) and v=sin−1(3x−4x3) Now, put x = sin θ ⇒ θ = sin−1(x), then u=tan−1(
sin‌θ
√1−sin2θ
) and v=sin−1(3‌sin‌θ−4sin3θ) ⇒ u=tan−1(
sin‌θ
cos‌θ
) and v=sin−1(sin‌3‌θ) u=tan−1(tan‌θ) and v=sin−1(sin‌3‌θ) u=θ and v=3θu=sin−1x and v=3sin−1x On differentiating both sides w.r.t. x, we get