令u = ln(x^2 + 1) + x^2,则y = sin(u)
则y' = cos(u) * u'
= cos(u) * {2x * [1 / (x^2 + 1)] + 2x}
= cos[ln(x^2 + 1) + x^2] * {2x * [1 / (x^2 + 1)] + 2x}