Answer
$$
2s^{2}+\sqrt {st}-4=3t
$$
The velocity $\frac{ds}{dt}$ is given by:
\begin{aligned}
\frac{d s}{d t}=\frac{-s+6 \sqrt{s t}}{8 s \sqrt{s t}+t}
\end{aligned}
Work Step by Step
$$
2s^{2}+\sqrt {st}-4=3t
$$
Now, we can calculate the velocity $\frac{ds}{dt}$ by implicit differentiation, use the product rule as follows:
$$
\begin{aligned}
\frac{d }{d t}(2s^{2}+\sqrt {st}-4 )&=\frac{d}{d t}(3t) \\
4s \frac{d s}{d t}+\frac{1}{2}(s t)^{-1 / 2}\left(s+t \frac{d s}{d t}\right) &=3\\
4 s \frac{d s}{d t}+\frac{s+t \frac{d s}{d t}}{2 \sqrt{s t}}&=3 \\
\frac{8 s(\sqrt{s t}) \frac{d s}{d t}+s+t \frac{d s}{d t}}{2 \sqrt{s t}}&=3\\
\frac{(8 s \sqrt{s t}+t) \frac{d s}{d t}+s}{2 \sqrt{s t}}&=3\\
(8 s \sqrt{s t}+t) \frac{d s}{d t}=6 \sqrt{s t}-s \\
\frac{d s}{d t}=\frac{-s+6 \sqrt{s t}}{8 s \sqrt{s t}+t}.
\end{aligned}
$$
