Relation between two Riemannain connections Let $g$ be a Riemannian metric on $M$ and let $\tilde{g}=f^{2}g$ where $f$ is a smooth function that is never zero. let $\nabla$ and $\nabla'$ be the Riemannain connections of $g$ and $\tilde{g}$ on $M$, give the relation between $\nabla$ and $\nabla'$.
So I'm not really sure what to do here but I tryed to express each connection in terms of its Christoffel symbols and locall expression for a connection
What I get when I do that is the following  let $\Gamma^{k}_{i,j}$ and $'\Gamma^{k}_{i,j}$ denote the Christoffel of each connections then I get that
$$'\Gamma^{k}_{i,j}=\Gamma^{k}_{i,j}+\dfrac{1}{f}g^{k,l}(g_{i,j}\partial_{i}f+g_{i,l}\partial_{j}f-g_{i,j}\partial_{l}f) $$
Now when I plug this into the local expression for the connection of $\nabla'$ I get 
$$ \nabla'_{X}Y=\nabla_{X}Y + big ~mess$$
Is there a better way of doing this?
 A: The formula for the conformal rescaling of the Levi-Civita connection is an essential tool in Riemannian geometry, and its derivation is given in many sources.
As Isaac Solomon and Ted Shifrin have mentioned in the comments, a slick way to derive it is to consider $f = e^{\omega}$ and use the Koszul formula. The result will be in the form:
$$
\nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1}
$$
Proof. The Koszul formula (see e.g. here) gives the following expression for the Levi-Civita connection $\nabla$ of the metric $g$:
$$ 
\begin{align}
2 g(\nabla_X Y, Z) & = X \, g(Y,Z) + Y \, g(Z,X) - Z \, g(X,Y) \\ \tag{2}
&- g(X,[Y,Z]) +  g(Y,[Z,X]) + g(Z,[X,Y])
\end{align}
$$
Let $\nabla'$ be the Levi-Civita connection for the metric $g' = e^{2\omega}g$. Substituting these objects into (2)
$$ 
\begin{align}
2 e^{2 \omega} g(\nabla'_X Y, Z) & = X \left( e^{2 \omega} g(Y,Z) \right) + Y \left( e^{2 \omega} g(Z,X) \right) - Z \left( e^{2 \omega} g(X,Y) \right) \\ 
&- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y])
\end{align}
$$
and computing the derivatives using the product rule, we obtain
$$ 
\begin{align}
2 e^{2 \omega} g(\nabla'_X Y, Z) & = e^{2 \omega} X  g(Y,Z) + e^{2 \omega} Y  g(Z,X)  - e^{2 \omega} Z g(X,Y) \\ 
& + 2 e^{2 \omega} g(Y,Z) \, X \omega   + 2 e^{2 \omega} g(Z,X) \, Y \omega - 2  e^{2 \omega} g(X,Y) \, Z  \omega \\
&- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y])
\end{align}
$$
In the last display we can divide both sides of the equation by $e^{2 \omega}$, which is a strictly positive function, to get
$$ 
\begin{align}
2 g(\nabla'_X Y, Z) & = X  g(Y,Z) + Y  g(Z,X)  -Z g(X,Y) \\ 
& + 2 g(Y,Z) \, X \omega   + 2 g(Z,X) \, Y \omega - 2  g(X,Y) \, Z  \omega \\
&- g(X,[Y,Z]) + g(Y,[Z,X]) +  g(Z,[X,Y])
\end{align}
$$
Using the Koszul formula (2) again we rewrite the above expression as
$$ 
\begin{align}
2 g(\nabla'_X Y, Z) & = 2 g(\nabla_X Y, Z) + 2 g(Y,Z) \, X \omega   + 2 g(Z,X) \, Y \omega - 2  g(X,Y) \, Z  \omega
\end{align}
$$
which is equivalent to (1) because vector field $Z$ is arbitrary, $g$ is non-degenerate, and $Z \omega = \mathrm{d} \omega (Z)$. Recall also that $\operatorname{grad} \omega = (\mathrm{d} \omega)^{\sharp}$.

The version of this formula in terms of coordinates and the Christoffel symbols is obtained by a similar calculation, the result will be
$$
'\Gamma^{k}_{ij}=\Gamma^{k}_{ij} + \delta_{i}^{k} \partial_j \omega  + \delta_{j}^{k} \partial_i \omega - g_{i j} g^{k l} \partial_{l} \omega
$$
This can be also obtained as a consequence of (1).
