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. $(n-1)!$
• C. ${(-1)}^n(n-1)!$
• D. ${(-1)}^{n}n!$
• E. $(n+1)!$

Answer 1

Answer: (C)

$y=\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)....\left(1+\frac{n}{x}\right)$ $\frac{dy}{dx}=\left(-\frac{1}{x^2}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)....\left(1+\frac{n}{x}\right)+$ $\left(1+\frac{1}{x}\right)\left(-\frac{2}{x^2}\right)\left(1+\frac{3}{x}\right).......\left(1+\frac{n}{x}\right)+$ $.....+\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right).....\left(-\frac{n}{x^2}\right)$ $\therefore$ ${\frac{dy}{dx}|}_{x=-1}=(-1)(-1)(-2)(-3).....(1-n)$ $={(-1)}^n(1)(2)(3)....(n-1)$ $={(-1)}^n(n-1)!$