= Solution
At the current exact path point $v(\mu)$, stationarity is already zero and $Cy=\mu e$. Therefore $F_{\bar\mu}(v(\mu))=(0,(\mu-\bar\mu)e)^T$. Its <Newton correction> solves
$$
\boxed{J(v(\mu))\Delta v_N=\begin{pmatrix}0\\(\bar\mu-\mu)e\end{pmatrix}.}
$$
The matrix has no explicit dependence on the barrier parameter, so comparison with part (iv) gives
$$
\boxed{\Delta v_N=(\bar\mu-\mu)v'(\mu)}
$$
when $J$ is nonsingular. If it is singular, the displayed vector is still a solution of the correction equations, but uniqueness of the correction cannot be inferred from differentiability of the path alone.
This step is exactly the tangent predictor in the stated first-order <Taylor expansion>. For nearby positive $\bar\mu$, it approximates the new path point with error $o(|\bar\mu-\mu|)$; a bounded second path derivative improves this to $O((\bar\mu-\mu)^2)$. Continuity also preserves strict positivity for sufficiently small parameter changes. Larger changes may require damping to keep both the constraint values and multiplier estimates positive.
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