Question Detail

Assume 'log' refers to the natural logarithm 'ln'. So, y = e2x ln(x+6).We need to use the product rule for differentiation, which states that if y = f(x)g(x), then dy/dx = f'(x)g(x) + f(x)g'(x).Let f(x) = e2x and g(x) = ln(x+6).1. Find f'(x): d/dx(e2x) = e2x * d/dx(2x) = 2e2x (using the chain rule).2. Find g'(x): d/dx(ln(x+6)) = (1 / (x+6)) * d/dx(x+6) = 1/(x+6) (using the chain rule).Now, apply the product rule: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 to match the options:dy/dx = e2x [1/(x+6) + 2log(x+6)]This matches option A.