Answer
$ \dfrac{1.5\cos (3x)}{ [\sin (3x)]^{0.5}} $
Work Step by Step
Let us suppose that $f(x)=[\sin (3x)]^{0.5}$
Use $\dfrac {d}{dx} (\sin a)=\cos a \dfrac{da}{dx}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [ [\sin (3x)]^{0.5}] \\=0.5 [\sin (3x)]^{-0.5}) \cos(3x) \times 3 $
Simplify to obtain:
$f^{\prime}(x)=\dfrac{1.5\cos (3x)}{ [\sin (3x)]^{0.5}} $
