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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

We need to find the derivative of y = e2x log(x+6). Assume 'log' refers to the natural logarithm (ln).We use the product rule for differentiation: d/dx (uv) = u'v + uv'.Let u = e2x and v = ln(x+6).First, find the derivatives u' and v':u' = d/dx(e2x) = 2e2x.v' = d/dx(ln(x+6)) = 1/(x+6).Now apply 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)].This matches option A.

Hint

Apply the product rule for differentiation (d/dx(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 583 of 930
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