Question Detail

We need to find the derivative of $y = e^{2x} \ln(x+6)$ (assuming $\log$ means $\ln$). This requires the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.Let $u = e^{2x}$ and $v = \ln(x+6)$.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}(\ln(x+6)) = \frac{1}{x+6} \cdot \frac{d}{dx}(x+6) = \frac{1}{x+6}$Now apply the product rule:$\frac{dy}{dx} = (2e^{2x})(\ln(x+6)) + (e^{2x})\left(\frac{1}{x+6}\right)$Factor out $e^{2x}$:$\frac{dy}{dx} = e^{2x} \left[2\ln(x+6) + \frac{1}{x+6}\right]$This matches option A.