Question Detail

To find dy/dx for y = esin 2x / cos 2x, use the quotient rule: (u/v)' = (u'v - uv') / v2.Let u = esin 2x and v = cos 2x.u' = d/dx(esin 2x) = esin 2x * d/dx(sin 2x) = esin 2x * (cos 2x * 2) = 2 cos 2x esin 2x.v' = d/dx(cos 2x) = -sin 2x * d/dx(2x) = -2 sin 2x.Apply the quotient rule:dy/dx = [ (2 cos 2x esin 2x)(cos 2x) - (esin 2x)(-2 sin 2x) ] / (cos 2x)2dy/dx = [ 2 cos2 2x esin 2x + 2 sin 2x esin 2x ] / cos2 2xFactor out 2esin 2x from the numerator:dy/dx = 2esin 2x [ cos2 2x + sin 2x ] / cos2 2xSeparate the terms in the numerator:dy/dx = 2esin 2x [ (cos2 2x / cos2 2x) + (sin 2x / cos2 2x) ]dy/dx = 2esin 2x [ 1 + (sin 2x / cos 2x) * (1 / cos 2x) ]Using the identities tan x = sin x / cos x and sec x = 1 / cos x:dy/dx = 2esin 2x [ 1 + tan 2x sec 2x ].This matches option C.