Question Detail

We are given y = e2x log(x+6). In calculus, 'log' usually refers to the natural logarithm (ln) unless a different base is specified. So, we'll assume log(x+6) = ln(x+6).We need to use the product rule: d/dx (uv) = u'v + uv'.Let u = e2x and v = ln(x+6).First, find u':u' = d/dx (e2x) = e2x * (d/dx [2x]) = 2e2x.Next, find v':v' = d/dx (ln(x+6)) = (1/(x+6)) * (d/dx [x+6]) = 1/(x+6).Now, apply the product rule:dy/dx = u'v + uv'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:dy/dx = e2x [1/(x+6) + 2ln(x+6)]Therefore, the correct option is A.