求y=√([(x-1)(x-2)]/[(x-3)(x-4)])的導(dǎo)數(shù)吧,用對(duì)數(shù)求導(dǎo)法
兩邊求對(duì)數(shù)
lny=1/2[ln(x-1)+ln(x-2)-ln(x-3)-ln(x-4)]
兩邊對(duì)x求導(dǎo),記住y不是自變量（看作中間變量）
y'/y=1/2[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)]
∴y'=√[(x-1)(x-2)]/[(x-3)(x-4)]*[1/(x-1)+1/(x-2)-1/(x-3)-1/(x-4)]/2