Question

求导数y=arcsinx根号下1-x/1+x求导

Answer 1

y= arcsinx .√[(1-x)/(1+x)]

y' = (1/2)√[(1+x)/(1-x)] . [-2/(1+x)^2] . arcsinx + √[(1-x)/(1+x)] . [1/√(1-x^2)]

= -√[1/[(1-x)(1+x)^3] . arcsinx + 1/(1+x)

= [1/(1+x)] ( 1- arcsinx. √ [1/(1-x^2)] )