Derivative of $\sin (x) $ at end points of $[o,\pi]$ Can we define derivative of a function at the end points of the domain? If so how? My confusion is only onside limit exists. Is one side enough? or we define the derivative in the higher space in this case $\mathbb{R}$ as the derivative of the subspace?
Example: $\sin(x)$ where $x\in [0,\pi]$. What is the derivative at $0$ and $\pi$? 
On a related topic; I ask this question because I want to show $\sin(x)$ is not a contraction in the above interval. But I am looking for a answer to the above question of derivatives.
Please explain. Thank you.
 A: It depends on which definition of derivative you use. One definition where the question does not even arise is:
Definition: Let $A \subset \mathbb{R}$, and $f \colon A \to \mathbb{R}$ a function. $f$ is differentiable in $x \in A$, if there is at least one sequence $(x_n)$ of points in $A\setminus \{x\}$ with $x = \lim\limits_{n\to\infty} x_n$, and for every sequence $(y_n)$ of points in $A\setminus \{x\}$ converging to $x$, the limit
$$f'(x) = \lim_{n\to\infty} \frac{f(y_n) - f(x)}{y_n-x}$$
exists. $f'(x)$ is the derivative of $f$ in $x$.
A nice standard definition of differentiability and the derivative that does not tie you to functions whose domain is open. With that definition, there is no question that $\sin\lvert_{[0,\pi]}$ is differentiable in $0$ and in $\pi$, and that the derivative there is $1$ resp. $-1$.
A: We cannot define the derivative of a function at the end points of its domain because it is only right-continuous at the left endpoint and left-continuous at the right endpoint. Therefore, it is only right-differentiable at the left endpoint and left-differentiable at the right endpoint, and hence not wholesomely differentiable at the two endpoints. Hence the derivative is not defined at the two endpoints.
A: In a sense it is difficult to answer your question, because anyone can define anything, so I think I can assume that when you ask if is possible to define derivatives in the end points of $[a,b]$, I believe that what you want is a definition which matchs with the standard definition in the usual case, i.e. you want to extend the usual definition that Works only for open sets. 
If you open some books of Partial Differential Equation, you will see that there exist a space $C^1(\overline{\Omega})$. Well, $\overline{\Omega}$ is not open, so how do we define the derivative in $\partial\Omega$? In our case this can be done by just considering one side derivatives. You could say something like this: a function $f:[a,b]\to\mathbb{R}$ is differetiable if it is differentiable in the usual sense in $(a,b)$ and if in the points $a,b$ it does have one side derivatives. Note that this make sense geometrically.
