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If you grant that $c'=0$ when $c$ is a constant, you can argue as follows:

EDIT: Actually $c'=0$ follows from $1'=0$ using the chain rule, since $c=c\circ 1$.

Let $I(x)=x$. Since $I=I\circ I$, the chain rule gives $(I')^2=I'$, so $I'$ is the characteristic function of a set $A$ of real numbers. Let $T_c(x)=x+c$. Then $T_c'=I'+c'=I'$, and the chain rule applied to $T_c=I\circ T_c$ then implies $I'=(I'\circ T_c)I'$. That means that the set $A$ is either all or nothing. But you can't have $I'$ identically $0$ because that would imply $f'=(f\circ I)'=0$ for all $f$. So $I'=1$ as expected.

Now consider linear functions. If $L_m(x)=mx$, then since $L_m(x+a)-L_m(x)$ is constant, $L_m'(x+a)-L_m'(x)$ is zero. Thus $L_m'$ is a constant depending on $m$, say $h(m)$. The map $h$ is an additive homomorphism, and by the chain rule it is also multiplicative. We also know $h(1)=1$. This implies that $h(m)=m$ for all $m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...)

So $f'$ is what it should be when $f$ is polynomial of degree at most one. Now let $S(x)=x^2$. From $S(x+t)=S(x)+2tx+t^2$ we get $S'(x+t)=S'(x)+2t$, therefore $S'(x)=S'(0)+2x$. But $S'(0)=0$ using $S(-x)=S(x)$. So the derivative of squaring is what it should be.

Now the special case of Leibniz that says $(f^2)'=2ff'$ follows by the chain rule. The general case follows by expressing $fg$ in terms of $f^2$, $g^2$, and $(f+g)^2$.

EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are: $f'$ is defined when $f$ is a $C^\infty$ real function whose domain is an open set $U\subset \mathbb R$, and $f'$ is another such function with the same domain. The axioms are

(U) $1_U'=0_U$ where $1_U$ is the constant function on $U$.

(A) $(f+g)'=f'+g'$ where $f$ and $g$ (and $f+g$) have the same domain.

(C) $(f\circ g)'$ is the product of (the restriction to $V$ of) $f'\circ g$ and $g'$, g)'=(f'\circ g)g'$, when$f$has domain$U$and$g$has domain$V$and$g(V)\subset U$, so that$f\circ g$has and$f'\circ g$have domain$V$. (Z) For every nonempty$U$there is some$f$with domain$U$such that$f'$is not identically zero. The arguments that I gave above can be adapted to show then that:$c_U'=0$for any constant function on any$U$.$I_U'=1$where$I_U$with domain$U$is defined by$I_U(x)=x$. (Here you have to mess around with compositions$I_U\circ (I_V+c_V)$.) So in the end you get the desired localization property, too: the operator commutes with restriction from$U$to$V\subset U$by the chain rule, because restriction is composition with$I_V$. 3 added 1267 characters in body EDIT: Actually$c'=0$follows from$1'=0$using the chain rule, since$c=c\circ 1$. EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are:$f'$is defined when$f$is a$C^\infty$real function whose domain is an open set$U\subset \mathbb R$, and$f'$another such function with the same domain. The axioms are (U)$1_U'=0_U$where$1_U$is the constant function on$U$. (A)$(f+g)'=f'+g'$where$f$and$g$(and$f+g$) have the same domain. (C)$(f\circ g)'$is the product of (the restriction to$V$of)$f'\circ g$and$g'$, when$f$has domain$U$and$g$has domain$V$and$g(V)\subset U$so that$f\circ g$has domain$V$. (Z) For every nonempty$U$there is some$f$with domain$U$such that$f'$is not identically zero. The arguments that I gave above can be adapted to show then that:$c_U'=0$for any constant function on any$U$.$I_U'=1$where$I_U$with domain$U$is defined by$I_U(x)=x$. (Here you have to mess around with compositions$I_U\circ (I_V+c_V)$.) So in the end you get the desired localization property, too: the operator commutes with restriction from$U$to$V\subset U$by the chain rule, because restriction is composition with$I_V$. 2 deleted 1 characters in body If you grant that$c'=0$when$c$is a constant, you can argue as follows: Let$I(x)=x$. Since$I=I\circ I$, the chain rule gives$(I')^2=I'$, so$I'$is the characterisitic characteristic function of a set$A$of real numbers. Let$T_c(x)=x+c$. Then$T_c'=I'+c'=I'$, and the chain rule applied to$T_c=I\circ T_c$then implies$I'=(I'\circ T_c)I'$. That means that the set$A$is either all or nothing. But you can't have$I'$identically$0$because that would imply$f'=(f\circ I)'=0$for all$f$. So$I'=1$as expected. Now consider linear functions. If$L_m(x)=mx$, then since$L_m(x+a)-L_m(x)$is constant,$L_m'(x+a)-L_m'(x)$is zero. Thus$L_m'$is a constant depending on$m$, say$h(m)$. The map$h$is an additive homomorphism, and by the chain rule it is also multiplicative. We also know$h(1)=1$. This implies that$h(m)=m$for all$m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...) So$f'$is what it should be when$f$is polynomial of degree at most one. Now let$S(x)=x^2$. From$S(x+t)=S(x)+2tx+t^2$we get$S'(x+t)=S'(x)+2t$, therefore$S'(x)=S'(0)+2x$. But$S'(0)=0$using$S(-x)=S(x)$. So the derivative of squaring is what it should be. Now the special case of Leibniz that says$(f^2)'=2ff'$follows by the chain rule. The general case follows by expressing$fg$in terms of$f^2$,$g^2$, and$(f+g)^2\$.

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