求y=Tan(x+y)的二阶导数,最后那部怎么来的?

2024-11-25 13:47:14
推荐回答(1个)
回答1:

y=tan(x+y)

y'=tan'(x+y)
=sec^2(x+y)(x+y)'
=sec^2(x+y)*(1+y')

y'=sec^2(x+y)/[1-sec^2(x+y))
=-sec^2(x+y)/tan^2(x+y)
=-1/sin^2(x+y)
=-csc^2(x+y)

y''=-2csc(x+y)*[csc(x+y)]'
=-2csc(x+y)*[-csc(x+y)cot(x+y)](x+y)'
=2csc^2(x+y)cot(x+y)(1+y')
=2csc^2(x+y)cot(x+y)[1-csc^2(x+y)].
1-csc^2(x+y)=cot^2(x+y)
所以可得你要的结果。