Question

$\purple{ \rm{If \: y = (x - 1)(x - 2)(x - 3) \dots \dots(x - 2021)}} \\ \red {\rm{then \: find \: \frac{dy}{dx} \bigg|_{x = 2021}}}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = (x - 1)(x - 2)(x - 3) \dots \dots(x - 2021)$

On taking log on both sides, we get

$\rm \: logy = log[(x - 1)(x - 2)(x - 3) \dots \dots(x - 2021)]$

We know,

$\boxed{ \rm{ \:log(xy) = logx + logy \: }}$

So, using this result, we get

$\rm \: logy = log(x - 1) + log(x - 2) + \cdots + log(x - 2021)$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}logy = \dfrac{d}{dx}\bigg[log(x - 1) + log(x - 2) + \cdots + log(x - 2021)\bigg]$

We know,

$\boxed{ \rm{ \:\dfrac{d}{dx}logx \: = \: \frac{1}{x} \: }}$

So, using this result, we get

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{1}{x - 1} + \dfrac{1}{x - 2} + \cdots + \dfrac{1}{x - 2021}$

$\rm \:\dfrac{dy}{dx} = \dfrac{y}{x - 1} + \dfrac{y}{x - 2} + \cdots + \dfrac{y}{x - 2021}$

$\rm \:\dfrac{dy}{dx} = \dfrac{(x - 1)(x - 2)\cdots(x - 2021)}{x - 1} + \dfrac{(x - 1)(x - 2)\cdots(x - 2021)}{x - 2} + \cdots + \dfrac{(x - 1)(x - 2)\cdots(x - 2021)}{x - 2021}$

$\rm \:\dfrac{dy}{dx} =(x - 2)(x - 3)\cdots(x - 2021) + (x - 1)(x - 3)\cdots(x - 2021) + \cdots + (x - 1)(x - 2)\cdots(x - 2020)$

Now,

$\rm \: \dfrac{dy}{dx} \bigg|_{x = 2021}$

$\rm \: =0 + 0 + \cdots + (2021 - 1)(2021 - 2)\cdots(2021 - 2020)$

$\rm \: = \: (2020)(2019)\cdots\cdots(1)$

$\rm \: = \: 2020!$

Hence,

$\color{green} \: \rm\implies \:\dfrac{dy}{dx} \bigg|_{x = 2021} = \: 2020!$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {x}^{n}\\ \\ \sf {nx}^{n - 1} & \sf {e}^{x} \end{array}} \\ \end{gathered}$