1.

Sin^-1(1+x^2/1+x^2)

Answer»

y=sin(1-x^2)/(1+x^2)

Let x=tanA

dx/dA=sec^2A=1+tan^2A

dx/dA=1+x^2………………..(1)

y=sin[(1-tan^2A)/(1+tan^2A)]

y=sin(cos2A)

dy/dA=cos(cos2A).(-2sin2A)

=-2.cos[(1-tan^2A)/(1+tan^2A)].[2tanA/(1+tan^2A)]

dy/dA=-2[2x/(1+x^2)].cos[(1-x^2)/(1+x^2)]

dy/dA=-4x/(1+x^2).cos[(1-x^2)/(1-x^2)]………(2)

Divide eq.(2)by (1)

dy/dA÷dx/dA=-4x/(1+x^2).cos[(1-x^2)/(1+x^2)] ÷(1+x^2).

Therefore, Answer = dy/dx={-4x/(1+x^2)^2}.cos{(1-x^2)/(1+x^2)} 


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