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Question Number 70891 by Henri Boucatchou last updated on 09/Oct/19

f(x) = x(x−1)(x−2)....(x−10)  f ′(0) = ?

$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:=\:\boldsymbol{{x}}\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{2}\right)....\left(\boldsymbol{{x}}−\mathrm{10}\right) \\ $$$$\boldsymbol{{f}}\:'\left(\mathrm{0}\right)\:=\:? \\ $$

Commented by kaivan.ahmadi last updated on 09/Oct/19

let g(x)=(x−1)(x−2)...(x−10)⇒f(x)=xg(x)  ⇒f′(x)=g(x)+xg′(x)⇒  f′(0)=g(0)+0=g(0)=(−1)(−2)...(−10)=10!

$${let}\:{g}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)...\left({x}−\mathrm{10}\right)\Rightarrow{f}\left({x}\right)={xg}\left({x}\right) \\ $$$$\Rightarrow{f}'\left({x}\right)={g}\left({x}\right)+{xg}'\left({x}\right)\Rightarrow \\ $$$${f}'\left(\mathrm{0}\right)={g}\left(\mathrm{0}\right)+\mathrm{0}={g}\left(\mathrm{0}\right)=\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)...\left(−\mathrm{10}\right)=\mathrm{10}! \\ $$

Commented by Prithwish sen last updated on 09/Oct/19

Excellent Sir.

$$\mathrm{Excellent}\:\mathrm{Sir}. \\ $$

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