(1)y=(x+lnx)/x^2=1/x+lnx/x^2所以y'=-1/x^2 +(x-2xlnx)/x^4(2)y=xlnx/(1+lnx^2)y'=[(lnx+1)(1+lnx^2) - 2lnx]/(1+lnx^2)^2(3)y=cosce^(x+1)^2=[-sine^(x+1)^2][e^(x+1)^2][(x+1)^2]'=[-sine^(x+1)^2][e^(x+1)^2][2(...(x+lnx)/x^2=1/x+lnx/x^2这个等吗？等啊 很明显 通分相加即可恩恩懂了谢谢呀