Question Detail

To find the derivative of y = e2x log(x+6) (assuming log is natural logarithm ln), we use the product rule: d⁄dx(uv) = u'v + uv'. Let u = e2x and v = ln(x+6). First, find the derivatives of u and v: u' = d⁄dx(e2x) = e2x * 2 = 2e2x (using the chain rule). v' = d⁄dx(ln(x+6)) = 1⁄(x+6) * 1 = 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)]. This matches option A.