${\sin ^{ - 1}}(x\sqrt x ),0 \le x \le 1.$

Let y$= {\sin ^{ - 1}}(x\sqrt x )$ …(i)
Differentiating (i) w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = \cfrac{1}{{\sqrt {1 - {x^3}} }} \cdot \cfrac{d}{{dx}}x\sqrt x = \cfrac{1}{{\sqrt {1 - {x^3}} }}\left[ {x \cdot \cfrac{1}{{2\sqrt x }} + \sqrt x } \right]$

$= \cfrac{1}{{\sqrt {1 - {x^3}} }}\left[ {\cfrac{{\sqrt x }}{2} + \sqrt x } \right] = \cfrac{1}{{\sqrt {1 - {x^3}} }}\left[ {\cfrac{{\sqrt x + 2\sqrt x }}{2}} \right] = \cfrac{3}{2}\sqrt {\cfrac{x}{{1 - {x^3}}}}$