Let $y = f(x)$.

$$f'' = \frac{d^2y}{dx^2}$$

The explanation for this being that

$$ \Bigl(\dfrac{d}{dx}\Bigr)^2 y = \dfrac{d^2}{dx^2}\,y = \dfrac{d^2 y}{dx^2};$$

Since there are two $d$'s in the bottom of the fraction, why is it not written


Maybe it's because $dx$ needs to be thought of as a single thing. But notice that $d$ is used by itself and squared in the numerator..

Does my point here make sense, is it just a convenience to avoid the extra exponent, or is there a logical reason it's written in the form it is?

  • 1
    $\begingroup$ Perhaps the $dx^2$ is meant to be read as $(dx)^2$ (rather than $d(x^2)$), where one views "$dx$" as one symbol rather than two separate letters. But I am just speculating. $\endgroup$
    – angryavian
    Feb 14, 2022 at 22:06
  • 1
    $\begingroup$ $d$ isn’t a number, it more like operator. In particular, $(dx)^2\neq d^2x^2.$ It would be more approapriate to write $$\frac{d^2y}{(dx)^2},$$ but we abandon the standard rules of precedence here, and remove the parentheses. $\endgroup$ Feb 14, 2022 at 22:11
  • 1
    $\begingroup$ Duplicate? Why are the $d$'s not altered for higher-order derivatives? $\endgroup$
    – Blue
    Feb 14, 2022 at 22:14
  • $\begingroup$ In functions $f$ of multiple variables, we write $$\frac{d^2f}{dxdy},$$ which shows we are really talking $(dx)^2$ in the denominator, and we really use $dx^2$ as a shorthand. $d$ is not a “number-like” thing, but $dx$ is “number-like.” If $d$ were number-like, we’d get things like $$\frac{dy}{dx}=\frac{y}{x}.$$ That is wrong, and our notation should discourage treating $d$ as a number. $\endgroup$ Feb 14, 2022 at 22:23
  • $\begingroup$ It’s worth replacing $d$ with $\Delta.$ $\endgroup$ Feb 14, 2022 at 22:24

1 Answer 1


Its customary to write $dx^2$ to denote $(dx)^2$ in all common theories of calculus, then we set

$$ \frac{d}{d x}\circ \frac{d}{d x}=:\left(\frac{d}{d x}\right)^2=:\frac{d^2}{dx^2} $$

The last expression is a whole, that is, $d^2$ and $dx^2$ doesn't make sense by themselves (and the fraction is just a notation resembling the analytic definition of the derivative but it doesn't mean something). The second expression $\left(\frac{d}{d x}\right)^2$ is common for any linear operator to represent the composition of a linear operator with itself, and $\frac{d}{d x}$ is a linear operator in the space of real-to-real smooth functions.

  • $\begingroup$ Thanks. I don't understand what $d^2$ in the numerator is to be interpreted as.. $\endgroup$
    – Ben G
    Feb 15, 2022 at 4:01
  • $\begingroup$ @bgcode the elements alone in this notation have no meaning, its the whole $\frac{d}{d x}$ what makes sense. Then, when we apply this operator twice we use the exponents to denote it. $\endgroup$
    – Masacroso
    Feb 15, 2022 at 5:53
  • $\begingroup$ So the $^2$ should be read as "there are two of these" rather than implying anything is multiplied by anything else? $\endgroup$
    – Ben G
    Feb 15, 2022 at 6:21
  • $\begingroup$ @bgcode exactly. Mathematicians use many other notations to represent the same that $\frac{d^2}{d x^2}$, by example $D^2$ or $\partial ^2$ are very used (and makes more sense, I mainly use the last one). Think that the notation about you are asking is very old (it was the notation used by Leibniz, one of the creators of calculus!), prior to any formalization of mathematics $\endgroup$
    – Masacroso
    Feb 15, 2022 at 6:23
  • $\begingroup$ Ok cool. I've been using it now to solve Parametric Equations (khanacademy.org/math/ap-calculus-bc/bc-advanced-functions-new/…). I would maybe add one line to your answer addressing what we've said in the comment here about it meaning 2 of something not squared, and can grant the answer $\endgroup$
    – Ben G
    Feb 15, 2022 at 6:31

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