let
h(x) = f(x)/g(x),then h (x)在閉區(qū)間[a,b]上連續(xù),在開區(qū)間(a,b)內(nèi)可導(dǎo)
h'(x)= {f(x)g'(x) - f'(x)g(x)}/ [g(x)]^2
f(a)g(b)=g(a)f(b).
=> f(a)/g(a) = f(b)/g(b)
=> h(a) = h(b)
存在ξ,∈(a,b),令到
h'(ξ,) =0
=>f `(ξ)g(ξ)-f(ξ)g`(ξ)=0
=>f `(ξ)g(ξ)=f(ξ)g`(ξ)..