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MTH 102 MTH 102 Past Questions Objective Question

Question

If y = esin(2x) / cos(2x), find dy/dx.
Options
A <i>dy/dx = 2e<sup>sin(2x)</sup> [1 - sec(2x)tan(2x)]</i>
B <i>dy/dx = 2e<sup>sin(2x)</sup> [1 - sec(2x)cos(2x)]</i>
C <i>dy/dx = 2e<sup>sin(2x)</sup> [1 + sec(2x)tan(2x)]</i>
Correct Answer
D <i>dy/dx = 2e<sup>sin(2x)</sup> [1 + sec(2x)cos(2x)]</i>
Correct Answer

Option C is the correct answer.

Detailed Explanation

Use the quotient rule: (u/v)' = (u'v - uv') / v2.Let u = esin(2x) and v = cos(2x).1. Find u': u' = d/dx (esin(2x)) = esin(2x) * d/dx (sin(2x)) = esin(2x) * cos(2x) * 2.2. Find v': v' = d/dx (cos(2x)) = -sin(2x) * d/dx (2x) = -2sin(2x).Now, apply the quotient rule:dy/dx = ( (2esin(2x) cos(2x)) * cos(2x) - esin(2x) * (-2sin(2x)) ) / (cos(2x))2dy/dx = ( 2esin(2x) cos2(2x) + 2esin(2x) sin(2x) ) / cos2(2x)Factor out 2esin(2x) from the numerator:dy/dx = 2esin(2x) (cos2(2x) + sin(2x)) / cos2(2x)Separate 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))]dy/dx = 2esin(2x) [1 + tan(2x)sec(2x)].This matches option C.

Hint

Apply the quotient rule. Remember the chain rule for derivatives of eg(x) and trigonometric functions like sin(ax) and cos(ax).

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