Question

已知f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f'(1)=?求过程

Answer 1

f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)
所以，f'(x)=(x-1)'(x-2)(x-3)(x-4)(x-5)+(x-1)(x-2)'(x-3)(x-4)(x-5)+(x-1)(x-2)(x-3)'(x-4)(x-5)+(x-1)(x-2)(x-3)(x-4)'(x-5)+(x-1)(x-2)(x-3)(x-4)(x-5)'
=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-1)(x-2)(x-4)(x-5)+(x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-3)(x-4)
所以，f'(1)=(1-2)(1-3)(1-4)(1-5)+0+0+0+0
=(-1)(-2)(-3)(-4)
=24

追问

谢了！

Answer 2

已知f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f'(1)=?求过程.
令g（x）= (x-2)(x-3)(x-4)(x-5),
f(x)=(x-1)*g(x);
f'(x)=g(x)+(x-1)*g'(x)；则f'(1)=g(1)=24

追问

谢了！

Answer 3

把x=1代入方程，可得f（x）=（1一1）……。即为O

Answer 4

f(1)不就是0吗

追问

求f（1）的导数好吗？

追答

-_-b导数都忘了

追问

额，好吧！