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MTH 102 MTH 102 Past Questions Objective Question
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Question

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

Option A is the correct answer.

Detailed Explanation

This problem requires the product rule: d/dx (uv) = u'v + uv'.Let u = e2x and v = log(x+6). Assume 'log' here means natural logarithm (ln).Then u' = d/dx [e2x] = e2x * 2 = 2e2x.And v' = d/dx [ln(x+6)] = 1/(x+6).Substitute these into 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 bracket to match the option: dy/dx = e2x [1/(x+6) + 2log(x+6)].

Hint

Remember the product rule for differentiation and the derivative of a logarithmic function.

Question Info

Course: MTH 102
Topic: MTH 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 556 of 930
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