Can we multiply between two Derivative operator? I don't know what should be the tile of my question sorry for that ..
I will explain here clearly ...
Suppose this is my integral :
$\int \frac{d^2x}{dt^2}dx$
I have to show that the integral is equal to $(\frac{dx}{dt})^2$
My steps :
$=\int \frac{d}{dt}(\frac{dx}{dt})dx$
$=\int d\frac{d^2x^2}{d^2t^2}$  Note:  Here I am multiplying  between the two operator like $dx \cdot dx=d^2x^2 $ and $dt \cdot dt=d^2t^2 $
$=(\frac{dx}{dt})^2$
So now my question is can we multiply like the way I did according to my note . If my procedure is wrong how can i show $(\frac{dx}{dt})^2$ from the integral
 A: @WilliamBarnes has the correct answer, but I wanted to take some time to correct issues with your understandings of higher order differentials.  Additionally, I'll give a way to do this without the weird substitution.
The $d()$ is an operator, not a multiplier.  $dx$ is equivalent to saying $d(x)$, like you would say $\sin(x)$ or something.  $d^2(x)$ actually means applying the operator twice: $d(d(x))$.  Also, $dx^2$ actually means $(d(x))^2$.
However, the standard notation for the second derivative is actually misleading (or, I would say, wrong).  You shouldn't take the second derivative as meaning what it says.  Most people write it the second derivative of $x$ with respect to $t$ as you did: $\frac{d^2x}{dt^2}$.  However, this is fairly misleading as far as actual differentials go.  If you want manipulable differentials, you have to write the second derivative of $x$ with respect to $t$ as:
$$  \frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2} $$
Using this, you can now treat these as algebraically manipulable objects, keeping in mind the shorthands we are using (discussed in the first paragraph).  See my paper "Extending the Algebraic Manipulability of Differentials" for more information.
So, to do the integral, let's start (using the manipulable notation) by doing:
$$
I = \left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right)\,dx
$$
We can move one of the $dt$s to be with the $dx$, giving:
$$
I = \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)\,\frac{dx}{dt}
$$
The left side (in parenthesis) is the differential of the first derivative ($\frac{dx}{dt}$).  Therefore, we will set it to be $dv$ for integration by parts. $\frac{dx}{dt}$ will be $u$ as well as $v$.
This means we can rewrite this as:
$$
I = \frac{dx}{dt}\frac{dx}{dt} - \int \frac{dx}{dt} \left(\frac{d^2x}{dt} - \frac{dx}{dt}\frac{d^2t}{dt}\right)
$$
You'll notice that the right-hand side of this is equivalent to our previous definition of $I$.  Therefore, we can rewrite this as:
$$
I = \frac{dx}{dt}\frac{dx}{dt} - I
$$
Adding $I$ to both sides yields:
$$
2I = \frac{dx}{dt}\frac{dx}{dt} \\
I = \frac{1}{2} \left(\frac{dx}{dt}\right)^2
$$
A: You are aware that
$$\frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2},$$
because you used that identity in your first step.
But why isn't it
$$\frac{d}{dt}\left(\frac{dx}{dt}\right) \stackrel?= \frac{d^2x}{d^2t^2},$$
as implied by your note,
"Here I am multiplying  between the two operator like ... $dt \cdot dt=d^2t^2$"?
Also, consider the U-substitution $u = x^2$:
$$
\int du = \int d(x^2) = \int 2x\,dx.
$$
A factor of $2$ shows up that is not accounted for simply by taking factors on the left of the $d$ and multiplying them on the right instead.
Indeed, if we followed your operators-as-factors approach, we would have gotten
$\int x\,dx$ as the last step.
There are a lot of legitimate uses of the $\frac{dx}{dt}$ that look like simple multiplication and/or division of parts of the formulas, but the fact that we have a name for a fact like the Chain Rule,
$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt},$
should tip you off to the fact that none of these facts about derivatives are simple applications of multiplication and division of the pieces of the notation.
Whatever manipulation you do with derivatives, you need to be able to point back to a formula in a textbook that says you could do exactly what you're doing in exactly the way you're doing it, or you're liable to get zero credit on your homework or exam (and rightly so, because what you did here isn't mathematics, it's just wild guesswork).
The reason you're allowed to use the formula from the book is that someone has proved that the formula actually works that way.
And these proofs never involve just "multiplying  between the two operator" like anything.
A: Maybe this will help. Let $q = dx/dt$ such that your integral becomes $$I = \int \frac{d^2x}{dt^2} q \: dt \:.$$ Then, integrate by parts with: $$u = q \hspace{2.54cm} dv = \frac{d^2x}{dt^2} \: dt$$ $$du = dq \hspace{2.54cm} v = \frac{dx}{dt}$$
So then the standard recipe $I = uv \big| - \int v \: du$ tells us $$I = q \frac{dx}{dt} \bigg| - \int \frac{dx}{dt} \: dq \:,$$ or more simply, \begin{align*}I &= \left(\frac{dx}{dt}\right)^2 \bigg| - \int q \: dq \\ &= \left(\frac{dx}{dt}\right)^2 - \frac{q^2}{2} + C \\ &= \frac{1}{2} \left(\frac{dx}{dt}\right)^2 + C \:.\end{align*}
Note: The factor of $1/2$ belongs there, as your original integral is just Newton's second law applied across a distance - also known as work. The final answer is the kinetic energy plus a constant.
A: I thought of another, even simpler way of solving this.
The derivative, $\frac{d}{dx}\left(\right)$, is actually a combination of two separate operations: the differential $d()$ and a division by $dx$.  Most of the time people combine these, but they are in fact separable.  Integration is actually the opposite of differentiation, not the derivative itself.  As an example, $d(x^2) = 2x\,dx$ and $\int 2x\,dx = x^2 + C$.
So, the second derivative is just the derivative of the derivative.  So,
$$y'' = \frac{d\left(\frac{dy}{dx}\right)}{dx}$$
So, the integral you are interested in is $\int y''\,dy$.  This can be expressed as:
$$ \int \frac{d\left(\frac{dy}{dx}\right)}{dx}\,dy $$
Rearranging slightly, we get:
$$ \int d\left(\frac{dy}{dx}\right) \frac{dy}{dx} $$
So, if we set $u = \frac{dy}{dx}$, this just becomes:
$$\int d(u)\, u \quad\text{or, written another way,}\quad \int u\,du$$
$\int u\,du$ is obviously $\frac{u^2}{2} + C$, and since $u = \frac{dy}{dx}$, then the solution is $$\frac{1}{2}\left(\frac{dy}{dx}\right)^2 + C$$
