I am trying to understand the last few lines of the proof of Lemma 4.4 in [1]. In particular, a class K function $\alpha:[0,\alpha)\to [0,\infty)$ is continuous, has $\alpha(0) = 0$, and is strictly increasing. From here, the function $\eta$ is defined as $\eta(y) = -\int_b^y \frac{dx}{\alpha(x)}$ for $b\in (0,\alpha)$ and $\sigma$ is defined as
$$ \sigma(r,s) = \begin{cases} \eta^{-1}(\eta(r) + s) & r > 0 \\ 0 & r = 0\end{cases} \tag{1} \label{eq1} $$.
The proof then states that \begin{equation} \frac{\partial}{\partial r}\sigma(r,s) = \frac{\alpha(\sigma(r,s))}{\alpha(r)} > 0 \tag{2} \label{eq2} \end{equation} and \begin{equation} \frac{\partial}{\partial s}\sigma(r,s) = -\alpha(\sigma(r,s)) > 0 \tag{3} \label{eq3} \end{equation}
I am not sure how these partial derivatives are calculated. This works when $\alpha$ is the identity and $b=0$ since in this case $\eta(y) = -\int_1^y \frac{dx}{x} = -\ln y$ and $\eta^{-1}(y) = e^{-y}$ so $\sigma(r,s) = e^{-(-\ln r + s)} = re^{-s}$.
When $\alpha$ is not necessarily the identity, $\eta^{-1}$ seems like it should be $$ \eta^{-1}(y) = \alpha^{-1}\left[\left(-\frac{d}{dy} y\right)^{-1}\right] \tag{4} \label{eq4} $$ so that $\eta^{-1}(\eta(y)) = \alpha^{-1}\left[\left(-\frac{d}{dy}\left(-\int_b^y\frac{dx}{\alpha(x)}\right)\right)^{-1}\right] = \alpha^{-1}\left[\left(\frac{1}{\alpha(y)}\right)^{-1}\right] = y$.
My question is how do I use equation \eqref{eq4} to derive equations \eqref{eq2} and \eqref{eq3} from equation \eqref{eq1}? More broadly, is this even the right approach? An explicit form of $\eta^{-1}$ is not given in the proof so perhaps I am missing something. Thanks!
[1] H. Khalil, Nonlinear Systems. Prentice Hall, 2002.