Is Schwarz's theorem valid for regular derivatives?

Question

If $f$ and $y$ are functions of $x$, is it true that:

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

for any reasonably general condition, similar to Schwarz's theorem?

Answer 1

Let $f$ and $y$ be "well-behaved" functions of $x$ (for example, $C^2$-diffeomorphisms between open intervals of $\mathbb R$). It is not generally true that:

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

The argument is straightforward:

$$\begin{align} \frac{d}{dy}\left(\frac{df}{dx}\right) &= \frac{d}{dx}\left(\frac{df}{dx}\right)\left(\frac{dy}{dx}\right)^{-1} \tag{1} \\ \frac{d}{dx}\left(\frac{df}{dy}\right) &= \frac{d}{dx}\left[\frac{df}{dx}\left(\frac{dy}{dx}\right)^{-1}\right] \\ &= \frac{d}{dx}\left(\frac{df}{dx}\right)\left(\frac{dy}{dx}\right)^{-1} + \frac{df}{dx}\cdot\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right] \tag{2} \\ \frac{d}{dy}\left(\frac{df}{dx}\right) &= \frac{d}{dx}\left(\frac{df}{dy}\right) + \frac{df}{dx}\cdot\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right] \tag{3} \end{align}$$

The extra term:

$$ \frac{df}{dx}\cdot\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right] $$

only vanishes if $f$ or $dy/dx$ are constant, which is not true in general.

Example

Let's define $f$ and $y$ as: $$ \begin{cases} f = x^4 = y^2 \\ y = x^2 \end{cases} $$

The functions and their inverses are $C^2$ in $\mathbb R^+$. We have:

$$\begin{align} \frac{df}{dx} &= 4x^3 = 4y^{3/2} \\ \frac{df}{dy} &= 2y = 2x^2 \\ \frac{d}{dy}\left(\frac{df}{dx}\right) &= \frac{d}{dy}\left(4y^{3/2}\right) = 6\sqrt{y} = 6x \\ \frac{d}{dx}\left(\frac{df}{dy}\right) &= \frac{d}{dx}\left(2x^2\right) = 4x \\ \end{align}$$