Does every continuous function have a left and right derivative? I understand that differentiability implies continuity, whereas the converse isn't true. But must a continuous function have both a left and right derivative, not necessarily equal to one another?
 A: For $f(x)=x \sin{\frac{1}{x}}$, $f(0)=0$ (Just like Dave commented) you have:
$$\frac{f(0 + \Delta x) - f(0)}{ \Delta x} = \frac{(0 + \Delta x) \sin (\frac{1}{0 +\Delta x}) - 0}{\Delta x} = \sin (\frac{1}{\Delta x})$$
As $\Delta x$ approaches $0$ from the left/right you will just get $- \sin(x)$ / $\sin(x)$ as $x$ tends to $\infty$, i.e, no limit.  
But, let $\epsilon > 0$ and $ \delta = \epsilon$:
$$|x|=|x-0| < \delta \implies |x \sin (\frac{1}{x}) - 0| = |x \sin (\frac{1}{x})| \leq |x| < \delta =  \epsilon $$
Hence $f$ is continous at $0$.
A: If $f$ has a left and right derivative at every point, then it certainly has finite Dini derivatives at every point, and then it follows immediately from the Denjoy-Young-Saks Theorem that $f$ is differentiable away from a set of measure $0$.  But as pointed out in Ted Shifrin's answer, there are certainly continuous functions which are nowhere differentiable, e.g. Weierstrass's function.  
[Or in other words: it is relatively common to describe Weierstrass-like nowhere differentiable functions as "having corners at every point".  The above paragraph shows that this is very loose language: taken literally, there are no such functions!]
Thus the Weierstrass function is continuous everywhere and certainly has left and right hand derivatives only at a set of measure zero.  This checks a weak form of Ted's answer, but enough to use the Weierstrass function to answer the OP's question...lazily.
Really though this answer is almost a joke: I believe I am applying too big a theorem to deduce too small a result.  When I first read the question I was tempted to say that a function which has (finite!) left and right hand derivatives at every point must be differentiable on the complement of a countable set, just like a function which has left and right hand limits at every point must be continuous except at countably many points.  Is that actually true?  (Added: Hmm, I think I've changed my guess; I now doubt that's true.  But I'd still like to know.)
A: The famous Weierstrass function, which is continuous but nowhere differentiable, fails everywhere to have left- and right-derivatives, as well. This takes a bit of work to check.
