Relation between algebraic and analytic derivations in $\mathcal C(\mathbb R)$

Question

Let $A,B$ be two $\mathbb R$-algebras. By an $\mathbb R$-linear (algebraic) derivation from $A$ to $B$, we mean a group homomorphism $D:A\rightarrow B$ which contains $\mathbb R$ in its kernel and satisfies the Liebnitz rule, i.e., $D(ab)=aD(b)+bD(a)$ for all $a,b\in A$. Let $\mathcal C(\mathbb R)$ denote the $\mathbb R$-algebra of all real valued continuous functions defined on $\mathbb R$. If $f:\mathbb R\rightarrow \mathbb R$ is a continuous map, we know that $f$ is said to be (analytically) differentiable at $x\in \mathbb R$ if the limit $\lim_{a\to 0}\frac{f(x+a)-f(x)}{a}$ exists; and a function which is differentiable at every point of $\mathbb R$ is called a differentiable function. In this case, we denote the derivative of $f$ by $f'$. Let us call an $\mathbb R$-linear algebraic derivation $D:A\rightarrow \mathcal C(\mathbb R)$, where $A$ is an $\mathbb R$-subalgebra of $\mathcal C(\mathbb R)$, to be maximal if there does not exist any $\mathbb R$-subalgebra $A'$, properly containing $A$, and an $\mathbb R$-linear algebraic derivation $D':A'\rightarrow \mathcal C(\mathbb R)$ such that $D=D'|_A$. My question is that, if $D:A\rightarrow \mathcal C(\mathbb R)$ is a maximal $\mathbb R$-linear algebraic derivation such that $D(x)=1$, where $x:\mathbb R\rightarrow \mathbb R$ is the identity function, does it follow that $A$ is the ring of real differentiable functions (in the analytic sense), and $D(f)=f'$?