Chain rule characterizes the derivative?

To what extend does the chain rule $(f\circ g)'=(f'\circ g)\cdot g'$ characterize the derivative $(\ldots)'$?

More precisely: If a non-constant operator $T$ defined on all differentiable functions, satisfies the relation $$T(f\circ g)=((Tf)\circ g)\cdot T g\,,$$ does this imply that $T$ is the derivative?

Motivation: I am particularly interested in the following questions:

• What kind of operators are obtained when the defining relation is something like $$(T(f\circ g))(x)=((Tf)(g(x))^{(T g)(x)}\,?$$
• Is there a somewhat standard way to solve these equations for $T$?
• What branch of mathematics deals with these kinds of equations?
• Is there a standard representation we can fill in for $T$, to solve these equations? (like substituting a Taylor series into a differential equation)
• It's usually the product rule that is used to define differential operators. I have no idea what happens if we use the chain rule instead. – Arthur Feb 11 '18 at 11:02
• What if $T$ sends each differentiable function to $x\mapsto 1$? Or even $x\mapsto 0$. – Git Gud Feb 11 '18 at 11:09
• @GitGud, Thank you. I didn't think of x\mapsto 1. However, non-constant solutions for $T$, would be more interesting. – Carucel Feb 11 '18 at 12:54

I tried to show for the $V =\mathbb{R}[x]$ (or in general, for $V = k[x]$ with a characteristic 0 field $k$) first, and it seems that chain rule is sufficient to characterize the differentiation for this case. Before we start, we will assume that $T:\mathbb{R}[x]\to \mathbb{R}[x]\,$ is $\mathbb{R}$-linear and $T(1)=0$ for a constant function $f\equiv 1$. (I think that if we assume $T\not\equiv 0$, we can show this, but I don't know how to do it. It should be easy!)
Since the equation $T(f\circ g) = (Tf\circ g)\cdot Tg\,$ is linear on $f$, it is enough to consider $f=x^{n}$. First, put $g(x) = x$ and we get $Tf = (Tf)\cdot Tx$ for any $f$, so $T\equiv 0$ or $Tx=1$. Now assume that $T$ is nontrivial, i.e. $Tx=1$. Now put $f(x) = x^{2}$ and $g(x) = x+a$ for $a\in \mathbb{R}$, then $$Tx^{2} + 2a = T((x+a)^{2}) = T(f\circ g) = (Tf\circ g)\cdot Tg = (Tx^{2}\circ(x+a))\cdot 1 = Tx^{2}\circ(x+a)$$ If $T(x^{2}) =p(x)$, then we have $p(x)+2a = p(x+a)$, so $p(x)$ has degree 1 and $p(x) =Tx^{2} = 2x+c_{2}$ for some $c_{2}\in \mathbb{R}$. However, for $f=x^{2}$ and $g=ax$, we have $$a^{2}Tx^{2} = (Tx^{2}\circ ax)\cdot a\Leftrightarrow a\cdot p(x) = p(ax)$$ which implies that $p(x)$ is linear, i.e. $p(x) =Tx^{2} = 2x$. Similarly, if $Tx^{3} =p_{3}(x)$, then $$p_{3}(x) + 6ax+3a = p_{3}(x+a), \quad a^{2}\cdot p_{3}(x)=p_{3}(ax)$$ for any $a\in \mathbb{R}$, so $p_{3}(x)/x^{2} = p_{3}(ax)/(ax)^{2}\Rightarrow p_{3}(x) = \alpha_{3}x^{2}=3x^{2}$. By using induction, we can show that $Tx^{n} = nx^{n-1}$ for any $n$, hence $T = d/dx$ for $V = \mathbb{R}[x]$.
To show $T(1)=0$, let $T(1) =p_{0}(x)$ and put $g(x)=c$. Then $$T(f(c)) = f(c)p_{0}(x) = p_{0}(c)\cdot cp_{0}(x)$$ for any $c\in \mathbb{R}$. Now choose $f\in \mathbb{R}[x]$ with $f(c)\neq cp_{0}(c)$, then we have $p_{0}(x)\equiv 1$.