Question Detail

To find dy/dx for y = e^(2x) log(x+6), we use the product rule for differentiation: (uv)' = u'v + uv'.Let u = e^(2x) and v = log(x+6) (assuming log refers to the natural logarithm, ln).1. Find u': u' = d/dx(e^(2x)) = 2e^(2x) (by chain rule)2. Find v': v' = d/dx(log(x+6)) = 1/(x+6) (by chain rule, since d/dx(ln(f(x))) = f'(x)/f(x))Now, apply the product rule:dy/dx = u'v + uv'dy/dx = (2e^(2x)) log(x+6) + e^(2x) (1/(x+6))Factor out e^(2x):dy/dx = e^(2x) [2log(x+6) + 1/(x+6)]Rearranging the terms inside the bracket:dy/dx = e^(2x) [1/(x+6) + 2log(x+6)]Thus, option A is correct.