f(x)=(x-1)(x-2)²(x-3)³(x-4)^4求导
分部分求导吧,先把(x-1)(x-2)²(x-3)³看成一个整体
于是f'(x)=【(x-1)(x-2)²(x-3)³】'(x-4)^4+【(x-1)(x-2)²(x-3)³】【(x-4)^4】'
=【(x-1)(x-2)²(x-3)³】'(x-4)^4+【(x-1)(x-2)²(x-3)³】×4(x-4)³
再求【(x-1)(x-2)²(x-3)³】'=【(x-1)(x-2)²】'(x-3)³+(x-1)(x-2)²【(x-3)³】'
=【(x-1)(x-2)²】'(x-3)³+(x-1)(x-2)²×3(x-3)²
继续求【(x-1)(x-2)²】'=(x-2)²+(x-1)×2(x-2)=(x-2)(3x-4)
于是f'(x)=【{(x-2)(3x-4)(x-3)³+(x-1)(x-2)²×3(x-3)²}(x-4)^4+[(x-1)(x-2)²(x-3)³]×4(x-4)³】
但是题目应该不会这样吧
可能题目只要求f'(1)
那我们可以直接令g(x)=(x-2)²(x-3)³(x-4)^4
于是f(x)=(x-1)g(x)
那么
f'(x)=(x-1)'g(x)+(x-1)g'(x)=g(x)+(x-1)g'(x)
于是f'(1)=g(1)+(1-1)g'(1)=g(1)=(1-2)²(1-3)³(1-4)^4=648