fabian s. klinke

A fascinating differential equation: when the derivative equals the inverse function

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A fascinating differential equation: when the derivative equals the inverse function

Source: A fascinating differential equation: when the derivative equals the inverse function, Maths 505, 9:11, uploaded 2023-12-16, category Mathematics, playlist index 1234.

Maths 505 starts with a simple question that leads to a short derivation involving differentiation, inverse functions, and the golden ratio: which function ff has a derivative equal to its own inverse?

The equation is

f′(x)=f−1(x).f'(x)=f^{-1}(x).

The familiar candidates fail at once. The derivative of exe^x is exe^x, whilst its inverse is ln⁡x\ln x. Differentiating sin⁡x\sin x gives cos⁡x\cos x, which has no relation of this kind to sin⁡−1x\sin^{-1}x. The video therefore tests a power function, since differentiating and inverting powers keeps the same general form.

Take

f(x)=αxβ,f(x)=\alpha x^\beta,

where the video allows α\alpha and β\beta to be complex numbers. Differentiation gives

f′(x)=αβxβ−1.f'(x)=\alpha\beta x^{\beta-1}.

To find the inverse, write y=αxβy=\alpha x^\beta and solve for xx:

x=(yα)1/β=α−1/βy1/β.x=\left(\frac{y}{\alpha}\right)^{1/\beta}=\alpha^{-1/\beta}y^{1/\beta}.

The inverse function, expressed with xx as its input, is therefore

f−1(x)=α−1/βx1/β.f^{-1}(x)=\alpha^{-1/\beta}x^{1/\beta}.

The differential equation now compares two power functions:

αβxβ−1=α−1/βx1/β.\alpha\beta x^{\beta-1}=\alpha^{-1/\beta}x^{1/\beta}.

The video matches the exponents and coefficients separately. The exponents give

β−1=1β,\beta-1=\frac{1}{\beta},

which becomes

β2−β−1=0.\beta^2-\beta-1=0.

The quadratic formula produces

β=1+52orβ=1−52.\beta=\frac{1+\sqrt{5}}{2} \quad\text{or}\quad \beta=\frac{1-\sqrt{5}}{2}.

The coefficient equation is

αβ=α−1/β.\alpha\beta=\alpha^{-1/\beta}.

The first relation also says 1+1/β=β1+1/\beta=\beta and 1/β=β−11/\beta=\beta-1. Using those identities in the coefficient equation gives

αβ=1β=β−1,\alpha^\beta=\frac{1}{\beta}=\beta-1,

so the corresponding coefficient can be written as

α=(β−1)β−1.\alpha=(\beta-1)^{\beta-1}.

The two candidate functions

Let

φ=1+52\varphi=\frac{1+\sqrt{5}}{2}

be the golden ratio. The two roots for β\beta are φ\varphi and 1−φ1-\varphi. Substituting them into the coefficient gives the two power-function candidates

f1(x)=(φ−1)φ−1xφf_1(x)=(\varphi-1)^{\varphi-1}x^\varphi

and

f2(x)=(−φ)−φx1−φ.f_2(x)=(-\varphi)^{-\varphi}x^{1-\varphi}.

The second coefficient comes from raising a negative number to an irrational power, so the video treats it as complex. That is why the initial permission to use complex α\alpha and β\beta matters. The golden ratio appears because the exponent equation has the same quadratic relation as φ\varphi, rather than because it was added to the problem from outside.

Verification through the golden-ratio identity

The video checks the first candidate. Differentiating it gives

f1′(x)=φ(φ−1)φ−1xφ−1.f_1'(x)=\varphi(\varphi-1)^{\varphi-1}x^{\varphi-1}.

Its inverse is

f1−1(x)=(x(φ−1)φ−1)1/φ.f_1^{-1}(x)=\left(\frac{x}{(\varphi-1)^{\varphi-1}}\right)^{1/\varphi}.

At first the two expressions look unrelated. The identity

φ2−φ−1=0\varphi^2-\varphi-1=0

gives φ−1=1/φ\varphi-1=1/\varphi, and the same relation reduces the coefficient and exponent on both sides to the same powers of φ\varphi. The derivative and inverse then agree. The video leaves the second candidate as an exercise.

Limits of the result

The calculation searches within the family f(x)=αxβf(x)=\alpha x^\beta. It does not prove that every solution of f′=f−1f'=f^{-1} must have that form. The speaker says that the two candidates are probably the only solutions, then labels that conclusion as a guess and asks viewers to find others.

The complex expressions also need domain and branch choices before they become fully specified functions. The caption track misrecognises φ\varphi and several square-root and minus signs, so this note reconstructs those symbols from the displayed algebra and the relations the speaker states aloud. The source gives a formal derivation, with the classification question left open.

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