Bernoulli transformation in " A Close Form Solution to the Ramsey model" by Smith (2006)
If $z = k^{1-\alpha}$ then

\begin{eqnarray}
\dot{z} &amp;=&amp; (1-\alpha)k^{1-\alpha-1}\dot{k} = (1-\alpha)k^{-\alpha} \dot{k} \\ &amp;=&amp; (1-\alpha) k^{-\alpha} [k^{\alpha} - \delta k - c] \\
&amp;=&amp; (1-\alpha) [1 - \delta k^{1-\alpha} - c k^{-\alpha}\color{blue}{(k/k)}] \\
&amp;=&amp; (1-\alpha) [1 - \delta z - c(k^{1-\alpha})/k] \\
&amp;=&amp; (1-\alpha) [1 - \delta z - x z]
\end{eqnarray}

I will leave $\dot{c}/c$ for you to work out