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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 <i>dy/dx = e<sup>2x</sup> [1/(x+6) + 2log(x+6)]</i>
Correct Answer
B <i>dy/dx = e<sup>2x</sup> [1/(x+6) - 2log(x+6)]</i>
C <i>dy/dx = e<sup>-2x</sup> [1/(x+6) + 2log(x+6)]</i>
D <i>dy/dx = e<sup>-2x</sup> [-1/(x+6) - 2log(x+6)]</i>
Correct Answer

Option A is the correct answer.

Detailed Explanation

We need to use the product rule for differentiation: d/dx (uv) = u'v + uv'.Let u = e2x and v = log(x+6). We assume log means natural logarithm (ln).Then u' = d/dx (e2x) = 2e2x (using the chain rule).And v' = d/dx (ln(x+6)) = 1/(x+6) (using the chain rule).Applying the product rule:dy/dx = (2e2x) ∙ ln(x+6) + e2x ∙ (1/(x+6))Factor out e2x:dy/dx = e2x ∙ (2ln(x+6) + 1/(x+6))Rearranging the terms inside the parenthesis to match the option:dy/dx = e2x ∙ (1/(x+6) + 2ln(x+6)). This matches option A.

Hint

Apply the product rule (uv)' = u'v + uv'. Remember the derivative of eax is aeax and the derivative of ln(x+c) is 1/(x+c).

Question Info

Course: MTH 102
Topic: MTH 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 781 of 930
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