differentiate y= sin^(-1) (x/x+1)?

Let : y = sinֿ¹ [ x / ( x + 1 ) ] ....... (1)

By Chain Rule,

d/dx ( sinֿ¹ u ) = [ 1 / √(1-u²) ] · du/dx.

Hence, from (1),

dy/dx = { 1 / √[ 1 - ( x² / (x+1)²)] } • d/dx [ x / (x+1) ]

. . . . .= { (x+1) / √[ (x+1)² - x²] } • { [ (x+1)(1) - x(1) ] / (x+1)² }

. . .. . = { 1 / √[ (x²+2x+1) - x² ] } • { 1 / (x+1) }

. . . . . = 1 / [ (x+1)· √(2x+1) ] ....................... Ans.
..........................................................................................................

Happy To Help !
............................................................................................................

Let a = x, so a' = 1
Let b = x + 1, so b' = 1
Let c = a/b = x / (x + 1), so we use the Quotient Rule:
c' = (a'b - ab') / (b^2)
c' = ((1)(x + 1) - (x)(1)) / ((x + 1)^2)
c' = 1 / ((x + 1)^2)
Since y = arcsin(c), dy/dc = 1 / sqrt(1 - c^2), therefore:
dy/dc = 1 / sqrt(1 - (x / (x + 1))^2)
dy/dc = |x + 1| / sqrt(2x + 1)
By the Chain Rule:
dy/dx = dy/dc * dc/dx
dy/dx = (|x + 1| / sqrt(2x + 1)) * (1 / ((x + 1)^2))
dy/dx = 1 / (sqrt(2x + 1) * |x + 1|)