y'=lnx+x*1/x=1+lnx
x=1时,y'=1,经过(1,0),切线方程：y-0=1*(x-1),即y=x-1
y=sinx * 1/x
y'=(xcosx-sinx)/x^2
x=∏时,y'=-1/∏,所以切线方程：y-0=(-1/∏)*(x-∏)