Let y=sin−1(2x√1−x2).....(i) and z=sin−1(3x−4x3)....(ii) Now, putting x=cos‌θ in Eq. (i), we get y=sin−1(2‌cos‌θ‌√1−cos2θ) =sin−1(2‌cos‌θ‌sin‌θ) =sin−1(sin‌2‌θ) ⇒‌y=2θ ⇒‌y=2cos−1x Differentiating it w.r.t. θ, we get ‌
dy
dθ
=2...(iii) Also, putting x=sin‌theta in Eq. (ii), we get