0}) + f(x_{0}).$$
One way to actually visualize it is with Desmos.  Take a look at this example.  I included two functions (in red), one that is differentiable at zero and one that is not.  Click on the folder icons to switch them on/off.  As you move the slider for $h$ toward $0$ from the left and right observe the difference in the behavior of the secant line (in blue).  In the differentiable function the line approaches a horizontal line regardless of whether you approach $h=0$ from the left or the right. With the non-differentiable function, the secant line approaches a vertical line as $h\to 0$ from the right and a horizontal line as $h\to 0$ from the left.  A function which is differentiable everywhere will exhibit this sort of consistent behavior with respect to the secant lines at every point in its domain.
A: Intuitively, saying the derivative of a function $f$ exists at a point $x$ in its domain means that, if you zoom in 'enough', it will look more and more like a straight line. The tangent line of $f$ at $x$, then, is that line it resembles if you zoom in, with the slope of the tangent line being the derivative at $x$.
If a function $f$ is differentiable at its entire domain, that simply means that you can zoom into each point, and it will resemble a straight line at each one (though, obviously, it can resemble a different line at each point - the derivative need not be constant).
Not every function has this nice property, for instance $f(x) = |x|$; no matter how much you zoom in at the point $x = 0$, it will never resemble a line. (For all other $x$, of course, it is differentiable).
Depending on how advanced a course may be, this idea can be developed further, in that one can speak of a function not just resembling a line 'around' a given point $x$, but it may resemble a parabola, or a cubic equation, or an even higher order polynomial (this is what Taylor's theorem is about, intuitively speaking).
A: I think of "differentiable" as "smooth."  A differentiable function on the domain $(a,b)$ has no holes, points, cusps, vertical asymptotes or other things that might make the ride from $a$ to $b$ unpleasant.  The tangent line at every point is a really good approximation to the function, so "smooth" is a good word.
However, "smooth" is a technical term, so what I just wrote will eventually be misleading.  In higher math, a "smooth" function has to have a continuous derivative (or even several higher-order, continuous derivatives.)  But still, in layman's terms, "smooth" is a good way to think of "differentiable."
