Question

f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5)的导数怎么求

Answer 1

解析如下：

f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5)

lnf(x)=lnx+ln(x-1)+ln(x-2)+ln(x-3)+ln(x-4)+ln(x-5)

f'(x)/f(x)=1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)

∴f'(x)=f(x)[1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)]

=x(x-1)(x-2)(x-3)(x-4)(x-5)[1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)]。

四则运算的运算顺序：

1、如果只有加和减或者只有乘和除，从左往右计算。

2、如果一级运算和二级运算，同时有，先算二级运算。

3、如果一级，二级，三级运算（即乘方、开方和对数运算）同时有，先算三级运算再算其他两级。

4、如果有括号，要先算括号里的数（不管它是什么级的，都要先算）。

5、在括号里面，也要先算三级，然后到二级、一级。

Answer 2

令 f(x)=y=x(x-1)(x-2)(x-3)(x-4)(x-5)
两边取对数得：
lny = lnx + ln(x-1) + ln(x-2) + ln(x-3) + ln(x-4)+ ln(x-5)
两边求导得：
1/y * y′ = 1/x + 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4) + 1/(x-5)

f ′(x) = y′ = y { 1/x + 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4) + 1/(x-5) }
= x(x-1)(x-2)(x-3)(x-4)(x-5) * { 1/x + 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4) + 1/(x-5) }
= (x-1)(x-2)(x-3)(x-4)(x-5) + x(x-2)(x-3)(x-4)(x-5) + x(x-1)(x-3)(x-4)(x-5) + x(x-1)(x-2)(x-4)(x-5) + x(x-1)(x-2)(x-3)(x-5) + x(x-1)(x-2)(x-3)(x-4)

Answer 3

f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5)
lnf(x)=lnx+ln(x-1)+ln(x-2)+ln(x-3)+ln(x-4)+ln(x-5)
f'(x)/f(x)=1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)
∴f'(x)=f(x)[1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)]
=x(x-1)(x-2)(x-3)(x-4)(x-5)[1/x+1/(x-1)+1/(x-2)+1/(x-3)+1/(x-4)+1/(x-5)]