Answer
$y'=-\dfrac{1}{2}x^{-3/2}\sec(2x)^2 +8x^{1/2}\sec(2x)^2\tan(2x)^2 $
Work Step by Step
Given that $y=(x)^{-1/2}\sec(2x)^2 $
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[(x)^{-1/2}\sec(2x)^2 ]$
$=-\dfrac{1}{2x^{3/2}}\sec(2x)^2 +\dfrac{1}{x^{1/2}}\sec(2x)^2\tan(2x)^2 ]\dfrac{d}{dx}[(2x)^2 ]$
$=-\dfrac{1}{2x^{3/2}}\sec(2x)^2 +\dfrac{1}{x^{1/2}}\sec(2x)^2\tan(2x)^2 ](8x)$
Hence, $y'=-\dfrac{1}{2}x^{-3/2}\sec(2x)^2 +8x^{1/2}\sec(2x)^2\tan(2x)^2 $
