Question

Find the derivative of  $\sqrt{sin2x}$ by first principle.

Answer 1

y = sqrt(sin 2x)
y' = Lim dx->0  , D
where D = [sqrt( sin(2x + 2dx) - sqrt(sin 2x) ] / dx
D = [sqrt(sin 2x cos 2dx + cos 2x sin 2dx) - sqrt(sin 2 x)  ] / dx
D = [sqrt(sin 2x (1-dx) + 2 cos 2x dx) - sqrt(sin 2 x) ] / dx    for dx << 0
D = sqrt(sin 2x) [sqrt( 1 + 2 cot 2x dx) - 1 ] / dx     for   2 cot 2x dx << 1
D = sqrt(sin 2x) [ (1 + cot 2x dx) - 1 ] / dx
So  y' = sqrt(sin 2x) cot 2x 
        or,    Cos 2x / sqrt(sin 2x)