Taylor expansion of the metric Consider a coordinate change
$$
x^a\mapsto \tilde x^a=x^a+\epsilon y^a
$$
In the note I am reading, the author calculate the change of metric by
$$
g_{ab}(x) = \tilde g_{ab}(\tilde x)=\tilde g_{ab}(x^a+\epsilon y^a)=\tilde g_{ab}(x^a)+\epsilon\mathcal{L}_Y\tilde g_{ab}(x^a)+\cdots
$$
My question is, why do we use $\mathcal{L}_Y\tilde g$ rather than $\nabla_Y\tilde g$ in the first order term?

Update
Following is my newly try.
As Prahar pointed out, I should write
$$
g_{ab}(x)\mapsto\tilde g_{ab}(\tilde x)=\frac{\partial x^c}{\partial {\tilde x}^a}\frac{\partial x^d}{\partial {\tilde x}^b} g_{cd}(x)
$$
Then we have 
$$
g_{ab}(x)=\tilde g_{ab}(x^a+\epsilon y^a)=\tilde g_{ab}(x^a)+\epsilon\nabla_Y\tilde g_{ab}(x^a)+\cdots
$$
And I have to show
$$
\nabla_Y\tilde g_{ab}(x^a)=\mathcal{L}_Yg_{ab}(x^a)
$$
We have
$$
\begin{align}
\tilde g_{ab}(x^a)&=\frac{\partial x^c}{\partial {\tilde x}^a}\frac{\partial x^d}{\partial {\tilde x}^b} g_{cd}(x^a)\\
&=\left(\delta_a^c-\epsilon\frac{y^c}{\tilde x^a}\right)\left(\delta_b^d-\epsilon\frac{y^d}{\tilde x^b}\right)\left.\right|_{x^a-\epsilon y^a}g_{cd}(x^a-\epsilon y^a)\\
&=g_{ab}(x^a-\epsilon y^a)-\epsilon\left[\frac{y^d}{\tilde x^b}g_{ad}+\frac{y^c}{\tilde x^a}g_{cd}\right]\left.\right|_{x^a-\epsilon y^a}+O(\epsilon^2)\\
&=g_{ab}(x^a) - \epsilon\nabla_Yg_{ab}(x^a)-\epsilon\left[\frac{y^d}{\tilde x^b}g_{ad}+\frac{y^c}{\tilde x^a}g_{cd}\right]\left.\right|_{x^a} +O(\epsilon^2)
\end{align}
$$
So we get
$$
\nabla_Y\tilde g_{ab}(x^a)=\nabla_Yg_{ab}(x^a)-\epsilon\nabla_Y\left(\nabla_Yg_{ab}+\frac{y^d}{\tilde x^b}g_{ad}+\frac{y^c}{\tilde x^a}g_{cd}\right)\left.\right|_{x^a} + O(\epsilon^2)
$$
On the other hand, we have
$$
\mathcal{L}_Yg_{ab}(x^a)=\nabla_Yg_ab{x^a}+g_{ac}\nabla_bY^c+g_{cb}\nabla_aY^c
$$
I cannot see these two are identical. Is there anything wrong in my deduction?
 A: I'm not quite sure what you are doing in your post. It is not true that
$$
g_{ab}(x) = {\tilde g}_{ab}({\tilde x})
$$
The correct equality as I pointed out is
$$
{\tilde g}_{ab}({\tilde x}) = \frac{ \partial x^c}{ \partial{\tilde x}^a} \frac{ \partial x^d}{ \partial{\tilde x}^b} g_{cd}(x)
$$
where ${\tilde x}^a = x^a + \epsilon y^a(x) \implies x^a = {\tilde x}^a - \epsilon y^a(x) $. This implies
$$
\frac{ \partial x^c}{ \partial{\tilde x}^a}  = \delta^c_a - \epsilon \frac{\partial y^c(x)}{\partial {\tilde x}^a }   = \delta^c_a - \epsilon \frac{\partial y^c(x)}{\partial x^b } \frac{\partial x^b}{\partial {\tilde x}^a } = \delta^c_a - \epsilon\partial_a y^c  + {\cal O}(\epsilon^2)
$$
where we have introduced notation
$$
\partial_a y^c \equiv \frac{\partial y^c(x)}{\partial x^a}
$$
We therefore have
$$
{\tilde g}_{ab}({\tilde x}) = {\tilde g}_{ab} ( x  + \epsilon y ) = {\tilde g}_{ab}(x) + \epsilon y^c \partial_c {\tilde g}_{ab}(x) + {\cal O}(\epsilon^2)
$$
From the other side of the equality, we also 
\begin{equation}
\begin{split}
{\tilde g}_{ab}({\tilde x}) &= \left( \delta^c_a - \epsilon\partial_a y^c  + {\cal O}(\epsilon^2) \right) \left( \delta^d_b - \epsilon\partial_b y^d  + {\cal O}(\epsilon^2) \right) g_{cd}(x) \\
&= g_{ab}(x) - \epsilon \left( \partial_a y^c g_{cb} + \partial_b y^d g_{ad}  \right) + {\cal O}(\epsilon^2)
\end{split}
\end{equation}
We therefore have
$$
 {\tilde g}_{ab}(x) + \epsilon y^c \partial_c {\tilde g}_{ab}(x) + {\cal O}(\epsilon^2) = g_{ab}(x) - \epsilon \left( \partial_a y^c g_{cb} + \partial_b y^d g_{ad}  \right) + {\cal O}(\epsilon^2)
$$
and thus
$$
g_{ab}(x) =  {\tilde g}_{ab}(x)  + \epsilon \left(  y^c \partial_c {\tilde g}_{ab}  + \partial_a y^c g_{cb} + \partial_b y^d g_{ad}  \right) + {\cal O}(\epsilon^2)
$$
Finally, since ${\tilde g}_{ab} = g_{ab}+ {\cal O}(\epsilon)$, we have
$$
g_{ab}(x) =  {\tilde g}_{ab}(x)  + \epsilon \left(  y^c \partial_c g_{ab}  + \partial_a y^c g_{cb} + \partial_b y^d g_{ad}  \right) + {\cal O}(\epsilon^2)
$$
The quantity in the bracket above is precisely the Lie derivative of $g_{ab}$. 
