If $y = \left( 1 + \frac{1}{x} ight)\left( 1 + \frac{2}{x} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( 1 + \frac{n}{x} ight)$ and $x
eq 0,\ then\frac{dy}{dx}\ when\ x = - 1\ is\$

Question

If $y = \left( 1 + \frac{1}{x} ight)\left( 1 + \frac{2}{x} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( 1 + \frac{n}{x} ight)$ and $x 
eq 0,\ then\frac{dy}{dx}\ when\ x = - 1\ is\$

Dr. Abubacker · Accepted Answer

(c)
$y = \left( 1 + \frac{1}{x} ight)\left( 1 + \frac{2}{x} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( 1 + \frac{n}{x} ight)$
On differentiating w.r.t. $x,\$we get
$\ \frac{dy}{dx} = \left( - \frac{1}{x^{2}} ight)\left( 1 + \frac{2}{x} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( 1 + \frac{n}{x} ight)$
$+ \left( 1 + \frac{1}{x} ight)\left( - \frac{2}{x^{2}} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( 1 + \frac{n}{x} ight)$
$+ \ldots + \left( 1 + \frac{1}{x} ight)\left( 1 + \frac{2}{x} ight)\left( 1 + \frac{3}{x} ight)\ldots\left( - \frac{n}{x^{2}} ight)$
$	herefore\ \ \ \ \ \ \left. \ \frac{dy}{dx} ight|_{x = - 1} = ( - 1)( - 1)( - 2)( - 3)\ldots(1 - n)$
$= ( - 1)^{n}(1)(2)(3)\ldots(n - 1)$
$= ( - 1)^{n}(n - 1)!\ \$

Q. If $y = \left( 1 + \frac{1}{x} \right)\left( 1 + \frac{2}{x} \right)\left( 1 + \frac{3}{x} \right)\ldots\left( 1 + \frac{n}{x} \right)$ and $x \neq 0,\ then\frac{dy}{dx}\ when\ x = - 1\ is\$

Explanation

Reviewed by Dr. Abubacker

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