y=arcsin根号1+x分之1-x求导数

2025-04-16 19:03:50
推荐回答(2个)
回答1:

y=arcsin√[(1-x)/(1+x) ]
siny =√[(1-x)/(1+x) ]
(siny)^2 = (1-x)/(1+x)
2(siny).(cosy) . dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 1 - (1-x)/(1+x) ] .dy/dx = -1/(1+x)^2
2√[(1-x)/(1+x) ] . √ [ 2x/(1+x) ] .dy/dx = -1/(1+x)^2
{ 2√[2x(1-x)]/(1+x) } .dy/dx = -1/(1+x)^2
dy/dx = -1/{ 2(1+x) .√[2x(1-x)] }

回答2:

y=arcsin√[(1-x)/(1+x)],
y'=1/√{1-(1-x)/(1+x)}*1/{2√[(1-x)/(1+x)]}*[-(1+x)-(1-x)]/(1+x)^2
=1/√[2x/(1+x)]*1/{2√[(1-x)/(1+x)]}*(-2)/(1+x)^2
=-1/{(1+x)√[2x(1-x)]}.