Suppose that $R$ is a prime ring and charR $\neq2$. $d_{1},d_{2}$ are derivations of $R$ such that for some non-zero ideal of $I$ of $R$ and some $c\in C$ ($C$: extended centroid of $R$).
$d_{1}d_{2}(x)=cx$ for all $x \in I$.
By computing in different ways $d_{1}d_{2}(xty)$ for $x,t,y\in I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $x\in I$ and some $c\in C$, one can easily obtain $$0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$$ My question is how can we show that $\hspace{0.1cm}$ $0=d_{1}(x)td_{2}(y)+d_{2}(x)td_{1}(y)$?
$\textbf{My attempt:}$
\begin{align} \begin{aligned} d_1d_2(xty) =& d_1[d_2(x)ty+xd_2(ty)]\\ =& d_1d_2(x)ty+d_2(x)[d_1(ty)]+d_1(x)[d_2(ty)]+xd_1[d_2(ty)]\\ =& d_1d_2(x)ty+d_2(x)[d_1(t)y+td_1(y)]+d_1(x)[d_2(t)y+td_2(y)]+xd_1[d_2(t)y+td_2(y)] \\ =& d_1(x)td_2(y)+d_2(x)td_1(y)+d_1d_2(x)ty+d_2(x)d_1(t)y + x[d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)] \\ =& d_1(x)td_2(y)+d_2(x)td_1(y)+3cxyt+[d_2(x)d_1(t)+d_1(x)d_2(t)]y+x[d_2(t)d_1(y)+d_1(t)d_2(y)] \end{aligned} \end{align}
Now,$cty=d_1d_2(ty)=d_1[d_2(t)y+td_2(y)]=d_1d_2(t)y+d_2(t)d_1(y)+d_1(t)d_2(y)+td_1d_2(y)$.
By hypothesis $cty=cty+d_2(t)d_1(y)+d_1(t)d_2(y)+cty$ Hence, we obtain $d_2(t)d_1(y)=-cty-d_1(t)d_2(y)$ and $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.