第4个回答 2019-04-11
(3) 代公式: (secu)^2 = 1+(tanu)^2, 代原题 y = tan(x+y)
y' = [sec(x+y)]^2 (1+y') = {1+[tan(x+y)]^2} (1+y') = (1+y^2)(1+y')
y'' = 2y'/y^3 (上式代入) = 2(-1/y^2-1)/y^3 = -2(1+y^2)/y^5 , 代原题 y = tan(x+y)
y'' = -2{1+[tan(x+y)]^2}/[tan(x+y)]^5 = -2{1+[tan(x+y)]^2}[cot(x+y)]^5
= -2{[cot(x+y)]^2+1}[cot(x+y)]^3 = -2[csc(x+y)]^2 [cot(x+y)]^3本回答被提问者采纳