Question

求详解y=tan(x+y)

求隐函数的二阶导数d²y/dx²

Answer 1

你要求什么？ dy/dx?

1*dy/dx= sec²(x+y) *(dy/dx+1)
dy/dx(1-sec²(x+y))=sec²(x+y)
dy/dx(-tan²(x+y))=sec²(x+y)
dy/dx=-(1+tan²(x+y))/tan²(x+y)
dy/dx=-cot²(x+y)-1

d²y/dx²=cot(x+y)csc(x+y)(1+dy/dx)
=cot(x+y)csc(x+y)(-cot²(x+y))
=-cot³(x+y)csc(x+y)

追问

二导。

追答

刚刚有个地方错了cot²(x+y)处理失误，重来
d²y/dx²=-cot(x+y)csc(x+y){-2cot(x+y)}[1+dy/dx]
=2cot²(x+y)csc(x+y)[-cot²(x+y)]
=-2cot^4(x+y)csc(x+y)

Answer 2

tan (x +y )=(tan x +tan y )/(1-tan x tan y )

追问

看题。。