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

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Answer 1 · 2016-12-12T02:28:24

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

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

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

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

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

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