Question

If y=(x-1)(x-2)(x-3)..,(x-2019)(x-2020) hen dy/dx at x=2020

Answer 1

$\displaystyle\Large\boxed {\sf {\quad2019!\quad}}$

Consider,

$\displaystyle\longrightarrow\sf{y(x)=(x-1)(x-2)(x-3)\dots(x-n)}$

$\displaystyle\longrightarrow\sf{y(x)=\prod_{k=1}^n(x-k)}$

So that for $\displaystyle\sf {x=n,}$

$\displaystyle\longrightarrow\sf{y(n)=0}$

The derivative of $\displaystyle\sf {y}$ wrt $\displaystyle\sf {x}$ is, by product rule,

$\displaystyle\begin{aligned}\longrightarrow\sf{y'(x)}&=\sf{[(x-2)(x-3)\dots(x-n)]}\\&+\sf{[(x-1)(x-3)\dots(x-n)]}\\&+\sf{\,\dots\dots\,}\\&+\sf{[(x-1)(x-2)\dots(x-n+2)(x-n)]}\\&+\sf{[(x-1)(x-2)\dots(x-n+2)(x-n+1)]}\end{aligned}$

$\displaystyle\longrightarrow\sf{y'(x)=\dfrac {y(x)}{x-1}+\dfrac {y(x)}{x-2}+\dots+\dfrac {y(x)}{x-n}}$

$\displaystyle\longrightarrow\sf{y'(x)=\sum_{k=1}^n\dfrac {y(x)}{x-k}}$

Then for $\displaystyle\sf {x=n,}$

$\displaystyle\longrightarrow\sf{y'(n)=\sum_{k=1}^n\dfrac {y(n)}{n-k}}$

$\displaystyle\longrightarrow\sf{y'(n)=\sum_{k=1}^n\dfrac {0}{n-k}}$

$\displaystyle\longrightarrow\sf{y'(n)=\dfrac {0}{n-1}+\dfrac {0}{n-2}+\dots\,+\dfrac {0}{n-n}}$

Then the derivative leaves us with,

$\displaystyle\longrightarrow\sf{y'(n)=\dfrac {y(n)}{n-n}}$

Because each other terms are removed since the numerator is zero, but this one is not cancelled as it shows indeterminate form.

So it's better to write,

$\displaystyle\longrightarrow\sf{y'(n)=\lim_{x\to n}\dfrac {y(x)}{x-n}}$

Again,

$\displaystyle\longrightarrow\sf{y'(n)=\lim_{x\to n}\dfrac {(x-1)(x-2)(x-3)\dots(x-n+1)(x-n)}{x-n}}$

$\displaystyle\longrightarrow\sf{y'(n)=\lim_{x\to n}(x-1)(x-2)(x-3)\dots(x-n+1)}$

$\displaystyle\longrightarrow\sf{y'(n)=(n-1)(n-2)(n-3)\dots(n-n+1)}$

$\displaystyle\longrightarrow\sf{y'(n)=(n-1)!}$

Thus we got a formula!

$\displaystyle\Large\boxed {\sf {y(x)=\prod_{k=1}^n(x-k)\ \implies\ y'(n)=(n-1)!}}$

Let $\displaystyle\sf {n=2020.}$ Then,

$\displaystyle\longrightarrow\sf{y(x)=(x-1)(x-2)\dots(x-2020)}$

Therefore,

$\displaystyle\longrightarrow\sf {\underline {\underline {y'(2020)=2019!}}}$

Answer 2

Given:

y=(x-1)(x-2)(x-3)..,(x-2019)(x-2020)

To Find:

• $\bf{\dfrac{dy}{dx}\:at\:x=2020}$

Product Rule of Differentiation:

$\sf{\dfrac{d}{dx}(a.b.---.z)=\bigg(\dfrac{d}{dx}(a)\bigg)(b--z)+\bigg(\dfrac{d}{dx}(b)\bigg)(a.c--z)---\bigg(\dfrac{d}{dx}(z)\bigg)(a--y)}$

Solution:

So,

$\sf{\dfrac{dy}{dx}=\bigg(\dfrac{d}{dx}(x-1)\bigg)(x-2)..(x-2020)+..+\bigg(\dfrac{d}{dx}(x-2020)\bigg)(x-1)..(x-2019)}$

$\sf{\dfrac{dy}{dx}=(1)(x-2)..(x-2020)+..+(1)(x-1)..(x-2019)}$

$\sf{\dfrac{dy}{dx}=(x-2)..(x-2020)+..+(x-1)..(x-2019)}$

On putting x=2020 in above equation, all terms excluding last term will vanish

Because all terms except last term are having term (x-2020) in product, which on putting x=2020 will become (2020-2020=0).

That means only last term will left

Now,

$\sf{\dfrac{dy}{dx}=(2020-1)(2020-2)...(2020-2018)(2020-2019)}$

On reversing the order, we get

$\sf{\dfrac{dy}{dx}=(2020-2019)(2020-2018)...(2020-2)(2020-1)}$

$\sf{\dfrac{dy}{dx}=(1)(2)...(2018)(2019)}$

$\sf\pink{\dfrac{dy}{dx}=2019!}$

Hence, value $\sf{\dfrac{dy}{dx}\:at\:x=2020}$ is 2019! or factorial of 2019.