# 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.

• You should interpret the symbol $dx$ as a "single quantity"... Oct 14, 2014 at 18:11
• The quantity $d^2 y / dx$ doesn't make sense. Oct 14, 2014 at 18:13
• @user The OP is showing work that leads up to the use of the notation: $\dfrac{d^2y}{dx}\cdot \dfrac 1{dx}=\dfrac{d^2 y}{dx^2}$. Oct 14, 2014 at 18:15
• ...and then you can simplify further by cancelling the $d\,$s, right? Oct 14, 2014 at 18:37
• I have never liked this notation because I would like to interpret $\frac{dx^2}{dx}$ as the derivative of $x^2$. But in the second derivative notation $\frac{d^2y}{dx^2}$, $dx^2$ means $(dx)^2$. Oct 14, 2014 at 21:16

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 $(dx)^2$.

• It is useful to note that one means $(dx)^2$ when one writes $dx^2$. Oct 14, 2014 at 18:16

$(dx)^2$ 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 $x$.

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.

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 $x$.

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

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.