復(fù)合函數(shù)求導(dǎo)
d f(√x) /dx = f '(√x) * 1/(2√x)
f '(x) = (x cosx - sinx) / x^2
f '(√x)= (√x cos√x - sin√x) / x
d f(√x) /dx = (√x cos√x - sin√x) / (2 x √x)可是答案是sinx/2√x啊不可能的。g(x)= f(√x) = sin (√x) / √x g '(x) = [ cos (√x) * 1/(2√x) ] / √x+ sin (√x) * (-1/2) / (x √x ) = (√x cos√x - sin√x) / (2 x √x)這是道選擇題，四個答案A sin√x/√x B sinx/2√x C sinx/√x D2sinx/√x函數(shù)沒給錯？ sinx / x是同學(xué)把題抄錯了，應(yīng)該是f ' (x)=sinx / xd f(√x) /dx=f '(√x) * 1/(2√x)= （sin√x / √x）* 1/(2√x) = sin√x /(2x)