How do you differentiate y=(sin^-1x)/(1+x)y=sin−1x1+x?

1 Answer
Alan N.
Dec 12, 2016

dy/dx=1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2dydx=1(1+x)√1−x2−sin−1x(1+x)2

Explanation:

y=sin^-1x/(1+x)y=sin−1x1+x

Apply the quotient rule

dy/dx= ((1+x)* d/dx sin^-1x - sin^-1x*d/dx(1+x))/(1+x)^2dydx=(1+x)⋅ddxsin−1x−sin−1x⋅ddx(1+x)(1+x)2

= ((1+x)* 1/sqrt(1+x^2) - sin^-1x* 1)/(1+x)^2=(1+x)⋅1√1+x2−sin−1x⋅1(1+x)2 (Standard differential)

= 1/((1+x)sqrt(1-x^2)) - sin^-1x/(1+x)^2=1(1+x)√1−x2−sin−1x(1+x)2

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