Given equation of curve y=(1+x)2y+cos2(sin−1x) at [x=0] y=(1+0)2y+cos2(sin−10) y=1+1[y=2] So we have to find the normal at (0,2) Now y=e2y‌ln(1+x)+cos2(cos−1√1−x2) y=e2y‌ln(1+x)+(√1−x2)2 y=e2y‌ln(1+x)+(1−x2) ......(1) Now differentiate w.r.t. x y′=e2y‌ln(1+x)[2y.(
1
1+x
)+ln(1+x).2y′]−2x Put x=0&y=2 y′=e2×2ln1[2×2(
1
1+0
)+ln(1+0).2y′]−2×0 y′=e0[4+0]−0 y′=4= slope of tangent to the curve so slope of normal to the curve =−
1
4
{m1m2=−1} Hence equation of normal at (0,2) is y−2=−