Is there a function whos derivative of its inverse equals the inverse of its derivative? This may also hold on a specific interval only. $$\frac{d}{dx}f^{\langle-1\rangle}(x)=\left(\frac{d}{dx}f(x)\right)^{\langle-1\rangle}$$ Considering that $f(x)=x$ is its own inverse and $e^x$ is its own derivative, neither of them satisfies the equation, but both are the only functions with the respective properties (over the reels). Is this already a proof that there is no solution to the above equation?

Also, I dont know much about complex analysis but if a complex-valued function meets the criteria I'd be happy to know too :)

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    $\begingroup$ A little bit of progress, plugging in $f'(x)$ on both sides gives the equation $$\frac{1}{f'(f^{-1}(f'(x)))} = x$$ $\endgroup$ – Ninad Munshi Feb 17 at 10:49
  • $\begingroup$ Take a look at math.stackexchange.com/q/2117928 $\endgroup$ – Jean Marie Feb 17 at 11:25
  • $\begingroup$ See as well math.stackexchange.com/q/2617954 $\endgroup$ – Jean Marie Feb 17 at 11:26
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    $\begingroup$ Plug the Ansatz $f(x):=a x^\beta$ into your identity, and you will obtain equations for (complex) $\beta$ and $a$. $\endgroup$ – Christian Blatter Feb 17 at 15:39
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    $\begingroup$ I don't think it's possible to have a very nice solution. In order to have the inverse of both the function and the derivative of the function exist, it must be true that both $f'(x)$ and $f''(x)$ don't change sign (and aren't $0$ for any interval). Then I think it's impossible for the sign of the first derivative of the LHS and the RHS to be equal to each other given the prior restrictions. $\endgroup$ – Varun Vejalla Feb 24 at 21:33

Edits: I leave here the preliminary lines of thoughts/failed attempts. The answer based on monotonicity of the functions is probably correct and one may try to give a more abstract justification, cf. part 2.

  • To begin with it is clear that the OP denotes $f^{\langle-1\rangle}$ the "reciprocal" function and not the inverse as in $\frac{1}{f}$. Hence if ever one wants to formulate the problem in algebraic terms, we need to introduce an algebra $\mathcal{A}$ of functions with two products: $\times$ the pointwise product and $\circ$ the composition. The notion of derivation is readily generalized as a linear map $D:\mathcal{A} \to \mathcal{A}$ which satisfies the Leibniz rule w.r.t. $\times$. One can define the group $G$ of $\circ$-invertible elements and study already its image by $D$, or stable subsets etc...
  • The immediate relation we get for $\mathcal{A}:= \mathcal{C}^{\infty}(\mathbb{R})$ are (to avoid confusion $g$ is the inverse of $f$ for $\circ$, while $h$ is the inverse of $f'$ for that same law, Id unit for $\circ$ while the constant $1$ function is the unit fo $\times$) $$ f\circ g=\mathrm{Id}_{\mathbb{R}}\quad \Longrightarrow\quad g' \times f'\circ g = 1\quad \label{1}\tag{1}$$ $$ f'\circ h=\mathrm{Id}_{\mathbb{R}}="x"\quad \Longrightarrow\quad \frac{1}{x} \times f'\circ h = 1 \quad \label{2}\tag{2}$$ with these notations, OP's equation reads $g'\overset{!}{=} h$ and by (\ref{1}): $\quad g' = \frac{1}{f'\circ g}=h$. (I assume the $\circ$-inverse are left and right inverse, as in the usual definition). What Ninad Munshi wrote on his first comment is the composition on the right of this equality with $f'$: $\quad \frac{1}{f'\circ g \circ f} = h\circ f =\mathrm{Id}_{\mathbb{R}}$. I did play with similar equalities but to no avail unfortunately... (@Christian by the way your "Feb 18" comment is correct, but for clarity you should for example use the variable $x$ for the domain and $y$ for the target space...)
  • Let us examine more abstractly the behavior of $\circ$ and of $f\mapsto f^{\langle -1\rangle}$ w.r.t. $+$ and $\times$. For the derivation of a composition there is that Faà di Bruno formula for derivation of composition but I was thinking more about a characterize the set $\mathcal{S}$ of solutions of the equation: does it define a subgroup/subvector space/ideal of $\mathcal{A}$. Unfortunately no, although (if not empty) $\mathcal{S}$ is stable under taking the inverse which we may denote $\mathrm{Inv}$: indeed if $f\in \mathcal{S}$ then so is $ g:=\mathrm{Inv} f$ $$\frac{d}{dx}\ g = \mathrm{Inv} \circ \frac{d}{dx}\circ\mathrm{Inv}\ g \quad\Longrightarrow\quad \mathrm{Inv} \circ \frac{d}{dx}\ g = \mathrm{Inv} \circ \mathrm{Inv} \circ \frac{d}{dx}\circ\mathrm{Inv}\ g = \frac{d}{dx}\circ\mathrm{Inv}\ g $$ The equation can in fact be understood as commutation of the derivation operator $\frac{d}{dx}$ and that of taking the inverse $$\frac{d}{dx}\circ \mathrm{Inv}\ f = \mathrm{Inv} \circ \frac{d}{dx}\ f \label{Eq}\tag{Eq}$$ The contradiction I had hoped to get if $(\mathcal{S},\circ)$ were a group, was to exhibit a stable subvector space, and show that a group for $\circ$ could not contain a vector space, but $\mathcal{S}$ is not stable under composition!!! To do so, I assumed $\mathcal{A}:=\mathcal{C}^{\infty}_c(\mathbb{R}) \hookrightarrow H^1(\mathbb{R})$ (Sobolev space, or any space on which one could "diagonalize" $\frac{d}{dx}$, by Fourier transform usually). As a consequence of (\ref{Eq}), $\mathrm{Inv}$ should preserve the eigenspaces of $\frac{d}{dx}$. If we really want to continue along this line, I think we have to define a new kind of algebraic structure...
  • "Reciprocal" function also appear when one does a change of variable in integration, so one could also get an integral version of the equation $g'=h$ in the case $\mathcal{A}:= \mathcal{C}^{\infty}(\mathbb{R})$ $$\begin{split} g(y)-g(0) &=\int_0^y g'(w)\, dw = \int_0^y h(w)\, dw =\genfrac{[}{]}{0pt}{0}{z:=h(w)}{dz = h'(w)\, dw = \frac{1}{f''\circ h(w)}\, dw}\\ &= \int_0^{h(y)} f''(z)\, z\, dz = \left[z\, f'(z) \right]^{h(y)}_0 - \int_0^{h(y)} f'(z)\, dz\\ &= h(y)\times y -\big(f\circ h(y) - f(0) \big) \end{split}$$
  • Could there be power series solution? From the comments, the attempt with $f(x):= a x^{b} $ involved functions of the form $\sqrt[b]{x}$ and $\sqrt[b-1]{x}$ so maybe instead of looking at power series, one may look at sums of functions of this form $\sqrt[n]{x}$.
  • Can't we adapt some fixed point techniques? (very explicitly, start with some function $f_0,\ \Delta(f_0):=\frac{d}{dx}f_0^{\langle-1\rangle} - \left(\frac{d}{dx}f_0\right)^{\langle-1\rangle}$ not necessarily $0$ and look for some transformation $T$ s.t. $\Delta \big(T(f_0)\big)$ is closer to $0$ and so on). In particular for the case $\mathcal{A}:=\mathcal{C}^{\infty}(\mathbb{R})$ where inversion is symmetry w.r.t the diagonal in the plane $\mathbb{R}\times \mathbb{R}$ where we represent the graph of the function.

Trick to give an algebraic formulation of the argument based on monotonicity: define the action of translation on functions. If we want to break it into small steps, we can define translation by a constant $h\in \mathbb{R}$ acting on $\mathcal{A}:= \mathcal{C}^{\infty}(\mathbb{R})$ $$\tau_h: \left\lbrace \begin{aligned} \mathcal{A}\quad & \longrightarrow \quad \mathcal{A}\\ "f(t)" & \longmapsto "f(t-h) "\end{aligned} \right. \in GL(\mathcal{A}) $$ More generally the action of the translation group is defined by the group morphism $$\left\lbrace \begin{aligned} \big(\mathbb{R},+\big) & \longrightarrow \big(GL(\mathcal{A}),\circ \big)\\ h\quad & \longmapsto\quad \tau_h \end{aligned} \right. \label{A}\tag{A}$$ To imitate the other reasoning, based on the formula for the derivative of a composition (\ref{1}-\ref{2}), we need to show that left hand side and r.h.s. of (\ref{Eq}) do not behave in the same way under (\ref{A}).


Disclaimer. In view of comments issued, and by growing insight, this answer must be considered as a partial one.

The easiest way is to start with considering the derivatives. Let $g(x)=f'(x)$. Then $g(x)$ must have an inverse $g^{-1}(x)$. This means that $y=g(x)$ is monotonically increasing or decreasing. $$ \mbox{Increasing:} \quad \begin{cases} x_1 \lt x_2 \quad \Longleftrightarrow \quad y_1 \lt y_2 \\ x_1 \gt x_2 \quad \Longleftrightarrow \quad y_1 \gt y_2 \end{cases} \\ \mbox{Decreasing:} \quad \begin{cases} x_1 \lt x_2 \quad \Longleftrightarrow \quad y_1 \gt y_2 \\ x_1 \gt x_2 \quad \Longleftrightarrow \quad y_1 \lt y_2 \end{cases} $$ The inverse $g^{-1}(x)$ is obtained (graphically) by mirroring $g(x)$ in the line $y=x$, thus by exchanging $x$ and $y$. From this it clear that $g(x)$ and $g^{-1}(x)$ must be both monotonically increasing or both be monotonically decreasing.
The same considerations are valid for $f(x)$ and $f^{-1}(x)$ as well, because these two are each others inverse too.
But $g(x)$ is also a derivative. If this derivative would become zero somewhere, then $f(x)$ would no longer be monotonic at that place. So $g(x)$ must be nonzero everywhere in its domain. For the inverse $g^{-1}(x)$ this means that there cannot be an intersection with the $y$-axis. Without loss of generality, we may assume that both $g(x)$ and $g^{-1}(x)$ are defined in the first quadrant, for $x \gt 0$ and $y \gt 0$. Other cases are covered by re-defining $g$ using symmetry: $g(x) := g(-x)$ , $g(x) := -g(x)$ , $g(x) := -g(-x)$.
Before proceeding, a few examples will be presented. First example: $$ f'(x) = g(x) = 1/x \quad ; \quad [f'(x)]^{-1} = g^{-1}(x) = 1/x \\ f(x) = \int_1^x\frac{dt}{t} = \ln(x) = y \quad ; \quad x = \ln(y) \; \Rightarrow \; f^{-1}(x) = e^x \quad ; \quad [f^{-1}]'(x) = e^x $$ It is noticed that $[f'(x)]^{-1}$ is monotonically decreasing while $[f^{-1}]'(x)$ is monotonically increasing. Second example: $$ f'(x) = g(x) = \sqrt{x} \quad ; \quad [f'(x)]^{-1} = x^2 \\ f(x) = \int_0^x\sqrt{t}\,dt = \frac{2}{3}x^{3/2} \quad ; \quad x = \frac{2}{3}y^{3/2} \; \Rightarrow \; f^{-1}(x) = \left(\frac{3x}{2}\right)^{2/3} \quad ; \quad [f^{-1}]'(x) = \left(\frac{3x}{2}\right)^{-1/3} $$ It is noticed that $[f'(x)]^{-1}$ is monotonically increasing while $[f^{-1}]'(x)$ is monotonically decreasing.
A pattern is emerging. Let's consider now the general case for $g(x)=f'(x)$ monotonically decreasing. It will be assumed that the reader is able to construct the analogue for monotonically increasing $g(x)$. Then $[f'(x)]^{-1}$ is monotonically decreasing as well. Define the function $f(x)$ as a definite integral in statu nascendi (a Riemann sum). For example as: $$ \sum_i g\left(x_i\right)\left(x_{i+1}-x_i\right) \approx \int_a^x g(t)\,dt = f(x) $$ And select the subdivisions in such a way that $$ g\left(x_i\right)\left(x_{i+1}-x_i\right) = \mbox{constant} = dA $$ Then we can construct the inverse function $f(x)$ by numerical approximation. It is clear thar if $g(x_i)$ is decreasing, then the area $dA$ can only kept constant by increasing the intervals $\left(x_{i+1}-x_i\right)$. This means that the intervals of $f^{-1}(x_i)$ must be increasing, because of $\left(x_{i+1}-x_i\right)\to\left(y_{i+1}-y_i\right)$. But then the derivative of $f^{-1}(x_i)$ is equal to $\left(y_{i+1}-y_i\right)/dA$, which means that $[f^{-1}]'(x)$ is increasing numerically, therefore analytically when taking limits. So here comes our end-result, apart from technicalities to filled in by the questioner eventually.

$[f'(x)]^{-1}$ is monotonically decreasing if and only if $[f^{-1}]'(x)$ is monotonically increasing.
$[f'(x)]^{-1}$ is monotonically increasing if and only if $[f^{-1}]'(x)$ is monotonically decreasing.
Therefore it is impossible to find (real valued) functions $f$ with $[f'(x)]^{-1}=[f^{-1}]'(x)$.


Below is a visualization of the technique employed, for the special example $f'(x)=[f'(x)]^{-1}=1/x$ , $\color{red}{f(x)=\ln(x)}$ , $\color{green}{f^{-1}(x)=[f^{-1}]'(x)=e^x}$ .
Further specifications: $-0.1 \le x \le 9$ ; $-0.1 \le y \le 9$ ; $dA = 1/10$ ;
thick colored lines numerically, thin colored lines analytically.

enter image description here


The domains of interest are much trickier than I thought.

Counter example. Consider $f(x)=1/x$, defined for $x\gt 0,y\gt 0$. Then $f'(x)=-1/x^2$ is defined for $x\gt 0,y\lt 0$; $[f'(x)]^{-1}=\sqrt{-1/x}$ is defined for $x\lt 0,y\gt 0$ and increasing; $f^{-1}(x)=1/x$ is defined for $x\gt 0,y\gt 0$; $[f^{-1}]'(x)=-1/x^2$ is defined for $x\gt 0,y\lt 0$ and increasing.
So we have a problem with the first two statements of our Theorem.

In general. Consider $f(x)$, defined for $x\gt 0,y\gt 0$ and monotonically decreasing. Then $f'(x)$ is defined for $x\gt 0,y\lt 0$ (i.e. negative) and monotonically increasing. $[f'(x)]^{-1}$, because of the mirroring in $y=x$, is defined for $x\lt 0,y\gt 0$ and monotonically increasing as well. $f^{-1}(x)$, because of the mirroring in $y=x$, is defined for $x\gt 0,y\gt 0$ and monotonically decreasing. So $[f^{-1}]'(x)$ is defined for $x\gt 0,y\lt 0$ and monotonically increasing.
However, because of the two different domains $[f^{-1}]'(x)$ cannot be identical to $[f'(x)]^{-1}$.
Conclusion: the last statement of our Theorem remains valid, but the first two statements, in general, are wrong.

Note. A better numerical approximation of the integral sometimes can be obtained with the trapezium rule: $$ \frac{g\left(x_i\right)+g\left(x_{i+1}\right)}{2}\left(x_{i+1}-x_i\right) = \mbox{constant} = dA = \frac{1}{n} $$ But it does work only if $g\left(x_{i+1}\right)$ can be predicted, given the value of $dA$. In case $g(x)$ is an orthogonal hyperbola, prediction is successful: $$ x_{i+1}-x_i=\frac{1}{n}\frac{2}{1/x_{i+1}+1/x_i}=\frac{1}{n}\frac{2x_{i+1}x_i}{x_{i+1}+x_i} \quad \Longleftrightarrow \\ n\left(x_{i+1}-x_i\right)\left(x_{i+1}+x_i\right)-2x_{i+1}x_i=nx_{i+1}^2-2x_ix_{i+1}-nx_i^2=0 \quad \Longleftrightarrow \\ x_{i+1} = \frac{x_i}{n}\pm\sqrt{\left(\frac{x_i}{n}\right)^2+x_i^2}\quad \Longrightarrow\quad x_{i+1}=x_i\left(\frac{1}{n}+\sqrt{1+\frac{1}{n^2}}\right) $$

  • $\begingroup$ How can we extend this train of thought to complex-valued functions? $\endgroup$ – Christian Feb 26 at 10:34
  • $\begingroup$ @Christian: Interesting, but I'm affraid that is beyond my competence. $\endgroup$ – Han de Bruijn Feb 26 at 12:49
  • $\begingroup$ I didn't read through but $x\mapsto x^3$ has a vanishing derivative at $0$ but is increasing. This function really is invertible on the whole $\mathbb{R}$ although not it's derivative.\\ If a differentiable function is complex valued and invertible, then it's variable should also be 2 $\mathbb{R}$-dimensional. Restricting both the domain and the target to a 1 dimensional subspace should give a solution to the initial problem. If one requires the function to be $\mathbb{C}$ differentiable then it is analytic, so we should have found a solution that expends as a series. $\endgroup$ – Noix07 Feb 26 at 19:32
  • $\begingroup$ @Noix07: $f'(x)=x^3$ has an inverse $[f'(x)]^{-1}=x^{1/3}$. Both are monotonically increasing for $x\ge 0$. $f(x)=\int_0^x t^3\,dt = x^4/4 \; \Rightarrow \; f^{-1}(x)=(4x)^{1/4} \; \Rightarrow \; [f^{-1}]'(x)=1/(4x)^{3/4}$. The latter function is monotonically decreasing, as it should. $\endgroup$ – Han de Bruijn Feb 26 at 20:35
  • $\begingroup$ @HandeBruijn I have not yet read most of the answer but it is the assertion that monotonic implies derivative non zero that is wrong.It is true that the derivative will not vanish on an interval but the example shows that $x^3$ is increasing in $\mathbb{R}$ although its derivative vanishes at 0 $\endgroup$ – Noix07 Mar 1 at 12:07

I foud this video about complex solutions to the problem of the form $Cx^N$:

$$\frac{d}{dx} f^{\langle -1 \rangle}(x) = \left(\frac{d}{dx} f(x) \right)^{\langle -1 \rangle} \quad \to \; \frac{ x^\frac{1-N}{N}}{NC^\frac{1}{N}} = \left(\frac{x}{CN} \right)^\frac{1}{N-1}$$

This only holds when $\quad \frac{1-N}{N} = \frac{1}{N-1} \qquad N^2-N+1=0 \qquad N = \frac{1 \, \pm \, i \sqrt{3}}{2} = e^{\pm \frac{i \pi}{3}}$

and $\quad NC^\frac{1}{N} = (CN)^\frac{1}{N-1} \qquad C = N^{(N-2)N}= \pm \, ie^{\pm \frac{\pi}{2\sqrt{3}}} \;$ or $\; e^{\pm \frac{\pi}{\sqrt{3}}}$

  • $\begingroup$ Excellent post (+1)! But alas, I can't reproduce the last simplification $C=N^{(N-2)N}=$ ?? $\endgroup$ – Han de Bruijn Apr 21 at 18:01
  • $\begingroup$ To be completely sure, I have formulated the discrepancy between your thoughts and mine as a question: Simplify $N^{N(N-2)}$ with $N=(1\pm i\sqrt{3})/2$. And so it seems that the very last part of your answer is not entirely correct. $\endgroup$ – Han de Bruijn Apr 24 at 14:54
  • $\begingroup$ The two solutions simplified correctly: $$ y = -i.e^{\pi/\sqrt{3}/2}.x^{1/2+i.\sqrt{3}/2} \\ y = +i.e^{\pi/\sqrt{3}/2}.x^{1/2-i.\sqrt{3}/2} $$ $\endgroup$ – Han de Bruijn May 13 at 10:06

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