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?

  • $\begingroup$ 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 Nov 29 '15 at 9:54

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 $dy$:

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

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

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

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

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

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

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

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

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

  • 1
    $\begingroup$ @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 Jan 30 '18 at 12:17

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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