expression for the pressure due to a molecule in state number $i$ $$P_{i}=-\frac{d \varepsilon_{i}}{d V}$$
To find pressure
$$\begin{aligned} P=\frac{N}{z} \sum_{i} P_{i} e^{-\varepsilon_{i} / k_{\mathrm{B}} T} &=-\frac{N}{z} \sum_{i}\left(\frac{d \varepsilon_{i}}{d V}\right) e^{-\varepsilon_{i} / k_{\mathrm{B}} T} \\ &=\frac{N k_{\mathrm{B}} T}{x} \sum_{i}\left(\frac{\partial}{\partial V} e^{-\varepsilon_{i} / k_{\mathrm{B}} T}\right)_{T} \quad \text{Eq.1} \end{aligned}$$
I dont understand how you come to the second line from the first line of Eq.1. I know there is somehow $\frac{\partial}{\partial \varepsilon_{i} } e^{-\varepsilon_{i} / k_{\mathrm{B}} T}$. A step by step derivation from 1st line to second line would be really helpful
This is just chain rule, $\varepsilon_i$ is a function of $V$
$$\begin{aligned} \frac{\partial}{\partial V} (e^{-\varepsilon_{i} / k_{\mathrm{B}} T}) = - \frac{1} {k_{\mathrm{B}} T}e^{-\varepsilon_{i} / k_{\mathrm{B}} T}\left(\frac{d \varepsilon_{i}}{d V}\right) \end{aligned}$$ therefore $$\begin{aligned} - e^{-\varepsilon_{i} / k_{\mathrm{B}} T}\left(\frac{d \varepsilon_{i}}{d V}\right) = k_{\mathrm{B}} T \left( \frac{\partial}{\partial V} (e^{-\varepsilon_{i} / k_{\mathrm{B}} T}) \right) \end{aligned}.$$