Cubic addition and differentiablility It came to my thought that if we define $a\oplus b = (a^3 + b^3)^{\frac13}$ then $\Bbb R$ endowed with $\oplus$ and $\cdot$ the latter being the usual multiplication is a field, with usual $0$ and $1$ being the zero and the unity of this field respectively. Not really knowing what to do next with this, I decided to check how will new derivatives be different from the old ones. If I am not mistaken, it holds that
$$
\lim_{h\to 0}\frac{f(x\oplus h)\ominus f(x)}{h} = \left(f'(x)\frac{f^2(x)}{x^2}\right)^{\frac13}
$$
if $x\neq 0$, while the direct evaluation of the limit when $x = 0$ gives that indeed it is infinite unless $f(0) = 0$. In the latter case we get the new derivative (namely, the limit) being exactly the old derivative $f'(0)$. This looks very odd to me: why $0$ should be any special point of this field?
So my questions are:

*

*Did I make a mistake somewhere? If yes, please point me to it.


*If not, why in this new field differentiability at $0$ is something special?
 A: $\def\D{\mathscr{D}}\def\H{\mathscr{H}}\def\R{\mathbb{R}}\DeclareMathOperator{\id}{id}\DeclareMathOperator{\sgn}{sgn}\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}$All the terms in this answer are in the classic sense unless specified otherwise.
Define\begin{align*}
\H &= \{φ: \R → \R \mid φ\ \text{is an increasing homeomorphism},\ φ(0) = 0\},\\
\D_0 &= \{φ \in \H \mid φ\ \text{is derivable on}\ \R^*,\ φ'(x) ≠ 0\ (\forall x ≠ 0)\},\\
\D &= \{φ \in \H \mid φ\ \text{is a diffeomorphism}\},
\end{align*}
and for any $φ \in \H$, define\begin{gather*}
x +_φ y = φ^{-1}(φ(x) + φ(y)),\\
D_φ f(x) = \lim_{h → 0} \frac{1}{h} (f(x +_φ h) -_φ f(x)).
\end{gather*}
Note that the metric induced by $+_φ$, i.e.$$
d_φ(x, y) = \left\{ \begin{array}{ll} x -_φ y; & x \geqslant y\\
y -_φ x; & x < y \end{array} \right\} = φ^{-1}(|φ(x) - φ(y)|),
$$
is equivalent to the classic metric $d(x, y) = |x - y|$, so $D_φ$ is well-defined.

It is natural to assume that $φ$ and $f$ are derivable to analyze $D_φ f$, but the cases where $φ \in \D$ yields the most intuitions. For $φ \in \D$ and any derivable $f$,\begin{align*}
D_φ f(x) &= \left. \frac{\partial}{\partial h}(f(x +_φ h) -_φ f(x)) \right|_{h = 0}\\
&=\bigl( (φ^{-1})'(φ(f(x +_φ h)) - φ(f(x))) · φ'(f(x +_φ h)) · f'(x +_φ h) ·\\
&\mathrel{\phantom=} (φ^{-1})'(φ(x) + φ(h)) · φ'(h) \bigr)\Bigr|_{h = 0}\\
&= \color{red}{ \frac{1}{φ'(0)} } · φ'(f(x)) · f'(x) · \color{blue}{ \frac{1}{φ'(x)} } · \color{red}{ φ'(0) } \tag{1}\\
&= \frac{φ'(f(x)) f'(x)}{φ'(x)} = \frac{(φ \circ f)'(x)}{φ'(x)},
\end{align*}
so $D_φ f = \dfrac{D(φ \circ f)}{Dφ}$, where $D$ is the classic derivative operator. Note that this formula also implies that diffeomorphism in the classic sense is equivalent to that in the sense of $D_φ$.
Now, in order to show that $(\R, +)$ cannot be differentiated from $(\R, +_φ)$ by how their derivative operators interact with other non-classic plus operations, define\begin{gather*}
x +_{φ_1, φ_2} y = φ_2^{-1}(φ_2(x) +_{φ_1} φ_2(y)),\\
D_{φ_1, φ_2}f(x) = \lim_{h → 0} \frac{1}{h} (f(x +_{φ_1, φ_2} h) -_{φ_1, φ_2} f(x))
\end{gather*}
for any $φ_1, φ_2 \in \D$. Since$$
x +_{φ_1, φ_2} y = φ_2^{-1}(φ_1^{-1}(φ_1(φ_2(x)) + φ_1(φ_2(y)))) = x +_{φ_1 \circ φ_2} y.
$$
then $+_{φ_1, φ_2} = +_{φ_1 \circ φ_2}$ and$$
D_{φ_1, φ_2}f = D_{φ_1 \circ φ_2}f = \frac{D(φ_1 \circ φ_2 \circ f)}{D(φ_1 \circ φ_2)} = \frac{D(φ_1 \circ φ_2 \circ f) / Dφ_1}{D(φ_1 \circ φ_2) / Dφ_1} = \frac{D_{φ_1}(φ_2 \circ f)}{D_{φ_1}φ_2}.
$$
This formula on $(\R, +_{φ_1})$ is just the same as $D_{\id, φ_2} f = \dfrac{D_{\id}(φ_2 \circ f)}{D_{\id} φ_2}$, or $D_{φ_2} f = \dfrac{D(φ_2 \circ f)}{Dφ_2}$, on $(\R, +)$.

The situation gets much more complicated even if the condition $φ \in \D$ is simply relaxed as $φ \in \D_0$. Now the definition of $\D_0$ ensures that the blue term in (1) is always defined for $x ≠ 0$, but the red terms suggest that the simple chain rule fails if $φ'(0) = 0$ or $φ'(0)$ does not exist. This suggests that $0$ is a special point for $D_φ f$.
More specifically, for $x ≠ 0$, since\begin{align*}
&\mathrel{\phantom=} \left. \frac{\partial}{\partial k}φ(f(φ^{-1}(φ(x) + k))) \right|_{k = 0}\\
&= \bigl( φ'(f(φ^{-1}(φ(x) + k))) · f'(φ^{-1}(φ(x) + k)) · (φ^{-1})'(φ(x) + k) \bigr)\Bigr|_{k = 0}\\
&= φ'(f(x)) · f'(x) · \frac{1}{φ'(x)} = \frac{φ'(f(x)) f'(x)}{φ'(x)},
\end{align*}
then\begin{gather*}
\frac{1}{h} (f(x +_φ h) -_φ f(x)) = \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) }\ (h → 0). \tag{2}
\end{gather*}
Therefore $D_φf(x)$ exists if the limit of the RHS of (2) exists. Even though (2) is not valid for $x = 0$, the factor $φ'(f(x))$ in (2) implies that whether $f(0) = 0$ or $f(0) ≠ 0$ makes a difference for $D_φ f(0)$ if $φ'(0) = 0$ since there is a factor $φ'(x)$ in the denominator and derivatives are Darboux functions. Below are two explicit examples.
Example 1: $φ(x) = \sgn(x) |x|^a$ ($a > 0$, $a ≠ 1$). Since $φ$ is multiplicative, then for $x ≠ 0$, (2) implies that\begin{align*}
D_φf(x) &= \lim_{h → 0} \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) }\\
&= \lim_{h → 0} φ^{-1}\paren{ \frac{1}{φ(h)} \paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } }\\
&= \lim_{h → 0} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} + o(1) }\\
&= φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} } = \sgn(f'(x)) \abs{ \frac{f(x)}{x} }^{\frac{a - 1}{a}} |f'(x)|^{\frac{1}{a}}.
\end{align*}
Again $0$ can be seen to be a special point because of the factor $|f(x)|^{\frac{a - 1}{a}}$ in the above expression.
Example 2: $φ(x) = \begin{cases} x\exp\paren{-\dfrac{1}{x^2}}; & x ≠ 0 \\ 0; & x = 0 \end{cases}$. Note that for any $c \in (0, 1)$, $φ(ch) = o(φ(h))$ ($h → 0$). For $x ≠ 0$, if $f(x) ≠ 0$ and $f'(x) ≠ 0$, then for any $ε \in (0, 1)$,$$
φ((1 - ε)h) = o(φ(h)),\quad φ(h) = o(φ((1 + ε)h)),\quad (h → 0)
$$
thus there exists $δ > 0$ such that for $h \in (-δ, δ) \setminus \{0\}$,$$
|φ((1 - ε)h)| < \abs{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } < |φ((1 + ε)h)|,\\
$$
so$$
1 - ε < \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } < 1 + ε.
$$
Therefore (2) implies that$$
D_φf(x) = \lim_{h → 0} \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } = 1.
$$
If $f(x) = 0$, then\begin{align*}
D_φf(x) &= \lim_{h → 0} \frac{1}{h} φ^{-1}(φ(f(φ^{-1}(φ(x) + φ(h)))) - φ(0))\\
&= \lim_{h → 0} \frac{1}{h} f(φ^{-1}(φ(x) + φ(h)))\\
&= \left. \frac{\partial}{\partial h}(f(φ^{-1}(φ(x) + φ(h)))) \right|_{h = 0}\\
&= \bigl( f'(φ^{-1}(φ(x) + φ(h))) · (φ^{-1})'(φ(x) + φ(h)) · φ'(h) \bigr)\Bigr|_{h = 0}\\
&= f'(x) · \frac{1}{φ'(x)} · φ'(0) = 0.
\end{align*}
This example shows that if $φ$ varies too slowly near $0$, then $D_φ$ will almost wipe out any information of the functions it applies to.
