lnu^v=vlnu两端对x求导得(vlnu)'=v'lnu+v/u*u'=cosxlnlnx+sinx/lnx*(lnx)'=cosxlnlnx+sinx/(xlnx)(u^v)'=e^[cosxlnlnx+sinx/(xlnx)]dz=(x+u^v)'dx={1+e^[cosxlnlnx+sinx/(xlnx)]}dx
