Does differentiation and (conformal) normal ordering commute? In Polchinski's String theory Volume I, chapter 2, he writes down the following OPE, 
$$
X^\mu (z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2) = -\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z_{12}|^2 + :X^\nu X^\mu(z_2,\bar{z}_2): + \sum_{k=1}^{\infty}\frac{1}{k!} \bigg[ (z_{12})^k :X^\nu\partial^kX^\mu(z_2,\bar{z}_2): + \ (\bar{z}_{12})^k :X^\nu\bar{\partial}^kX^\mu(z_2,\bar{z}_2): \bigg]
$$
which he gets by taylor expanding $:X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2): $ since it can be written as a sum of holomorphic and antiholomorphic functions (since it is harmonic in $(z_1,\bar{z}_1)$). 
But if we call the harmonic function $\ f(z_1,\bar{z}_1;z_2,\bar{z}_2) = :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$, and taylor expand $\ f$ in $(z_1,\bar{z}_1)$ around $(z_1,\bar{z}_1) = (z_2,\bar{z}_2)$, then we get terms that go like ~ $\partial^k f$ ~ $\partial^k :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$ evaluated at $(z_2,\bar{z}_2)$. Here the derivatives are outside the normal ordered expression while in Polchinski's expression, they are inside the normal ordered expression. Why are they equal? What am I misunderstanding?
Reference for the expression : Eq. 2.2.4 Pg 38, Polchinski Chapter 2 (hardcover) (2004 reprint I think). 
Edit 1: $z_{12} = z_1 - z_2$
Edit 2: All partial derivatives are with respect to $z_1,\bar{z}_1$.
 A: Yes, differentiation and normal ordering commute. You can prove this using the definition (2.2.7) of the normal ordering: 
$$:\mathcal{F}: = \exp \left( \frac{\alpha '}{4} \int d^2z_1 d^2z_2 \ln |z_{12}|^2 \frac{\delta}{\delta X^{\mu}(z_1 , \bar{z}_1)} \frac{\delta}{\delta X^{\nu}(z_2 , \bar{z}_2)} \right) \mathcal{F}$$
Using this expression, the commutativity of the normal order and differentiation boils down to commutativity of functional derivative and differentiation. Let us consider any functional $F$ of $X^{\mu}(z )$. Then 
$$\frac{\delta F[X^{\mu}(z )]}{\delta X^{\mu}(z_1)} = \lim\limits_{\epsilon \rightarrow 0} \frac{F[X^{\mu}(z) + \epsilon \delta(z-z_1) ]-F[X^{\mu}(z)]}{\epsilon} $$ 
(see for instance eq A.28 here). 
Then apply this for $F[X^{\mu}(z , \bar{z})] = \partial X^{\mu}(z)$: 
$$\frac{\delta \partial X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta ' (z-z_1)$$
On the other hand, 
$$\partial \frac{\delta  X^{\mu}(z)}{\delta X^{\mu}(z_1)} =  \delta ' (z-z_1)$$ 
since $\frac{\delta  X^{\mu}(z)}{\delta X^{\mu}(z_1)} =  \delta (z-z_1)$. The two expressions are equal, which proves the claim. 
