> 数学 >
y=tan(x+y)的隐函数调y的二阶导数
人气:459 ℃ 时间:2019-12-23 14:18:13
解答
两边对x求导:
y'=[sec(x+y)]^2*(1+y')
即y'=(sec(x+y))^2/[1-(sec(x+y))^2]
化简得:y'=-(sec(x+y))^2/(tan(x+y))^2=- [csc(x+y)]^2
两边再对x求导得:
y"=2csc(x+y)*csc(x+y)*ctg(x+y)*(1+y'),再代入y',得:
=2(csc(x+y))^2*ctg(x+y)*[1-(csc(x+y))^2]
=-2(csc(x+y))^2*(ctg(x+y))^2
=-2[csc(x+y)*ctg(x+y)]^2
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版