Question Detail

To find the derivative of y = esin 2x ⁄ cos 2x, we use the quotient rule: (f⁄g)' = (f'g - fg')⁄g2. Let f = esin 2x and g = cos 2x. f' = d⁄dx(esin 2x) = esin 2x * cos 2x * 2 (using chain rule). g' = d⁄dx(cos 2x) = -sin 2x * 2 (using chain rule). Now, substitute into the quotient rule: dy⁄dx = [ (esin 2x * cos 2x * 2) * cos 2x - (esin 2x) * (-sin 2x * 2) ] ⁄ (cos 2x)2. Simplify: dy⁄dx = [ 2 * esin 2x * cos2(2x) + 2 * esin 2x * sin(2x) ] ⁄ cos2(2x). Factor out 2 * esin 2x from the numerator: dy⁄dx = 2 * esin 2x * [ cos2(2x) + sin(2x) ] ⁄ cos2(2x). Separate the terms in the bracket over the denominator: dy⁄dx = 2 * esin 2x * [ cos2(2x)⁄cos2(2x) + sin(2x)⁄cos2(2x) ]. dy⁄dx = 2 * esin 2x * [ 1 + (sin 2x⁄cos 2x) * (1⁄cos 2x) ] = 2 * esin 2x * [ 1 + tan 2x * sec 2x ]. This matches option C.