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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 $\frac{dy}{dx}$. (Assume $\log$ denotes the natural logarithm).
Options
A $\frac{dy}{dx} = e^{2x} \left[ \frac{1}{x+6} + 2\log(x+6) \right]$
Correct Answer
B $\frac{dy}{dx} = e^{2x} \left[ \frac{1}{x+6} - 2\log(x+6) \right]$
C $\frac{dy}{dx} = e^{-2x} \left[ \frac{1}{x+6} + 2\log(x+6) \right]$
D $\frac{dy}{dx} = e^{-2x} \left[ -\frac{1}{x+6} - 2\log(x+6) \right]$
Correct Answer

Option A is the correct answer.

Detailed Explanation

To differentiate $y = e^{2x} \log(x+6)$, we use the product rule $\frac{d}{dx}(uv) = u'v + uv'$.Let $u = e^{2x}$ and $v = \log(x+6)$.Then $u' = \frac{d}{dx}(e^{2x}) = 2e^{2x}$.And $v' = \frac{d}{dx}(\log(x+6)) = \frac{1}{x+6}$.Applying the product rule:$\frac{dy}{dx} = (2e^{2x}) \log(x+6) + e^{2x} \left( \frac{1}{x+6} \right)$Factor out $e^{2x}$:$\frac{dy}{dx} = e^{2x} \left[ 2\log(x+6) + \frac{1}{x+6} \right]$.

Hint

Apply the product rule. Remember that $\frac{d}{dx} e^{ax} = ae^{ax}$ and $\frac{d}{dx} \log(f(x)) = \frac{f'(x)}{f(x)}$.

Question Info

Course: MTH 102
Topic: MTH 102 Past Questions
Type: Objective
Difficulty: Not specified
Question 669 of 930
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