If y = 1 + 1 x 1 + 2 x 1 + 3 x ⋯ ⋯ ⋯ 1 + n x and x ≠ 0 . When x = - 1 , d y d x =

Taking log on both sides we get

$\log y=\log \left(1+\frac{1}{x}\right)+\log \left(1+\frac{2}{x}\right).............\log \left(1+\frac{n}{x}\right)$

Differentiating,

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{\left(1+{\displaystyle \frac{1}{x}}\right)}\frac{-1}{x^2}+\frac{1}{\left(1+{\displaystyle \frac{2}{x}}\right)}\frac{-2}{x^2}...........\frac{1}{\left(1+{\displaystyle \frac{1}{n}}\right)}\frac{-n}{x^2}$

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

Now put $x=-1$

$\frac{dy}{dx}=\frac{\left(-1\right)(\left(-1\right)\left(-2\right)\left(-3\right)......\left(1-n\right))}{1}$

$\frac{dy}{dx}={\left(-1\right)}^n\left(1\right)\left(2\right).....\left(n-1\right)={\left(-1\right)}^n\left(n-1\right)!$