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