The derivative of the detector response with respect to the particle energy is
\[ \frac{dL}{dE_0} = S(E_0),\]and Birks' law can be linearized as
\[ \left( \frac{dL}{dE_0} \right)^{-1} = \frac{1+ kB \cdot \frac{dE}{dx}}{S_0}\]The data for the dependence $L(E_0)$ are not available in large quantity, so numerical differentiation is not feasible. As an alternative, we approximate the function $L(E_0)$
\[L = a E_0^4 + b E_0^3 + c E_0^2 + d E_0\]so that
\[\frac{dL}{dE_0} = 4 aE_0^3 + 3b E_0^2 + c E_0 + d \tag{1}\]

The fitting results are:
\[
\begin{cases}
a=5.218 \cdot 10^{-3} \text{MeV}^{-4} \\
b=-2.185 \cdot 10^{-1} \text{MeV}^{-3} \\
c=3.810 \text{MeV}^{-2} \\
d=14.62 \text{MeV}^{-1}
\end{cases}
\]

Now let us plot the linearized graph by calculating the values of $dL/dE$ using the equation (1). The value of $kB$ is the ratio of the slope to the offset.