Question

How do you differentiate `y=(sin^-1x)/(1+x)`?

Answer 1

`dy/dx=1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2`

Explanation:

`y=sin^-1x/(1+x)`

Apply the quotient rule

`dy/dx= ((1+x)* d/dx sin^-1x - sin^-1x*d/dx(1+x))/(1+x)^2`

`= ((1+x)* 1/sqrt(1+x^2) - sin^-1x* 1)/(1+x)^2` (Standard differential)

`= 1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2`