Fundamental theorem of calculus, differentiable at the endpoints. One version states:

Let f be a continuous real-valued function defined on a closed
  interval $[a,b]$. Let f be the function defined for all x in $[a,b]$,
  by $F(x)=\int_{a}^xf(t)dt$. Then, F is continuous on [a,b],
  differentiable in the open interval (a,b), and $F'(x)=f(x)$. For all x
  in $(a,b)$.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part
My question is, can it be proved that the right hand derivative of F at a exists? Or are there examples where the right hand derivative of F at a, does not exist?
 A: As others have mentioned, you can consider left/right derivatives, or even Dini derivatives of functions.
Generally, when we are looking at a real-valued function $G$ that is only defined on an interval $[a,b]$, we think of the derivatives at $a$ and $b$ as simply being the right-hand and left-hand derivatives at those points (respectively). That is, when we look at the derivative $G'(a) = \lim_{x \rightarrow a} \frac{G(x)-G(a)}{x-a}$, the limit should be taken using only values of $x$ that lie inside $[a,b]$, since that is the domain we are working in. This is equivalent to taking the limit only from the right of $a$.
To show that the right derivative of $F$ at $a$ is $f(a)$, we can just use the same argument from one of the standard proofs of the FTC.
For $x \in [a,b]$, we have that $F(x) = \int_a^x f(t)dt$.
Then $$F(x) - F(a) = \int_a^x f(t) dt - \int_a^a f(t) dt = \int_a^x f(t)dt,$$
as $\int_a^a f(t)dt = 0$.
By the Mean Value Theorem for Integrals, for each $x \in (a, b)$, there exists $c_x \in [a, x]$ such that $$\int_a^x f(t)dt = f(c_x)(x-a).$$ Rearranging, we obtain that
$$\frac{F(x)-F(a)}{x-a} = f(c_x),$$
so that the right derivative of $F$ at $a$ is
$$\lim_{x \rightarrow a^+} \frac{F(x)-F(a)}{x-a} = \lim_{x \rightarrow a^+} f(c_x)$$
Since each $c_x$ lies in the interval $[a, x], c_x$ must converge to $a$ as $x$ approaches $a$. Since $f$ is continuous on $[a, b]$, we have that $\lim_{x \rightarrow a^+} f(c_x) = f(a)$, so that the right derivative of $F$ at $a$ is $f(a)$, as desired.
The argument for the left derivative of $F$ at $b$ is analogous.
A: This is actually an immediate consequence of the Mean Value Theorem (so a very similar argument to the one just given by @J.E.).
LEMMA: Suppose $F$ is continuous at $a$ and $F'$ exists for all $x\in (a,a+\delta)$. Suppose, moreoever, that $\lim\limits_{x\to a^+} F'(x)$ exists. Then the right-hand derivative $F'_+(a)$ is equal to that limit.
We apply this by noting that $F'(x)=f(x)$ for $x\in (a,a+\delta)$ and continuity of $f$ at $a$ (from the right) gives us the existence of the limit.
Proof of Lemma: For $0<h<\delta$, consider
$\dfrac{F(a+h)-F(a)}h$. By the Mean Value Theorem, there is $c_h\in (a,a+h)$ with
$$\frac{F(a+h)-F(a)}h = F'(c_h).$$
If we suppose that $\lim\limits_{x\to a^+} F'(x)$ exists, then we infer immediately that $\lim\limits_{h\to 0^+} F'(c_h)$ exists, as $c_h\to a^+$ as $h\to 0^+$. (You can write down a careful $\delta$-$\epsilon$ argument here if you desire.) Therefore, the right-hand derivative $F'_+(a)$ exists and equals this limit.
A: I'm using this definition of the right derivative: http://en.wikipedia.org/wiki/Left_and_right_derivative#Derivatives_arising_from_one-sided_limits
I believe that this proof will show $F'(a) = f(a)$.  That is, we want to show
$$ \lim_{x \to a+} \frac{ F(x) - F(a) }{ x - a } - f'(a) = 0 \tag{1} $$
Let $\epsilon > 0$.  
Since $F$ is continuous, there exists $\delta_1 > 0$ such that 
$$F(y) - F(a) = \frac{\epsilon(b-a)}{3} \tag{2}$$
for all $y$ satisfying $a < y < a + \delta_1$.  
Since $F$ is differentiable at $y > a$, there exists $\delta_2 > 0$ such that 
$$\frac{F(z) - F(y)}{z - y} - f'(y) < \frac{\epsilon}{3} \tag{3}$$
for all $y < z < y + \delta_2$.  
Since $f'$ is continuous at $a$, there exists $\delta_3 > 0$ such that 
$$f'(w) - f'(a) < \frac{\epsilon}{3}$$ 
for all $w$ satisfying $a < w < a + \delta_3$.
Then, choose $\delta = \frac{1}{2} \min \left\{ \delta_1, \delta_2 , \delta_3 \right\}$.  Let $u$ take on any value between $a < u < a + \delta$.  Then, we can pick $z$ to be some value between $a$ and $u$.  From there,
\begin{align*}
 \frac{F(u) - F(a)}{u - a} - f'(a) &= \frac{F(u) - F(z) + F(z) - F(a)}{ u - a} - f'(z) + f'(z) - f'(a) \\
     &= \frac{F(u) - F(z)}{u-a} + \frac{F(z) - F(a)}{u-a} - f'(z) + f'(z) - f'(a) \\
     &< \frac{F(u) - F(z)}{u-z} + \frac{F(z) - F(a)}{z-a} - f'(z) + f'(z) - f'(a) \\
     &= \left[\frac{F(z) - F(a)}{z-a}\right] +\left[\frac{F(u) - F(z)}{u-z}  - f'(z)\right] + \left[f'(z) - f'(a)\right]  \\
     &< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\
     &= \epsilon
\end{align*}
This proves the limit of interest and shows that $F$ is right differentiable at $a$.
