I need help with DEERIVATIVES with Trig.?

Question

Find dy/dx if:
y = (sec^2)(sqrtsin2x)

^2 : squared
sqrt: square root

Answer 1

it's not a product - once again, it's the chain rule (many times), so:
dy/dx=2[sec(sqrtsin2x)(1/2*(sin2x)^-1/2*cos2x*2)]

i think...this one's really a mess...lol
hope it's right! :D

Answer 2

Use product rule.
derivative of sec^2(x) = derivative of sec(x) *sec(x)
= sec^2(x)tan^2(x)
derivative of sqrt(sin(2x)) = derivative of (sin(2x))^.5
=2/(cos2x)^.5
So you should now apply product rule and the answer is yours.