Question

If $y=\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)\dots .\left(1+\frac{n}{x}\right)$ and $x\ne 0$, then $\frac{dy}{dx}$ where $x=-1$ is:

• A. $n!$
• B. $\left(n-1\right)!$
• C. $-1^n\left(n-1\right)!$
• D. $-1^{n}n!$

Answer 1

The correct option is C

$-1^n\left(n-1\right)!$

Explanation for the correct option.

It is given that $y=\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)\dots .\left(1+\frac{n}{x}\right)$.

Differentiating w.r.t $x$, we get

$$\begin{gathered} \frac{dy}{dx}=\left(\frac{-1}{x^2}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)\dots .\left(1+\frac{n}{x}\right)+\left(\frac{-2}{x^2}\right)\left(1+\frac{1}{x}\right)\left(1+\frac{3}{x}\right)\dots .\left(1+\frac{n}{x}\right)+...+\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)...\left(\frac{-n}{x^2}\right)\left[by \\ product \\ rule\right] \end{gathered}$$

By substituting $x=-1$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& \left(-1\right)\left(-1\right)\left(-2\right)\dots .\left(1-n\right)+\left(-2\right)\left(0\right)\left(-2\right)\dots .\left(1-n\right)+...+\left(0\right)\left(-1\right)\left(-2\right)...\left(-n\right)\\ & =& \left(-1\right)\left(-1\right)\left(-2\right)\dots .\left(1-n\right)\left[\mathrm{as}\mathrm{all}\mathrm{other}\mathrm{terms}\mathrm{will}\mathrm{be}0\right]\\ & =& -1^n\left[1·2·3·......·\left(n-1\right)\right]\\ & =& -1^n\left(n-1\right)!\end{array}$

Hence, option C is correct.