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Use the extended-value convention for a normal family, which is essential here: every sequence in the holomorphic family has a subsequence converging uniformly on compact subsets either to a finite holomorphic function or to infinity. Convergence to infinity means ${\inf }_K|f_n|\to \infty$ for every compact $K\subset D$. Equivalently one uses local uniform convergence in the chordal metric. For holomorphic functions a nonconstant meromorphic limit cannot acquire a pole: near a supposed pole, the reciprocals of sufficiently large-index functions are holomorphic and nowhere zero, and converge to a holomorphic function with a zero. Hurwitz's theorem would force that reciprocal limit to vanish identically, giving the constant infinite limit instead.

Restriction proves that global normality implies normality near each point. For the converse, choose a countable cover by connected discs $U_j$ on which the family is normal, with smaller discs covering $D$ and having compact closure inside the corresponding $U_j$. Such a cover exists from the countable rational-disc basis and local normality. Starting with any sequence, extract a subsequence converging on $U_1$, then a further subsequence converging on $U_2$, and so on. The diagonal sequence converges on each $U_j$. On an overlap the limits agree; a finite holomorphic limit and infinity cannot both occur there. The overlap graph of this cover is connected, since otherwise its components would partition the connected domain into disjoint open subsets. Therefore all local limits are of the same type and patch to one holomorphic function or to infinity throughout $D$. Every compact set has a finite cover by the smaller discs, so convergence on their compact closures gives uniform convergence on that compact set. This proves that normality is a local property.

For the counterexample take the unit disc and the family $f_n(z)=n(1+z^2/2)$. Since $|1+z^2/2|\ge 1-|z{|}^2/2>1/2$, $f_n$ tends uniformly to infinity. Any sequence from this family either has a constant subsequence or has indices tending to infinity, so the family is normal. Its derivatives are $nz$. No subsequence of these derivatives can have a finite locally uniform limit, because their values at any fixed nonzero point grow without bound; nor can they tend locally uniformly to infinity, since every derivative vanishes at zero. Thus a holomorphic normal family need not have a normal derivative family.

Now impose the point bound at $z_0$. For any sequence of derivatives choose corresponding functions from the original family. Normality supplies a locally uniformly convergent subsequence, and the point bound excludes the infinite limit. Let its finite holomorphic limit be $f$. The Cauchy integral formula gives convergence of derivatives uniformly on compact subsets. Explicitly, for compact $K\subset D$ choose $\rho >0$ so its closed $\rho$-neighborhood lies in a compact $L\subset D$; thenEvery sequence of derivatives therefore has a locally uniform subsequential limit. This pointwise anchoring of a holomorphic normal family proves the derivative family is normal under the stated bound.