A question about the higher-order Weyl variation for the geodesic distance I have a question in deriving Eqs. (3.6.15b) and (3.6.15c) in Polchinski's string theory vol I p. 105. 
Given 

$$\Delta (\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2 (\sigma, \sigma') \tag{3.6.6}$$
  where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.

It is said for the Weyl variation $\delta_W$,

$$\partial_a \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1}{2} \alpha' \partial_a \delta \omega(\sigma) \tag{3.6.15a} $$
  $$ \partial_a \partial_b' \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1+ \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15b} $$
  $$ \nabla_a \partial_b \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= - \frac{  \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15c} $$
  Here $\gamma=-\frac{2}{3}$.

I know the reason for the factor of $1/2$ in the RHS of Eq. (3.6.15a) and I can derive it. But, putting $\nabla$ on LHS and RHS of Eq. (3.6.15a) will not lead to (3.6.15c). I guess it is because there is some higher-order terms in the geodesic distance where the treatment of (3.6.9)-(3.6.11) omitted. But I don't know how to get it. I tried
$ds^2= g_{ab} dx^a dx^b= (\eta_{ab} + h_{ab})dx^a dx^b $ to separate the deviation of Minkowski metric, but I cannot get the factor $\gamma$ 
My question is, how to derive Eqs. (3.6.15b) and (3.6.15c) (especially there is a prime, $'$, namely $\partial'_b$ in the LHS of (3.6.15b) but not in the RHS) 
 A: Assuming the distance $d$ ($z$ is a formal notation) is:
$$d(\sigma, \sigma') = \int_\sigma^{\sigma'} dz ~~e^{w(z)} \tag{1}$$
We have, developping at order 2, $w(z)$:
$$d(\sigma, \sigma') = e^{w(\sigma)} \int_\sigma^{\sigma'} dz ~ e^{(z-\sigma).\partial w(\sigma) + \frac{1}{2}(z-\sigma)^a(z-\sigma)^b \partial_a \partial_b w(\sigma)}\tag{2}$$
That is :
$$d(\sigma, \sigma') = e^{w(\sigma)} \int_\sigma^{\sigma'} dz ~ ( 1 + (z-\sigma).\partial w(\sigma) + \frac{1}{2}(z-\sigma)^a(z-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)) \tag{3}$$
where $Q(w)$ represents quadratic quantities in $\partial_a w \partial_b w$. 
Because the metrics is diagonal, Christophel symbol are of king $\partial_x w$, so the difference between $\nabla_x \partial_y w$ and  $\partial_x \partial_y w$ are precisely these quadratic quantities. We can only choose a inertial frame, so that $\partial_x w=0$, so we can neglect theses quantities $Q$, and working with standard derivatives (more details at the end of the answer).
After formal integration, we get :
$$d(\sigma, \sigma') = e^{w(\sigma)} |\sigma' - \sigma|(1 + \frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) +  \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_{ab}w(\sigma)+Q(w))\tag{4}$$
Reexpressing with an exponential, we have:
$$d(\sigma, \sigma') = e^{w(\sigma)} |\sigma' - \sigma| e^{\frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) +  \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)} \tag{5}$$
So, now, we get, with $\Delta (\sigma, \sigma') = \frac{\alpha'}{2} \ln d^2(\sigma, \sigma')$:
$$\Delta (\sigma, \sigma') =\alpha'(w(\sigma) + \frac{1}{2}(\sigma'-\sigma).\partial w(\sigma) +  \frac{1}{6}(\sigma'-\sigma)^a(\sigma'-\sigma)^b\partial_a \partial_bw(\sigma)+Q(w)+F(\sigma, \sigma'))\tag{6}$$
where $F(\sigma, \sigma')$ is a function of $\sigma, \sigma'$ which does not depend on $w$, so we don't care by looking at variations of $\Delta$ relatively to $w$. Choosing a inertial frame means that we don't care about the quadratic quantities $Q$ too.
We have an other problem, because here $\Delta (\sigma, \sigma')$ is not symmetric in $\sigma, \sigma'$, so we need to symmetrise it, so finally the relevant part of $\Delta$ is :
$$\Delta (\sigma, \sigma') =\alpha'( \frac{1}{2} [w(\sigma) + w(\sigma') ] + \frac{1}{4}(\sigma'-\sigma).[\partial w(\sigma) - \partial^{'} w(\sigma')] +  \frac{1}{12}(\sigma'-\sigma)^a(\sigma'-\sigma)^b[\partial_a \partial_bw(\sigma)+\partial^{'}_a \partial_b^{'}w(\sigma')])\tag{6'}$$ 
So, taking variations, and  derive relatively to $\sigma'_b$:
$$\partial^{'}_b \delta_W \Delta (\sigma, \sigma') =\alpha'( \frac{1}{2}\partial^{'}_b \delta w(\sigma') +  \frac{1}{4}[\partial_b w(\sigma) - \partial^{'}_b w(\sigma')] \\ - \frac{1}{4}(\sigma' - \sigma)^a \partial'_a \partial_b'w(\sigma')+ \frac{1}{6}(\sigma' - \sigma)^a[\partial_a \partial_bw(\sigma)+\partial^{'}_a \partial_b^{'}w(\sigma')] + O(\sigma - \sigma')^2\tag{7}$$
Then, we derive relatively to $\sigma_a$:
$$\partial_a \partial^{'}_b \delta_W \Delta (\sigma, \sigma') =\alpha'( \frac{1}{4}[\partial_a \partial_b \delta w(\sigma)+ \partial^{'}_a \partial^{'}_b \delta w(\sigma')] -  \frac{1}{6}[\partial_a \partial_b \delta w(\sigma)+ \partial^{'}_a \partial^{'}_b \delta w(\sigma')]+ O(\sigma - \sigma')\tag{8}$$
So, finally, when $\sigma' \rightarrow \sigma$, we have: 
$$\partial_a \partial^{'}_b \delta_W \Delta (\sigma, \sigma')= \alpha' \frac{1}{6}\partial_a \partial_b\delta w(\sigma)\tag{9}$$
This is precisely the expression (3.6.15b), remembering that we calculate in a intertial frame, so $\nabla_x \partial_y w = \partial_x \partial_y w $
The same method, beginning from $(6')$ and applying two derivates $\partial_a \partial_b$ gives $3.16.15c$ (a $\frac{1}{3}$ term)
[REMARK]
The quadratic quantities $Q$  could be written : 
$$(\sigma'-\sigma)^a(\sigma'-\sigma)^b (\partial_a w)(\partial_b w)$$
When you derive 2 times this expression, you get :
$$ O(\sigma'-\sigma) + O(\partial_a w)$$
So, when $\sigma' \rightarrow \sigma$ and when we are in a inertial frame ($\partial_a w=0$), these quantities are not relevant.
