Problem in the second-derivative symbol.

Question

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.

Answer 1

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 $(dx)^2$.

Answer 2

$(dx)^2$ 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 $x$.

Answer 3

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.

Answer 4

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 $x$.

Answer 5

we have per definition $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{d^2y}{dx^2}$

Answer 6

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.