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Assuming 'log' denotes the natural logarithm (ln), the function is y = e2x ln(x+6).Use the product rule for differentiation: (uv)' = u'v + uv'.Let u = e2x and v = ln(x+6).Then u' = d/dx (e2x) = 2e2x.And v' = d/dx (ln(x+6)) = 1/(x+6).So, dy/dx = (2e2x)ln(x+6) + e2x(1/(x+6)).Factor out e2x:dy/dx = e2x [2ln(x+6) + 1/(x+6)].Rearranging the terms inside the bracket gives:dy/dx = e2x [1/(x+6) + 2log(x+6)], which is option A.