f(x)=x(x+1)(x+2)(x+3)(x+4)/24,求F(x)在x=0处导数
1个回答

f'(x)=[x'*(x+1)(x+2)(x+3)(x+4)+x(x+1)'*(x+2)(x+3)(x+4)+x(x+1)(x+2)'*(x+3)(x+4)+x(x+1)(x+2)(x+3)'*(x+4)+x(x+1)(x+2)(x+3)(x+4)']/24

=[(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+x(x+1)(x+2)(x+4)+x(x+1)(x+2)(x+3)]/24

后面每项都有x

则x=0时等于0

所以f'(0)=(0+1)(0+2)(0+3)(0+4)/24=1