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求y=Tan(x+y)的二阶导数,最后那部怎么来的?
 
人气:460 ℃ 时间:2020-05-03 22:58:30
解答
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.
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