Clean form for derivative of symmetric polynomial?

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Let $m_n(x_1^M)$ be the elementary symmetric polynomial, the sum of all distinct products of $n$ distinct variables $x_i$, where $n \ll M$. That is to say, $m_n(x_1^M) = \sum_{j_1 \le \cdots \le j_n} x_{j_1}x_{j_2}\cdots x_{j_n}$. I want to find a clean form for $\frac{d}{dx_i}\left(\frac{m_n(x_1^M)}{m_{n-1}(x_1^M)}\right)$ for some fixed $x_i$. So far, I derived that the derivative of one elementary symmetric polynomial is given by $\frac{d\left(m_n(x_1^M)\right)}{dx_i} = \sum_{S \in \binom{J_i}{n-1}}\left(\prod_{s\in S} \lambda_s\right)$ for $J_i = \{ j_k \in \{j_1, \cdots, j_n\} : j_k > i \}$. Intuitively, this represents how the derivative is just the sum of products of the terms that contain $x_i$. I wonder if there is a more interpretable/cleaner form for this derivative, or for the derivative of the quotient $\frac{d}{dx_i}\left(\frac{m_n(x_1^M)}{m_{n-1}(x_1^M)}\right)$, which is what I'm really interested in to begin with.

EDIT: the first comment makes it clear that $\frac{d}{dx_i}\left(m_n(x_1^M)\right) = m_{n-1}(x_{1\neq i}^M)$, but my question about the quotient still stands.

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Assuming $M$ is the number of variables $\,x_1,x_2,\ldots,x_M$, and with the convention that $\,m_0 \equiv 1\,$, by the definition of the elementary symmetric polynomials:

$$ m_n(x^M) = m_{n}(x^{M-1}) + x_M \cdot m_{n-1}(x^{M-1}) $$

Let $\,\displaystyle\frac{\operatorname{d}f}{\operatorname{d} x_M}{} = f'\,$ to simplify the notations, then the above implies:

$$ m_n'(x^M) = m_{n-1}(x^{M-1}) $$

It follows that:

$$ \require{cancel} \begin{align} \left(\frac{m_n(x^M)}{m_{n-1}(x^M)}\right)' &= \frac{m'_n(x^M)m_{n-1}(x^M) - m_n(x^M)m'_{n-1}(x^M)}{m^2_{n-1}(x^M)} \\ &= \frac{m_{n-1}(x^{M-1})m_{n-1}(x^M) - m_n(x^M)m_{n-2}(x^{M-1})}{m^2_{n-1}(x^M)} \\ &= \frac{1}{m^2_{n-1}(x^M)} \cdot \Bigg(m_{n-1}(x^{M-1})\big(m_{n-1}(x^{M-1}) + \cancel{x_M \cdot m_{n-2}(x^{M-1})}\big) - \big(m_n(x^{M-1})+ \cancel{x_M \cdot m_{n-1}(x^{M-1})}\big)m_{n-2}(x^{M-1})\Bigg) \\ &= \frac{m^2_{n-1}(x^{M-1}) - m_n(x^{M-1})m_{n-2}(x^{M-1})}{m^2_{n-1}(x^M)} \end{align} $$

The latter is a rational function in $\,x_M\,$ of the form $\,\dfrac{a}{(b \cdot x_M + c)^2}\,$ where $\,a,b,c\,$ do not depend on $\,x_M\,$, since the symmetric polynomials in $\,x^{M-1}\,$ do not include $\,x_M\,$.