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Lebniz's notation for ordinary derivatives as quotients of differentials is a convenient abuse of notation, since it lets you express things like the chain rule and the derivative of the inverse function in a suggestive form:

$$\frac{dz}{dx} = \frac{dy}{dx} \cdot \frac{dz}{dy}$$

$$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$$

where $x,y,z$ are interdependent variables. This approach completely breaks down with second and higher order derivatives, since for example the second derivative of the inverse is

$$\frac{d^2x}{dy^2} =-\frac{d^2y}{dx^2} \left(\frac{dy}{dx} \right)^{-3} \neq \frac{1}{\frac{dy^2}{d^2x}}$$

where the left hand side isn't even defined.

I know that the concept of differential can be formalized, for example as infinitesimal variables in nonstandard analysis, and that this, in a sense, explains why these formal manipulations work. I know the concept of second degree differential exists, that's why I suspect that the reason they don't work in the case of higher degree derivatives is because the notation must be "wrong". My question is:

Is it possible to modify Leibniz's notation for second and higher order derivatives, so that the corresponding "differentiation rules" can be obtained by formal algebraic manipulation of the differentials $dx$, $d^2x$, etc. involved?

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    $\begingroup$ Here is a partial derivation $$ \frac{d^2x}{dy^2} = \frac{d\frac{dx}{dy}}{dy} = \frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right)\\ \overset{\text{Chain rule}}{=} -\left(\frac1{\frac{dy}{dx}}\right)^2\frac{d\frac{dy}{dx}}{dy} $$I can't figure out $\dfrac{d\frac{dy}{dx}}{dy}$ right now, but at least it's a bit closer to your goal. $\endgroup$
    – Arthur
    Commented Nov 29, 2015 at 9:54

3 Answers 3

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I think you just did the math wrong. When using Leibniz notation, always treat it as two operations - differential followed by division. So, we are going to take the differential of $\frac{1}{\frac{dy}{dx}}$ and then divide it by $dy$:

$$\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right) = \frac{d\left(\frac{1}{\frac{dy}{dx}}\right)}{dy}$$

So, let's solve for $d\left(\frac{1}{\frac{dy}{dx}}\right)$ using $u$-substitution: $$ u = \frac{dy}{dx} \\ d\left(\frac{1}{\frac{dy}{dx}}\right) = d(u^{-1}) \\ d(u^{-1}) = -u^{-2}du $$

Now we have the basic form of the differential, so let's find out $du$:

$$ u = \frac{dy}{dx} \\ du = d\left(\frac{dy}{dx}\right) \\ du = \frac{dx\cdot d^2y - dy\cdot d^2x}{dx^2} \\ $$

Now, going back, the basic form of the differential was $u^{-2}du$. So, substituting our $u$s and $du$s we get:

$$ d\left(\frac{1}{\frac{dy}{dx}}\right) = -\left(\frac{dy}{dx}\right)^{-2}\cdot\frac{dx\cdot d^2y - dy\cdot d^2x}{dx^2} \\ = -\left(\frac{dx^2}{dy^2}\right)\cdot\frac{dx\cdot d^2y - dy\cdot d^2x}{dx^2} \\ = - \frac{dx\cdot d^2y - dy\cdot d^2x}{dy^2} \\ = \frac{dy\cdot d^2x - dx\cdot d^2y}{dy^2} \\ = \frac{d^2x}{dy} - dx\frac{d^2y}{dy^2} \\ = \frac{d^2x}{dy} - dx\cdot 0 \\ $$

Now you may be surprised that $\frac{d^2y}{dy^2}$ reduces to 0. However, think of it this way. This is read as the "second derivative of y with respect to itself". The first derivative of y with respect to y is $\frac{dy}{dy} = 1$. So if the first derivative is a constant, then the second derivative must be zero. (For more info on this, see my blog post here)

Now, this reduces to: $$ d\left(\frac{1}{\frac{dy}{dx}}\right) = \frac{d^2x}{dy} \\ $$

This is the differential - to get the derivative we divide by $dy$:

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

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    $\begingroup$ @pregunton - I just wanted to note that I no longer believe my answer to be completely true. I do think it is true when $y$ is the independent variable. However, I am working on a paper that yields a different general answer for the second derivative notation: arxiv.org/abs/1801.09553 $\endgroup$
    – johnnyb
    Commented Jan 30, 2018 at 12:17
  • $\begingroup$ When representing derivatives as algebraic quantities, $d^2y/dy^2$ does not reduce to zero because it isn't the second derivative of $y$ with respect to itself because $d^2y/dy^2$ is not the algebraic double derivative $\endgroup$
    – eschavez
    Commented Jun 28, 2020 at 20:23
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You may find Faà di Bruno's formula interesting for computing the $n_{th}$ derivative of a composition

$$ {\left( {f \circ g} \right)^{(n)}}=\sum_{m_1+2m_2+\dots+nm_n=n} \frac{n!}{\prod_{i=1}^n m_i! (i!)^{m_i}}(\prod_{j=1}^n(g^{(i)})^{m_i}){(f^{(m_1+\ldots+m_n)}\circ g)} $$

In fact, you are asking to right down this complicated ugly formula by a modified Leibniz notation so that it becomes suggestive like for the case $n=1$. However, I don't think that it is possible from the nature of this complicated formula! If we right $(1)$ in original Leibniz notation it becomes

$$ {\frac{d^n f \circ g}{dx^n}}=\sum_{m_1+2m_2+\dots+nm_n=n} \frac{n!}{\prod_{i=1}^n m_i! (i!)^{m_i}} (\prod_{j=1}^n(g^{(i)})^{m_i}){(\frac{d^{(m_1+\ldots+m_n)} f}{dx^{(m_1+\ldots+m_n)}} \circ g)} $$

I cannot see anything suggestive! :)

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    $\begingroup$ Ugly formula indeed Hosein $\endgroup$
    – Mikasa
    Commented Jun 28, 2020 at 20:31
  • $\begingroup$ @mrs: Yeah! :D Super Ugly! $\endgroup$ Commented Jun 29, 2020 at 6:11
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The traditional "$d^2y/dx^2$" is not an algebraic quantity but the second and higher order derivatives can be represented as algebraic quantities. the most readable yet still correct way of representing an algebraic double derivative for $y=f(x)$ is

$$ f''=\frac{d\left[\frac{dy}{dx}\right]}{dx} $$

This can be expanded using the quotient rule

$$ f''=\frac{d^2y}{dx^2}-\frac{dy\ d^2x}{dx^3} $$

for the inverse $x=f^{-1}(y)$, the double derivative becomes $$ (f^{-1})''=\frac{d^2x}{dy^2}-\frac{dx\ d^2y}{dy^3} $$

From this we can see through algebra that $(f^{-1})''=-f''\frac{dx^3}{dy^3}=-f''/f'^3$, which is the second derivative rule for the inverse function.

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