differentiation of fractional part of $x$ What is the differentiation of fractional part of $x$?
Since the slope of $\{x\}$ is $1$ so that derivative of $\{x\}$ should be $1$. Is it correct or not
 A: Your function $\{x\}$ has derivative $1$ as you note, except $\{x\}$ has jumps (of $-1$) at each integer.  In the theory of distributions, the derivative of a unit jump at $0$ is a measure called $\delta$.  So
$$
\frac{d}{dx}\{x\} = 1 - \sum_{n \in \mathbb Z} \delta(x-n)
\tag{1}$$
(This is a simple example of a Lebesgue decomposition of a signed measure.)
What does it mean?  For example, we may write a Stieltjes integral like this
$$
\int_{-\infty}^{+\infty} \varphi(x)\;d\{x\} = \int_{-\infty}^{+\infty} \varphi(x)\;dx - \sum_{n \in \mathbb Z} \varphi(n)
\tag{2}$$
for nice enough functions $\varphi$.  For example, this works when $\varphi$ is continuous with compact support.
A: Is the function continuous? No
So the derivative is undefined for $x \in \mathbb{Z}$ and $1$ everywhere else.
A: It is not possible to differentiate the fractional part of x when $x \in \mathbb{Z}$. This is because the graph of $\{x\}$ is not continuous. So its derivative does not exist.
Why its derivative does not exist? If you look at the right-hand derivative and left-hand derivative of the integral values of $x$, they are not the same. For a function to be differentiated, the left-hand derivative and right-hand derivative must be the same. Hence, its derivative does not exist at $x \in \mathbb{Z}$.
However, other than the integral values of $\{x\}$ the derivatives exist. The derivative for other values of $x$ is evidently 1. To generalise the derivative of ($\frac{d}{dx}\{x\}, x \in \mathbb{R-Z} = 1$). 
A: You can differentiate this only at points $x\notin\mathrm Z,$ since exactly at such points $[x]$ is discontinuous and thus perforce not differentiable. With this proviso, since you have $\{x\}=x-[x],$ then we have that $$\left(\{x\}\right)'=\left(x-[x]\right)'=1-0=1,$$ provided $x\notin\mathrm Z.$
