Answer
$y'=8x^{1/2}sec(2x)^2tan(2x)^2-\frac{1}{2}x^{-3/2}sec(2x)^2$
Work Step by Step
Take the derivative of the equation using Power Rule, Trigonometric derivatives, Product Rule, and Chain Rule:
$y'=x^{-1/2}\times sec(2x)^2tan(2x)^2\times2(2x)\times2-\frac{1}{2}x^{-3/2}sec(2x)^2$
$=8x^{1/2}sec(2x)^2tan(2x)^2-\frac{1}{2}x^{-3/2}sec(2x)^2$
