Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 178250 by zaheen last updated on 14/Oct/22

how is the solution of this qustion f(x)=x(x−1)(x−2)(x−3)(x−4)∙.......∙(x−100) f^′ (x)=? f′(1)=?

$${how}\:{is}\:{the}\:{solution}\:{of}\:{this}\:{qustion} \\ $$$${f}\left({x}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\centerdot.......\centerdot\left({x}−\mathrm{100}\right) \\ $$$${f}^{'} \left({x}\right)=?\:\:\:\:\:\:\:\:\:\:{f}'\left(\mathrm{1}\right)=? \\ $$$$\:\: \\ $$

Answered by CElcedricjunior last updated on 14/Oct/22

f(x)=x(x−1)(x−2)(x−3)....(x−n) f(x)=Π_(k=1) ^(100) x(x−k)=x(x−100)! f′(x)=((x(x−1)(x−2)....(x−n))/x)+ ((x(x−1)(x−2)...(x−n))/(x−1))+((x(x−1)...(x−n))/(x−2)) +.....+((x(x−1)(x−2)...(x−100))/(x−100)) f′(x)=Σ_(k=0) ^(100) ((x(x−100)!)/(x−k)) f′(1)=1×(−1)×(−2)×.....(1−100) f′(1)=−99! ........ ....le celebre cedric junior.......

$${f}\left({x}\right)=\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)\left(\boldsymbol{{x}}−\mathrm{3}\right)....\left(\boldsymbol{{x}}−\boldsymbol{{n}}\right) \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{100}} {\prod}}\boldsymbol{{x}}\left(\boldsymbol{{x}}−\boldsymbol{{k}}\right)=\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{100}\right)! \\ $$$$\boldsymbol{{f}}'\left(\boldsymbol{{x}}\right)=\frac{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)....\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{n}}\right)}{\boldsymbol{\mathrm{x}}}+ \\ $$$$\frac{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)...\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{n}}\right)}{\boldsymbol{\mathrm{x}}−\mathrm{1}}+\frac{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)...\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{n}}\right)}{\boldsymbol{\mathrm{x}}−\mathrm{2}} \\ $$$$+.....+\frac{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)...\left(\boldsymbol{\mathrm{x}}−\mathrm{100}\right)}{\boldsymbol{\mathrm{x}}−\mathrm{100}} \\ $$$$\boldsymbol{{f}}'\left(\boldsymbol{{x}}\right)=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}\frac{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}−\mathrm{100}\right)!}{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{k}}} \\ $$$$\boldsymbol{\mathrm{f}}'\left(\mathrm{1}\right)=\mathrm{1}×\left(−\mathrm{1}\right)×\left(−\mathrm{2}\right)×.....\left(\mathrm{1}−\mathrm{100}\right) \\ $$$$\boldsymbol{{f}}'\left(\mathrm{1}\right)=−\mathrm{99}! \\ $$$$........\:....{le}\:{celebre}\:\:{cedric}\:{junior}....... \\ $$

Answered by Sheshdevsahu last updated on 14/Oct/22

f′(x)={1.(x−1)(x−2).....(x−100)} + {x.1.(x−2)....(x−100)} +....... {x(x−1)(x−2)....(x−99).1} f′(1)=(0) + {1.1(1−2)(1−3)....(1−100)}+.....(0)....+(0) f′(1)=1(−1)(−2)(−3).....(−98)(−99) f′(1)= −99! (Ans) f′(x)=((x!)/x)+((x!)/(x−1))+.......((x!)/(x−100))

$${f}'\left({x}\right)=\left\{\mathrm{1}.\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right).....\left({x}−\mathrm{100}\right)\right\}\:+\:\left\{{x}.\mathrm{1}.\left({x}−\mathrm{2}\right)....\left({x}−\mathrm{100}\right)\right\}\:+.......\:\left\{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)....\left({x}−\mathrm{99}\right).\mathrm{1}\right\} \\ $$$${f}'\left(\mathrm{1}\right)=\left(\mathrm{0}\right)\:+\:\left\{\mathrm{1}.\mathrm{1}\left(\mathrm{1}−\mathrm{2}\right)\left(\mathrm{1}−\mathrm{3}\right)....\left(\mathrm{1}−\mathrm{100}\right)\right\}+.....\left(\mathrm{0}\right)....+\left(\mathrm{0}\right) \\ $$$${f}'\left(\mathrm{1}\right)=\mathrm{1}\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right).....\left(−\mathrm{98}\right)\left(−\mathrm{99}\right) \\ $$$${f}'\left(\mathrm{1}\right)=\:−\mathrm{99}!\:\left({Ans}\right) \\ $$$$ \\ $$$${f}'\left({x}\right)=\frac{{x}!}{{x}}+\frac{{x}!}{{x}−\mathrm{1}}+.......\frac{{x}!}{{x}−\mathrm{100}} \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com