When is the derivative of $f(g(x))$ equal to $g(f'(x))$?

By the chain rule, we know that the derivative of $f(g(x))$ is $f'(g(x))g'(x)$.

Question: When is the derivative of $f(g(x))$ equal to $g(f'(x))$?

Trivial solutions include the following:

• Let $f$ be any differentiable function and $g$ be the identity function ($g(x)=x$) or the zero function ($g(x)=0$).
• Let $f$ be any constant function and $g$ be any differentiable function that fixes zero.

On the other hand, $f(x)=x^2$ and $g(x)=\sin x$ form a nontrivial solution (by the double angle formula for sine).

If $g(x)=x+a$, where $a$ is a constant, then $f'(x+a)=f'(x)+a$, so $f(x)=\frac{1}{2}x^2+bx+c$ would work for any two constants $b$ and $c$.

Finally, if $f(x)=mx+b$, where $m$ and $b$ are constants, then $mg'(x)=g(m)$, so $g'(x)=\frac{g(m)}{m}$ and $g(x)=\frac{g(m)}{m}x+c$ for some constant $c$. But this must in particular be true for $x=m$, so $c$ must be zero and $g(x)=nx$ would then work for any constant $n$.

But I do not know any other solutions. Can anyone help find one?

Note that $f$ may be replaced by any other function with the same derivative (i.e. differing from $f$ by a constant) without changing the validity of the equation, so we may assume without lost of generality that $f$ fixes zero (assuming, of course, that zero is in the domain of $f$).

Solution 1:

Not an answer, but too long for a comment: Let $F(x) = f'(x)$. Then you are trying to solve $$F(g(x))g'(x) = g(F(x))$$

Let $g(x)$ be any differentiable function. This becomes a functional equation for $F(x)$. For $g(x) = x^n$, where $n > 0$, it is $$F\left(x^n\right)nx^{n-1} = F(x)^n$$

The solution (found using Mathematica) is $$F(x) = n^{\frac{1}{n-1}}x^{c - \frac{(n-1)\ln(\ln(x))}{n\ln(n)}}$$

$f(x)$ would then be the integral of $F(x)$.

I think it may be possible to show that given "sufficiently nice" $g$, an $F$ exists that solves the functional equation. I'm not sure how to go about this though.