Question Detail

We use the quotient rule: d/dx (u/v) = (u'v - uv')/v2.Let u = esin2x and v = cos2x.u' = d/dx (esin2x) = esin2x ∙ cos(2x) ∙ 2 (using chain rule).v' = d/dx (cos2x) = -sin(2x) ∙ 2 (using chain rule).Now apply the quotient rule:dy/dx = [ (esin2x ∙ 2cos2x) ∙ cos2x - esin2x ∙ (-2sin2x) ] / (cos2x)2dy/dx = [ 2esin2x cos2(2x) + 2esin2x sin(2x) ] / cos2(2x)Factor out 2esin2x from the numerator:dy/dx = 2esin2x [ cos2(2x) + sin(2x) ] / cos2(2x)Separate the terms in the numerator:dy/dx = 2esin2x [ cos2(2x)/cos2(2x) + sin(2x)/cos2(2x) ]dy/dx = 2esin2x [ 1 + (sin(2x)/cos(2x)) ∙ (1/cos(2x)) ]dy/dx = 2esin2x [ 1 + tan(2x) sec(2x) ].Note: The options provided have esinx instead of esin2x. Assuming this is a typo and should be esin2x to match the question, option C is the correct structure.Therefore, the answer is 2esin2x [1 + sec2x tan2x]. This matches option C, assuming the typo in the options.