f(x)=x(x
f(x)=x(x-1)(x-2)…(x-100) 求f'(x)RT~
f(x)=x(x-1)(x-2)……(x-100)两边取对数得 lnf(x)=lnx+ln(x-1)+ln(x-2)+……+ln(x-1) 两边求导数得 f'(x)/f(x)=1/x+1/(x-1)+1/(x-2)+……+1/(x-100) 所以 f'(x)=f(x)[1/x+1/(x-1)+1/(x-2)+……+1/(x-100)……(*) =(x-1)(x-2)……(x-100)+x(x-2)……(x-100)+x(x-1)……(x-100)+……+x(x-1)(x-2)……(x-99) (*)式是最容易掌握的形式。
应该是求 f'(1) 或 。。。 吧??? 当然()里边的数不一定是1,但应该是0,1,2,3,。。。
100中的某个数吧 比如 求 f'(1),那么: f(x) = (x-1) * x(x-2)(x-3)…(x-100) f'(x) = (x-1)'*x(x-2)(x-3)…(x-100) + (x-1)*[x(x-2)(x-3)…(x-100)]' = x(x-2)(x-3)…(x-100) + (x-1) * [x(x-2)(x-3)…(x-100)]' f'(1) = 1(-1)(-2)…(-99) + 0 = -99! 比如 求 f'(3),那么: f(x) = (x-3) * x(x-1)(x-2)(x-4)(x-5)…(x-100) f'(x) = (x-3)'*x(x-1)(x-2)…(x-100) + (x-3)*[x(x-1)(x-2)…(x-100)]' = x(x-1)(x-2)…(x-100) + (x-3) * [x(x-1)(x-2)…(x-100)]' f'(3) = 3(2)(1)(-1)(-2)(-3)…(-97) + 0 = -97!*6 其它的会了吧??? 。
f'(x) =f(x)*[1/x +1/(x-1) +1/(x-2) +… +1/(x-100)]
答:详情>>