Identity of Operator Product Expansion (OPE) I have one more s****d question in Polchinski's string theory book, Eqs. (2.3.14a)
$$ j^{\mu}(z) :e^{ik \cdot X(0,0)}:~ \sim~  \frac{k^{\mu}}{2 z} :e^{ik \cdot X(0,0)}:,$$
where $j^{\mu}_a =\frac{i}{\alpha'} \partial_a X^{\mu}$, $::$ is normal ordered, defined as 
$$:X^{\mu}(z,\bar{z}): = X^{\mu} (z,\bar{z})$$
$$:X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2): = X^{\mu}(z_1,\bar{z}_1) X^{\nu}(z_2,\bar{z}_2) + \frac{\alpha'}{2} \eta^{\mu \nu} \ln |z_{12}|^2 $$. $\sim$ means equal up to nonsingular terms.
I thought I have derived it in analogy of integration by part, but it turns out that not as I thought. Actually how to derive Eq. (2.3.14a)? Eq.(2.3.14b) will be expected in analogous...
 A: We can use
\begin{equation}
\begin{split}
: F : : G: = \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)}  \right) :F G:
\end{split}
\end{equation}
This gives
\begin{equation}
\begin{split}
: \frac{i}{\alpha'} \partial X^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: &= \exp \left( - \frac{\alpha'}{2} \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\frac{\delta }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta }{\delta X_{G\mu}(z_2, {\bar z}_2)}  \right) \\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~: \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}: \\
&= : \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}:   \\
&~~~~~~~~~~~~~~~~  - \frac{i}{2} : \int d^2 z_1 d^2 z_2\log|z_{12}|^2 \frac{\delta( \partial X^\mu(z) ) }{\delta X_F^\mu(z_1, {\bar z}_1)} \frac{\delta  (  e^{i k \cdot X(w,{\bar w})})  }{\delta X_{G\mu}(z_2, {\bar z}_2)}: \\
&= : \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}:   \\
&~~~~~ - \frac{i}{2} : \int d^2 z_1 d^2 z_2 \log|z_{12}|^2\partial \left( \delta^\mu{}_\alpha \delta^2(z_1, z) \right) i k^\alpha \delta^2(z_2, w)  e^{i k \cdot X(w,{\bar w})}  \\
&= : \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}:   \\
&~~~~~ +\frac{k^\mu}{2} : \partial \left(  \int d^2 z_1 d^2 z_2 \log|z_{12}|^2 \delta^2(z_1, z) \delta^2(z_2, w)  e^{i k \cdot X(w,{\bar w})}\right)  : \\
&= : \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}:  +\frac{k^\mu}{2} : \partial \left(  \log|z-w|^2 e^{i k \cdot X(w,{\bar w})}\right)  : \\
&= : \frac{i}{\alpha'} \partial X^\mu(z)   e^{i k \cdot X(w,{\bar w})}:  + \frac{k^\mu}{2(z-w)} :  e^{i k \cdot X(w,{\bar w})} :  \\
\end{split}
\end{equation}
Therefore
\begin{equation}
\begin{split}
: j^\mu(z) : : e^{i k \cdot X(w,{\bar w})}: \sim   \frac{k^\mu}{2(z-w)} :  e^{i k \cdot X(w,{\bar w})} :
\end{split}
\end{equation}
A: An other way is to begin with (2.2.1.4)
$$:e^{i k_1.X(z,\bar z)}:~ :e^{i k_2.X(0,0)}: ~\sim ~|z|^{\alpha' k_1.k_2} ~:e^{i (k_1 + k_2).X(0,0)}:$$
Now, derive this expression relatively to $k_1^\mu$, then doing $k_1=0$, we get :
$$:i X_\mu(z, \bar z):~:e^{i k_2.X(0,0)}: ~\sim(\alpha' (k_2)_{\mu} \ln|z| + i :X_\mu(0,0):):e^{i k_2.X(0,0)}:$$ 
Now, we derive relatively to $z$ and divide by $\alpha'$, and we get : 
$$: \frac{i}{\alpha'} \partial _zX_\mu(z, \bar z):~:e^{i k_2.X(0,0)}: ~\sim \frac{k_2^\mu}{2z}  :e^{i k_2.X(0,0)}:$$
