= Solution
Put $\widetilde\xi_t=\phi_t(\xi_t)$ and $p_t=\phi_t'(\xi_t)$. The given <conformal change of half-plane capacity> makes the image mapping-out equation
$$
\partial_t\widetilde g_t(w)
=\frac{2p_t^2}{\widetilde g_t(w)-\widetilde\xi_t}.
$$
Indeed its capacity speed is $2p_t^2$, and the stated <Loewner local growth property> identifies the indicated single driver. For $t<T$, differentiate the conjugacy identity
$$
\widetilde g_t\circ\phi=\phi_t\circ g_t
$$
at a fixed original point. The <Chordal Loewner equation> then gives the \b[<conformal conjugacy derivative for the chordal Loewner equation>]
$$
\boxed{\dot\phi_t(z)
=\frac{2\phi_t'(\xi_t)^2}{\phi_t(z)-\phi_t(\xi_t)}
-\frac{2\phi_t'(z)}{z-\xi_t}.}
$$
The change of variable $z=g_t(w)$ covers the whole upper half-plane. The <conformal automorphism of the upper half-plane> $\phi_t$ is a real <Möbius transformation>. Its pole is the image of the still-unswallowed boundary point $-1$, so it is holomorphic in a neighbourhood of $\xi_t$ before $T$.
To evaluate the apparent singularity, write $h=z-\xi_t$, $p=\phi_t'(\xi_t)>0$ and $q=\phi_t''(\xi_t)$. The <Taylor series> is
$$
\phi_t(\xi_t+h)-\phi_t(\xi_t)=ph+\tfrac12qh^2+O(h^3),
\qquad\phi_t'(\xi_t+h)=p+qh+O(h^2).
$$
Thus the first term in the derivative is $2p/h-q+O(h)$, while the second is $2p/h+2q+O(h)$. Their poles cancel and their constant terms differ by $-3q$. Therefore
$$
\boxed{\dot\phi_t(\xi_t)=-3\phi_t''(\xi_t).}
$$
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