求隐函数y=cos(x+y)求y的一阶导数和二阶导数

y=cos(x+y)两边同时对x求导得
y'=-sin(x+y)(1+y') (*)
得y'=-sin(x+y) /[1+sin(x+y)] (#)
(*)式两边同时对x求导得
y''=-sin(x+y)y'' -cos(x+y)(1+y')(1+y')
即y''=-cos(x+y)(1+y')² / [1+sin(x+y)]
将(#)代入上式得
y''=-cos(x+y) / [1+sin(x+y)]^3

y=cos(x+y)
所以，y'=-sin(x+y)·(1+y') ==> [1+sin(x+y)]y'=-sin(x+y) ==> y'=-sin(x+y)/[1+sin(x+y)]
且：y''=-cos(x+y)·(1+y')²-sin(x+y)·y''
==> [1+sin(x+y)]y''=-cos(x+y)·(1+y')²
==> [1+sin(x+y)]y''=-cos(x+y)·{1-[sin(x+y)/(1+sin(x+y))}²
==> [1+sin(x+y)]y''=-cos(x+y)·{1/[1+sin(x+y)]²}
==> y''=-cos(x+y)/[1+sin(x+y)]³