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MTH 102 MTH 102 Past Questions Objective Question
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Question

If y = e2x log (x+6), find dy/dx.
Options
A dy/dx = e<sup>2x</sup> [ 1/(x+6) + 2log(x+6) ]
Correct Answer
B dy/dx = e<sup>2x</sup> [ 1/(x+6) - 2log(x+6) ]
C dy/dx = e<sup>-2x</sup> [ 1/(x+6) + 2log(x+6) ]
D dy/dx = e<sup>-2x</sup> [ 1/(x+6) - 2log(x+6) ]
Correct Answer

Option A is the correct answer.

Detailed Explanation

This problem requires the product rule: (uv)' = u'v + uv'. Let u = e2x and v = log(x+6) (assuming log is natural logarithm, ln). First, find the derivatives of u and v:u' = d/dx(e2x) = e2x * d/dx(2x) = 2e2x.v' = d/dx(log(x+6)) = 1/(x+6).Now apply the product rule: dy/dx = (2e2x)log(x+6) + e2x(1/(x+6)).Factor out e2x: dy/dx = e2x [ 2log(x+6) + 1/(x+6) ].This matches option A.

Hint

Use the product rule for differentiation, where u = e2x and v = log(x+6).

Question Info

Course: MTH 102
Topic: MTH 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 635 of 930
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