如何用导数求函数的极值？

Question

如题所述

Answer 1

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)