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SwitchingThese are zeros of $g'(x)$, where
$$g(x) :=\frac{(x+n_1)\cdots(x+n_k)}{x^{k-1}}.$$
Or, switching to $y:=\frac1x$, the problem reducesthey correspond to finding zeroes of $f'(y)$, where
$$f(y) := \frac{(1+n_1y)\cdots(1+n_ky)}y.$$
Switching to $y:=\frac1x$, the problem reduces to finding zeroes of $f'(y)$, where
$$f(y) := \frac{(1+n_1y)\cdots(1+n_ky)}y.$$
These are zeros of $g'(x)$, where
$$g(x) :=\frac{(x+n_1)\cdots(x+n_k)}{x^{k-1}}.$$
Or, switching to $y:=\frac1x$, they correspond to zeroes of $f'(y)$, where
$$f(y) := \frac{(1+n_1y)\cdots(1+n_ky)}y.$$
