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I have often seen "the second derivative of y with respect to x" written as $${d^2y\over dx^2},$$ but I don't understand the reason for this notation. I have always seen it written as $${d^2y\over dx^2},$$ and never as $${d^2y\over dx}.$$ Would $d^2y\over dx$ be a valid notation as well, and would it have a different meaning from $d^2y\over dx^2$?

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    $\begingroup$ Your alternative notation is not valid. The second derivative is written that way because that's how Leibniz wrote it. $\endgroup$ Aug 27, 2013 at 17:15
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    $\begingroup$ I posted this answer to a very similar question. It's not only the fact that it's $\left(\dfrac{d}{dx}\right)^2 y$, but also: the form you suggest is not dimensionally correct. $\endgroup$ Aug 27, 2013 at 19:38

4 Answers 4

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The notation you've often seen for the second derivative is the notation Leibniz used:

$$\dfrac {d^2 y}{dx^2} = \dfrac{d}{dx}\left(\frac {dy}{dx}\right) \tag{1}$$

Regarding the second question, no, $\dfrac{d^2y}{dx} = d\left(\dfrac {dy}{dx}\right)$ is not valid, and is rather meaningless.

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  • $\begingroup$ Does "d^2y/dx" have any meaning at all? $\endgroup$ Aug 27, 2013 at 17:13
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    $\begingroup$ No, unless you created a meaning to assign to that notation, and clearly stated your definition, but I suspect that doing so would simply confuse your audience. $\endgroup$
    – amWhy
    Aug 27, 2013 at 17:16
  • $\begingroup$ @amWhy $\huge{+}$ $\endgroup$
    – Software
    Aug 27, 2013 at 17:27
  • $\begingroup$ @amWhy: Maybe this will get to a nice answer! :-) +1 $\endgroup$
    – Amzoti
    Aug 28, 2013 at 11:40
  • $\begingroup$ I would be grateful if you express your opinion on the ideas expressed in my answer. $\endgroup$
    – zkutch
    Dec 20, 2023 at 2:52
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The derivative of $y$ with respect to $x$ can be written as $$\frac{\operatorname{d}}{\operatorname{d}\!x} \, y $$ Usually, we bring the $y$ up into the numerator to shortern the notation. So, what is the second derivative, well - it's the derivative of the derivative:

$$\frac{\operatorname{d}}{\operatorname{d}\!x} \left( \frac{\operatorname{d}}{\operatorname{d}\!x} \, y \right) = \frac{\operatorname{d}}{\operatorname{d}\!x} \, \frac{\operatorname{d}}{\operatorname{d}\!x} \, y = \frac{\operatorname{d^2}}{\operatorname{d}\!x^2} y $$

Again, we usually bring the $y$ up into the numerator.

I have never seen the notation $\tfrac{\operatorname{d}^2\!y}{\operatorname{d}\!x}$ in my life. I would suggest that it is meaningless. It's tempting to try to derive some meaning for it by thinking of it as the exterior derivative $\operatorname{d}\tfrac{\operatorname{d}\!y}{\operatorname{d}\!x}$, but that seems to be mixing up two different types of notation, and is best to be avoided.

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  • $\begingroup$ I would be grateful if you express your opinion on the ideas expressed in my answer. $\endgroup$
    – zkutch
    Dec 20, 2023 at 2:52
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As in the other answers, the main response to the question is that probably something is wrong... since the basic Leibniz notation would never write ${d^2y \over dx}$.

The only possible sense would be an "infinitesimal" change in the derivative ${dy\over dx}$, in Leibniz' notation indeed $d{dy \over dx}$, though I've seen this only very rarely, and it would rarely be useful.

Nevertheless, there are some manipulations to solve differential equations that do exactly make use of similar ideas. For example, in one dimension, Newton'e inverse square law ${d^2y\over dx^2}={-1\over y^2}$ can be rewritten (usefully!) as $$ {-1\over y^2} = {d^2 y\over dx^2} = {d\over dx}{dy\over dx} = {d(y')\over dx} = {dy'\over dy}{dy\over dx} = {dy'\over dy}\,y' $$ Then ${-dy\over y^2}=y'\,dy'$ gives ${1\over y}={1\over 2}(y')^2$, and so on. :)

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  • $\begingroup$ I would be grateful if you express your opinion on the ideas expressed in my answer. $\endgroup$
    – zkutch
    Dec 20, 2023 at 2:52
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My two cents.

For history buffs how, where and when is introduced "$d$" for derivative let me link with excellent book by Florian Cajori - A history of mathematical notations-Dover Publications, 1993, pages 203-206.

Let's get to work now:

A function $f:E\to \mathbb{R}$ defined on set $E\subset \mathbb{R}$ is called differentiable at point $x\in E$, that is limit point of $E$, if exists constant $A$ such that holds representation $$f(x+h)-f(x)=A\cdot h + o(h), h\to 0, x+h\in E.$$ Usually we denote constant $A$ by $f'(x)$ and call it derivative of $f$ at point $x$.

Differential at point $x$ for considered function $f$, is linear function $A\cdot h$ from vaiable $h$ and is denoted by $df(x)$ i.e. de facto differential depends on two variables $x$ and $h$ and is linear by second

$$(df)(x,h)=(df(x))(h)=A\cdot h= f'(x)\cdot h\quad(1) $$

So, $(df(x))$ is name for function from variable $h$ and I'll omit extra brackets farther. If we write $y=f(x)$, then differential can be written as $dy=df(x)$, but using only $dy$ we quietly mean appropriate $x$. In this case $dy(h)=A\cdot h= f'(x)\cdot h$.

Now, if we have identical function $I(x)=x$, then it has derivative in all points from domain with same derivative $I'(x)=1$, so, we obtain $$dI(x)(h)=dx(h)=1\cdot h= h$$ Using obtained $(1)$ can be written as

$$df(x)(h)=A\cdot h= f'(x)\cdot h=f'(x)dx(h)\quad(2) $$ Ommiting argument $h$ in both sides of $(2)$ we get $$dy=df(x)=f'(x)dx\quad(3) $$ and this is what we have in majority of books, though, formally we should write $(2)$.

Coming back to your question ${d^2y\over dx^2}$ we need to understand what is $d^2y $. From $(2)$ we can consider $df(x)(h_1)=f'(x)dx(h_1)$ as function from $x$ (I especially change $h$ to $h_1$ and consider it fixed), so, if function $f'$ have derivative (is differentiable) we can write $$(d^2f)(x,h_1, h_2) =d(df(x)(h_1))(h_2)=d(f'(x)\cdot dx(h_1))(h_2)= f''(x)\cdot dx(h_1)\cdot dx(h_2)=f''(x)\cdot h_1 \cdot h_2$$ Taking $h_1=h_2=h$ gives $$d^2f(x)(h)= f''(x)\cdot (dx(h))^2$$ again omitting argument $h$ gives well known $$d^2f(x)= f''(x)\cdot (dx)^2\quad(4)$$ Now from $(4)$ we can get formula for $\frac{d^2 y}{dx}$, but, of course, it is not second derivative.

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