Question Detail

This problem requires the product rule for differentiation: (uv)' = u'v + uv'.Let u = e2x and v = log(x+6) (assuming log means natural logarithm, ln).First, find the derivatives of u and v:u' = d/dx(e2x) = 2e2x (using the chain rule)v' = d/dx(ln(x+6)) = 1/(x+6) (using the chain rule)Now apply the product rule:dy/dx = u'v + uv' = (2e2x)log(x+6) + e2x(1/(x+6))Factor out e2x:dy/dx = e2x [2log(x+6) + 1/(x+6)]Rearranging the terms inside the bracket gives:dy/dx = e2x [1/(x+6) + 2log(x+6)]This matches option A.