Consider $M$ be the mass of rocket at any time $t$ and $v_1$ the velocity of rocket at the same time $t$.
$\Delta m=$ mass of gas ejected in time interval $\Delta t$.
Relative speed of gas ejected $=u$.
Now (K.E.) at $(t+\Delta t)$ is :
$(\mathrm{KE})(t+\Delta t)=\mathrm{KE}$ of rocket $+\mathrm{KE}$ of gas
$=\frac{1}{2}(M-\Delta m)(v+\Delta v)^2+\frac{1}{2} \Delta m(v-u)^2$
$=\frac{1}{2}\left[(M-\Delta m)\left(v^2+\Delta v^2+2 v \Delta v\right)\right.$
$\left.+\quad+\Delta m\left(v^2+u^2-2 u v\right)\right]$
$=\frac{1}{2}\left[M v^2+M \Delta v^2+2 M v \Delta v-\Delta m v^2\right.$
$\quad-\Delta m \Delta v^2-2 v \Delta m \Delta v+\Delta m v^2$
$$
\begin{aligned}
\Delta K &=(\mathrm{KE})_t+\Delta t-(\mathrm{KE})_t \\
&=(M \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^2
\end{aligned}
$$
Since, action-reaction forces are equal.
Hence, $M \frac{d v}{d t}=\frac{d m}{d t}|u|$
$$
\begin{aligned}
&\Rightarrow M \Delta v=\Delta m u \\
&\Delta K=\frac{1}{2} \Delta m u^2
\end{aligned}
$$
Now, by work-energy theorem,
$$
\begin{aligned}
&\Delta K=\Delta W \\
&\Rightarrow \Delta W=\frac{1}{2} \Delta m u^2
\end{aligned}
$$