Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda$ in $\mathbb{C}$, let $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X-1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$. Using formula $$\forall B \in \mathbb{C}_{2n-1}[X], \quad B(X) = \sum_{k=1}^{2n} B(\alpha_k) \frac{A(X)}{(X - \alpha_k) A'(\alpha_k)} \tag{I.1}$$ show that $$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n}+1}{X - \omega_k} \omega_k$$ then deduce that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1-\omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1-\omega_k)^2}.$$
Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda$ in $\mathbb{C}$, let $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X-1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$.
Using formula
$$\forall B \in \mathbb{C}_{2n-1}[X], \quad B(X) = \sum_{k=1}^{2n} B(\alpha_k) \frac{A(X)}{(X - \alpha_k) A'(\alpha_k)} \tag{I.1}$$
show that
$$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n}+1}{X - \omega_k} \omega_k$$
then deduce that
$$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1-\omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1-\omega_k)^2}.$$