1: (1) y = xlnx This is the derivative of a product, and y'=x'lnx+xlnx'=lnx+ 1 is obtained by the derivative operation of the product; (2) Bring x= 1 into the derivative function equation and get y' = 0+1=1; Therefore, the slope of the image at X= 1 is1; Then bring X= 1 into the original function equation and get y = 0；; Therefore, the function image passes (1, 0), and the slope is1; The tangent obtained by the point oblique method is y = x-1; 2. The original curve equation is y = sinx/x; Derivative, according to the derivative algorithm of quotient, y' = (xcosx-sinx)/(x * x); Simplify to get y' = cosx-sinx/(x * x); As above, substitute x=π into the derivative function equation to get y' = 0；; That is, the slope of the original curve image at point m is 0; It is easy to get that the tangent line is y = 0； by point skew; To sum up, this question focuses on your proficiency in derivative operation. If the derivative function can be found, the rest is nothing more than finding the slope of the representative point and then finding the tangent of the representative point. If you are not familiar with the basic derivation operation, it is easy to make mistakes. It is recommended that the landlord do more basic derivative topics and be familiar with them. This kind of topic is very dead, it is simply a subtitle.