When we take time derivative of a function of time, then is the result another function of time, again? (I'll try to explain my question by one known example), for example where the velocity is a function of time v(t) then its time derivative (which is acceleration: $a=\frac {dv}{dt}$) is another function of time a(t)?
(because according to defination of integral it must be another function of time: $v=\int a(t)dt$)
and then what will be time derivative of acceleration?
 A: You've obviously seen the equation $v = \int a(t)dt$ somewhere and been mislead by it. It's entirely possible for the acceleration to be a constant in which case we get $v = \int k \space dt$ for some constant $k$, and therefore $dv/dt$ is a constant, $k$, and not a function of time. We write the acceleration as $a(t)$ because this is the most general form and covers all eventualities.
The time derivative of acceleration is known as jerk. In the example above, where $a$ is constant, the jerk will be zero.
A: In general the acceleration would be a function of time, i.e. $ d^2x/dt^2 = a(t) $.
A: You've probably seen the equation for position when the acceleration is the constant $a$:
$x(t) = x_0 + v_0  t + \frac{1}{2} a t^2$
Differentiating once yields:
$\dot x(t) = v(t) = v_0 + at$
Differentiating again yields:
$\ddot x(t) = a(t) = a$
Now, it may seem funny to write $a(t) = $ constant  but it's perfectly acceptable because, in general, acceleration is a function of time.  In this case, it just happens to be a function of $t^0$.
A: The derivative of acceleration with respect to time is called jolt, or jerk. Sometimes, this jolt (jerk) may be non-differentiable so that the jounce (derivative of jolt/jerk) is not continuous. In such cases, the fourth, or more generally, the $n^{\operatorname{th}}$ derivative of the position may not be continuous. Some times, it is differentiable an arbitraryily large number of times, and then, for the position $x(t)$, a polynomial function of time $t$, differentiating it $n$ times results in a constant value, where $n$ is the degree of the polynomial function of time $t$ which the position $x$ is. 
So, it is easier to write $\frac{d^nx}{dt^n}$, and solve the differential equation you get. Hope it helps!
(If you believe I have misinterpreted your question, kindly mention so in the comments)
