y=cos(x+y)隱函數(shù)求二次導(dǎo)y"能用反函數(shù)做嗎
求一階導(dǎo)數(shù)時可以用反函數(shù)作,待一階求出后再求二階時就不能再用反函數(shù)來求.
我用兩種方法求一階,可看出結(jié)果是一致的.
用“隱函數(shù)求導(dǎo)法”：
F(x,y)=y-cos(x+y)=0
dy/dx=-(∂F/∂x)/(∂F/∂y)=sin(x+y)/[1+sin(x+y)]
用“反函數(shù)求導(dǎo)法”：
x+y=arccosy,故x=arccosy-y,
dx/dy=-1/√(1-y²) - 1=-1/√(1-cos²(x+y)-1=-1/sin(x+y)-1=-[1+sin(x+y)]/sin(x+y)
∴dy/dx=y′=1/(dx/dy)=-sin(x+y)/[1+sin(x+y)]
顯然,兩種方法,結(jié)果相同.
但求二階時,已無反函數(shù)可用,因此不能再用“反函數(shù)求導(dǎo)法”.可直接對x求導(dǎo),但要記?。?br/>要把y看作中間變量,遇到y(tǒng)時要用復(fù)合函數(shù)求導(dǎo)法.
d²y/dx²=dy′/dx=-{[1+sin(x+y)][cos(x+y)](1+y′)-[sin(x+y)cos(x+y)](1+y′)}/[1+sin(x+y)]²
=-(1+y′){[1+sin(x+y)][cos(x+y)]-[sin(x+y)cos(x+y)]}/[1+sin(x+y)]²
=-(1+y′)cos(x+y)/[1+sin(x+y)]²
再將y′=-sin(x+y)/[1+sin(x+y)]代入,化簡,即得：
y″=-cos(x+y)/[1+sin(x+y)]³.