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

1 Answer

Dec 12, 2016

#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#

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