Problem in the second-derivative symbol. 
The second derivative of this symbol according to the rules that we have learned the correct mathematical, I wish to know why this symbol is not used.
 A: $(dx)^2$ is not $d^2x^2$ because there is no quantity called $d$.  Rather $dx$ can be thought of as an infinitely small increment of the variable $x$.
A: Really, the standard second derivative symbol $d^2\over dx^2$ should be considered an abuse of notation in its own right.  The first derivative symbol $d\over dx$ is already a "single symbol", so iterating it twice should yield $({d\over dx})^2$.  Without parentheses that yields ${d\over dx}^2$, which is confusingly like $d^2\over dx$, which is very very wrong ($d$ isn't really a separate thing you can square).  So, by convention, it is permitted to re-write it as $d^2\over dx^2$ --- but that's an abuse of notation.  Abuses of notation are only permitted when they are conventional or useful, and further simplify to $d^2\over d^2x^2$ is neither.
A: Note what 
$$
\frac{dy}{dx}=\frac{d}{dx}[y]=g(x)
$$
where $y=y(x)$. So, it is convenient define
$$
\frac{dg}{dx}=\frac{d}{dx}\left[\frac{d}{dx}[y]\right]=\frac{d^2}{dx^2}[y]=\frac{d^2 y}{dx^2}
$$
to designate the second derivative of $y$ with respect to $x$.
A: The mnemonic is the following. The operator "d" is applied  twice to $y$, so $d(dy)=d^2y$. But to get the second derivative we have to divide by $dx$ twice namely the operator $d$ is applied once but the result is squared. So $dx\cdot dx=(dx)^2= dx^2$. This is not $d$ applied to $x^2$ ,  this is $(dx)^2$. 
A: we have per definition $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{d^2y}{dx^2}$
A: Please note tha $\frac{d}{dx}$ is an operator and not a quantity, so it makes sense to write
$$ \frac{d}{dx} \circ \frac{d}{dx} = \frac{d^2}{dx^2} $$
which means applying the operator twice.
