text: To differentiate x^x:\n\nset y = x^x\n\nthen ln y = ln x^x = x ln x.\n=> y' / y = ln x + x(1/x) = ln x + 1, so y' = x^x(ln x + 1).\n\nTherefore f'(x) = 2 cos(x) + 3(x^x)(ln x + 1), so\nf'(3) = 2cos 3 + 3(3^3)(ln 3 + 1).\n\nTo evaluate this, make sure your calculator is in radians, not degrees.