If y=sec^-1(2x / 1+x²) + sin^-1(x-1 / x+1), 0<x<1 then calculate dy/dx.

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y=u+v

(dy/dx)=(du/dx)+(dv/dx)

u=sec^-1(2x/1+x²) put x=tan

u=sec^-1(sin2)

du/dx=1/((sin2)(root(1-sin²2))2cos2.2d/dx

=(2/sin4)2cos2.2d/dx

=2(2/sin4(tan^-1x))2cos2(tan^-1x).(1/1+x²)

v=-sin^-1(1-x/1+x) put x=cos2

v=sin^-1(tan²)

dv/dx=1/root(1-tan⁴).2tan.sec².d/dx

dv/dx=1/root(1-tan⁴((1/2)cos^-1x)).2tan((1/2)cos^-1x).sec²((1/2)cos^-1x).(1/2)(-1/root(1-x²))