Answer
$y'=-2x^{-3} \sin^2 (x)^3+6\sin (x^3) \cos (x^3)$
Work Step by Step
Given that $y=x^{-2} \sin^2 (x)^3$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[x^{-2} \sin^2 (x)^3]$
$=-2x^{-3} \sin^2 (x)^3+2x^{-2}(\sin (x^3)) \dfrac{d}{dx}[\sin(x)^3]$
$=-2x^{-3} \sin^2 (x)^3+2x^{-2}(\sin (x^3)) \cos (x^3) \dfrac{d}{dx}[(x)^3]$
$=-2x^{-3} \sin^2 (x)^3+2x^{-2}(\sin (x^3)) \cos (x^3) (3x^2)$
Hence, $y'=-2x^{-3} \sin^2 (x)^3+6\sin (x^3) \cos (x^3)$
