Question
If $$y = \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right)$$ and $$x \ne 0,$$ then $$\frac{{dy}}{{dx}}$$ when $$x = - 1$$ is:
A.
$$n!$$
B.
$$\left( {n - 1} \right)!$$
C.
$${\left( { - 1} \right)^n}\left( {n - 1} \right)!$$
D.
$${\left( { - 1} \right)^n}n!$$
Answer :
$${\left( { - 1} \right)^n}\left( {n - 1} \right)!$$
Solution :
$$\eqalign{
& y = \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) \cr
& \frac{{dy}}{{dx}} = \left( { - \frac{1}{{{x^2}}}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) + \left( {1 + \frac{1}{x}} \right)\left( { - \frac{2}{{{x^2}}}} \right)\left( {1 + \frac{3}{x}} \right).....\left( {1 + \frac{n}{x}} \right) + ... + \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right).....\left( { - \frac{n}{{{x^2}}}} \right) \cr
& \because \,\,{\left. {\frac{{dy}}{{dx}}} \right|_{x = - 1}} = \left( { - 1} \right)\left( { - 1} \right)\left( { - 2} \right)\left( { - 3} \right).....\left( {1 - n} \right) \cr
& = {\left( { - 1} \right)^n}\left( 1 \right)\left( 2 \right)\left( 3 \right).....\left( {n - 1} \right) \cr
& = {\left( { - 1} \right)^n}\left( {n - 1} \right)! \cr} $$