Question

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

Answer 1

解：
令g(x)=(x-2)(x-3)(x-4)(x-5)，h(x)=(x+1)(x+2)(x+3)(x+4)(x+5)
f'(x)={[(x-1)g(x)]'·h(x)-[(x-1)g(x)]·h'(x)}/h²(x)
={[g(x)+(x-1)g'(x)]·h(x)-[(x-1)g(x)]·h'(x)}/h²(x)
f'(1)={[g(1)+(1-1)g'(1)]·h(1)-[(1-1)g(1)]·h'(1)}/h²(1)
={[g(1)+0]·h(1)-0}/h²(1)
=g(1)·h(1)/h²(1)
=g(1)/h(1)
=(1-2)(1-3)(1-4)(1-5)/[(1+1)(1+2)(1+3)(1+4)(1+5)]
=(-1)(-2)(-3)(-4)/(2·3·4·5·6)
=(1·2·3·4)/(2·3·4·5·6)
=1/30

Answer 2

令g(x)=(x-1)(x-2)(x-3)(x-4)(x-5)，h(x)=(x+1)(x+2)(x+3)(x+4)(x+5)
f'(x)=[g'(x)h(x)-h'(x)g(x)]/h²(x)，
g'(x)=x·(x-2)(x-3)(x-4)(x-5)+(x-1)·[(x-2)(x-3)(x-4)(x-5)]',g'(1)=(-1)×(-2)×(-3)×(-4)﹢0=24
h(1)=2×3×4×5×6=720,g(1)=0
f'(1)=[g'(1)h(1)-h'(1)g(1)]/h²(1)=g'(1)/h(1)-0×h'(1)/h²(1)=24÷720=1/30