Computing a conformal block in 2D CFT The following comes from Di Francesco et al., section 6.6.4. The exercise is number 6.5.
A conformal block is given by
$$\mathcal{F}^{21}_{34}(p\mid x) = x^{h_p-h_3-h_4}\sum_{\{k\}}\beta^{p\{k\}}_{34}x^K\frac{\langle h_1|\phi_2(1)L_{-k_1}\cdots L_{-k_N}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}$$
where the denominator is equal to $(C_{12}^{p})^{1/2}$ and $K=\sum_i k_i$.
The blocks can be written as a power series:
$$\mathcal{F}_{34}^{21}(p\mid x)=x^{h_p-h_3-h_4}\sum_{K=0}^{\infty} \mathcal{F}_K x^K$$
I want to compute
$$\mathcal{F}_1=\beta_{34}^{p\{1\}}\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}=\frac12\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}$$
and in particular show that
$$ \mathcal{F}_1 = \frac{(h_p+h_2-h_1)(h_p+h_3-h_4)}{2h_p}$$
The authors suggest to compute the block by commuting the Virasoro generators over $\phi_2(1)$.
$$\langle h_1|\phi_2(1)L_{-1}|h_p\rangle=\langle h_1|L_{-1}\phi_2(1)|h_p\rangle-\langle h_1|[L_{-1},\phi_2(1)]|h_p\rangle$$
I think the second term is equal to $\langle h_1|\partial\phi_2(1)|h_p\rangle$ while the first is zero because $L_{-1}$ annihilates primary fields. The term $\langle h_1|\partial\phi_2(1)|h_p\rangle$ is calculated using $\mathcal{L}_{-1}$, so
$$\langle h_1|\partial\phi_2(1)|h_p\rangle = \mathcal{L}_{-1}\langle h_1|\phi_2(1)|h_p\rangle =-\partial\langle h_1|\phi_2(1)|h_p\rangle$$
But doesn't $\partial \phi_2(1)=0$ if $\phi_2$ is defined at $1$?
Where have I gone wrong? How do you compute $\mathcal{F}_{1}$?
 A: Firstly, $\beta_{34}^{p\{1\}}=\frac{1}{2}$ only when $h_3=h_4$, this is the case of (6.177). For general $h_3$ and $h_4$, I believe
$$
\beta_{34}^{p\{1\}}=\frac{h_p+h_3-h_4}{2h_p},
$$
because this is the only way $h_3$ and $h_4$ appear in $\mathcal{F}_1$ and gives you $\frac{1}{2}$ when $h_3=h_4$. So we only need to prove
$$
\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}
{\langle h_1|\phi_2(1)|h_p\rangle}=h_p+h_2-h_1.
$$
Commute $L_{-1}$ over $\phi_2$ we can get
$$
\langle h_1|\phi_2(1)L_{-1}|h_p\rangle=
-\langle h_1|\partial\phi_2(1)|h_p\rangle
$$
where the $\partial\phi_2(1)$ means there is a field $\partial\phi_2(z)$ whose value at $z=1$ is $\partial\phi_2(1)$.
Consult the 3-point function in (5.26)
$$
\langle h_1|\phi_2(z)|h_p\rangle\propto\frac{1}{z^{h_p+h_2-h_1}},
$$
we immediately know
$$
\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}
{\langle h_1|\phi_2(1)|h_p\rangle}
=-\left.\frac{\partial_z\langle h_1|\phi_2(z)|h_p\rangle}
{\langle h_1|\phi_2(1)|h_p\rangle}\right|_{z=1}=h_p+h_2-h_1.
$$
