Đáp án:
$f'(1)=55.2021^9$
Giải thích các bước giải:
$f(x)=(2020+x)(2019+2x)(2018+3x)+\cdots+(2011+10x)\\ f'(x)=(2020+x)'(2019+2x)(2018+3x)+\cdots+(2011+10x)+(2020+x)((2019+2x)(2018+3x)+\cdots+(2011+10x))'\\ =(2019+2x)(2018+3x)+\cdots+(2011+10x)+(2020+x)((2019+2x)'(2018+3x)+\cdots+(2011+10x)+(2019+2x)((2019+2x)(2018+3x)+\cdots+(2011+10x))')\\ =(2019+2x)(2018+3x)+\cdots+(2011+10x)+(2020+x)(2(2018+3x)+\cdots+(2011+10x)+(2019+2x)((2019+2x)(2018+3x)+\cdots+(2011+10x))')\\ \cdots\\ =(2019+2x)(2018+3x)+\cdots+(2011+10x)+(2020+x).2.(2018+3x)+\cdots+(2011+10x)+\cdots+(2020+x)(2019+2x)(2018+3x)+\cdots+(2012+9x).10\\ f'(1)=2021^9+2.2021^9+\cdots+10.2021^9\\ =2021^9(1+2+3+\cdots+10)\\ =55.2021^9$
