Question Detail

To find the derivative of $y = e^{2x} \log(x+6)$, we use the product rule $ (uv)' = u'v + uv' $.Let $u = e^{2x}$ and $v = \log(x+6)$.First, find the derivatives of $u$ and $v$:$u' = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$$v' = \frac{d}{dx}(\log(x+6)) = \frac{1}{x+6} \cdot \frac{d}{dx}(x+6) = \frac{1}{x+6}$Now, apply the product rule:$\frac{dy}{dx} = u'v + uv' = (2e^{2x})(\log(x+6)) + (e^{2x})\left(\frac{1}{x+6}\right)$Factor out $e^{2x}$:$\frac{dy}{dx} = e^{2x} \left[2\log(x+6) + \frac{1}{x+6}\right]$Rearranging the terms inside the bracket matches option A: $e^{2x} \left[\frac{1}{x+6} + 2\log(x+6)\right]$