If $D_1$ and $D_2$ are derivations of $A$, show that $D_1 ◦ D_2$ is not necessarily a derivation of $A$ 
If $D_1$ and $D_2$ are derivations of $A$, show that $D_1 ◦ D_2$ is not necessarily a derivation of $A$.

What I worked out:
The Leibniz rule insists
$(D_1 ◦ D_2)(a) · b + a · (D_1 ◦ D_2)(b) = (D_1 ◦ D_2)(ab)$
$= D_1D_2(ab)$
= $D_1(D_2a)b + a(D_2b)$
= $D_1(D_2a)b+ D_1aD_2b$
= $(D_1 ◦ D_2)(a) · b + D_2aD_1b + D_1aD_1aD_2b + a · (D_1 ◦ D_2)(b)$,
which is true if and only if $D_1aD_2b = −D_2aD_1b$, which clearly isn’t necessarily the case.
Could someone give me a specific counterexample for this?

Hint: Let $A$ be the polynomial ring $k\left[X\right]$ over a field $k$. Let $D$ be the map $A \to A$ which sends every polynomial $p$ to its derivative.
(1) Show that $D$ is a derivation of $A$.
(2) What must $k$ satisfy in order for $D \circ D$ to be a derivation of $A$ ? (Most of the time it will not be, but sometimes it will be.)
