 Solution

Differentiation satisfies $D(af+bg)=aD(f)+bD(g)$ for constant scalars $a,b$, so it is a linear map on suitable vector spaces of differentiable functions. An arbitrary set $S$ need not even be a vector space, and an endomorphism interpretation also requires $D(S)\subseteq S$.

For a noninvertible example, take $S=\mathrm{span}\{1,x\}$. Then $D(a+bx)=b$ and the kernel of a linear map is the nonzero linear span of $1$, so $D:S\to S$ is not injective. The matrix in this basis is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

For an invertible example, take $S=\mathrm{span}\{e^x,e^{2x}\}$. In this basis, $D$ has matrix $\mathrm{diag}(1,2)$, andThus differentiation is not always invertible; the domain and codomain determine the answer.