若f(x)=x(x-1)(x-2)(x-3)(x-4),则f'(0)=?
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若f(x)=x(x-1)(x-2)(x-3)(x-4),则f'(0)=?
解由f(x)=x(x-1)(x-2)(x-3)(x-4)
得
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-1)(x-2)(x-3)(x-4)]+x[(x-1)(x-2)(x-3)(x-4)]'
故f'(0)=(0-1)(0-2)(0-3)(0-4)+0×[(0-1)(0-2)(0-3)(0-4)]'
=(-1)(-2)(-3)(-4)
=24
得
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-1)(x-2)(x-3)(x-4)]+x[(x-1)(x-2)(x-3)(x-4)]'
故f'(0)=(0-1)(0-2)(0-3)(0-4)+0×[(0-1)(0-2)(0-3)(0-4)]'
=(-1)(-2)(-3)(-4)
=24
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