Question Detail

Use the product rule: ( \frac{d}{dx}(uv) = u'v + uv' ).Let ( u = e^{2x} ) and ( v = \log(x+6) ) (assuming natural logarithm, ( \ln )).Then ( u' = \frac{d}{dx}(e^{2x}) = 2e^{2x} ).And ( v' = \frac{d}{dx}(\ln(x+6)) = \frac{1}{x+6} ).Substitute these into the product rule formula:( \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] )Rearranging the terms inside the bracket:( \frac{dy}{dx} = e^{2x} \left[\frac{1}{x+6} + 2\log(x+6)\right] ).This matches option A.