设f(x)=x(x-1)(x-2)(x-3)(x-4),则f‘(0)=?
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设f(x)=x(x-1)(x-2)(x-3)(x-4),则f‘(0)=?
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)'
=(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)
后面每项都有x,则x=0时等于0
所以f'(0)=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)
后面每项都有x,则x=0时等于0
所以f'(0)=4!=24
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