Krdytkyu

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 $cin C$ ($C$: 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 $x,t,yin I$, making use of the facts that $d_{i}$'s are derivations and $d_{1}d_{2}(x)=cx$ for $xin I$ and some $cin C$, 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 $d_2(x)d_1(t)=-ctx-d_1(x)d_2(t)$.