Question

What is the derivative of $y=(1-x)(2-x)....(n-x)$ at $x=1$?

Answer 1

Find the derivative.

$y=(1-x)(2-x)....(n-x)$

We know that

$\frac{d\left(uvw\right)}{dx}=u\text{'}vw+uv\text{'}w+uvw\text{'}$

Now,

$\frac{dy}{dx}=-1(2–x)(3–x)\dots (n–x)+(1–x)(-1)(3–x)\dots (n–x)+\dots .+(n–1–x)(-1)$

At $1$

$$\begin{gathered} \frac{dy}{dx}=-1(2–1)(3–1)\dots (n–1)+(1–1)(-1)(3–1)\dots (n–1)+\dots .+(n–1–1)(-1) \\ =-1(2–1)(3–1)\dots (n–1) \\ =-1(n–1)! \end{gathered}$$

Hence, the derivative of $y=(1-x)(2-x)....(n-x)$ at $x=1$is $-1\left(n-1\right)!$.