Bài tập nâng cao · Bài 8
Bài tập nâng cao Các quy tắc tính đạo hàm · Bài 8
Tính đạo hàm của hàm số $f\left( x \right)=x\left( x-1 \right)\left( x-2 \right)...\left( x-2021 \right)$ tại điểm $x=0$?
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Lời giải
(*) Cách 1:
$f\left( x \right)=x.\left[ \left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right) \right]$
$f'\left( x \right)=\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)+x.\left[ \left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right) \right]'$
$f'\left( 0 \right)=\left( -1 \right)\left( -2 \right)\ldots \left( -2021 \right)=-\left( 1.2\ldots 2021 \right)=-2021!$
(*) Cách 2:
$f'\left( 0 \right)=\lim\limits_{x\to 0}\frac{f\left( x \right)-f\left( 0 \right)}{x}=\lim\limits_{x\to 0}\frac{x\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)}{x}$
$=\lim\limits_{x\to 0}\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)=\left( -1 \right)\left( -2 \right)\ldots \left( -2021 \right)=-\left( 1.2\ldots 2021 \right)=-2021!$
$f\left( x \right)=x.\left[ \left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right) \right]$
$f'\left( x \right)=\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)+x.\left[ \left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right) \right]'$
$f'\left( 0 \right)=\left( -1 \right)\left( -2 \right)\ldots \left( -2021 \right)=-\left( 1.2\ldots 2021 \right)=-2021!$
(*) Cách 2:
$f'\left( 0 \right)=\lim\limits_{x\to 0}\frac{f\left( x \right)-f\left( 0 \right)}{x}=\lim\limits_{x\to 0}\frac{x\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)}{x}$
$=\lim\limits_{x\to 0}\left( x-1 \right)\left( x-2 \right)\ldots \left( x-2021 \right)=\left( -1 \right)\left( -2 \right)\ldots \left( -2021 \right)=-\left( 1.2\ldots 2021 \right)=-2021!$