But then the result == 0 is wrong, since I get:   
      
   d   
   -- (x^2+y^2)  == 2⋅x   
   dx   
      
   d   
   -- (x+y)  == 1   
   dx   
      
   Therefore we should have:   
      
     d   
   ------ (x^2+y^2)  == 2⋅x   
   d(x+y)   
      
   But what I get is:   
      
     d   
   ------ (x^2+y^2) == 0   
   d(x+y)   
      
   P.S.: I don't think 2⋅x is the correct answer, I   
   only wanted to point out, that I have doubts that   
   your system does d(f,g)=d(f,x)/d(g,x).   
      
   Am Dienstag, 16. Januar 2018 23:04:54 UTC+1 schrieb Dr Huang:   
   > yes d(f,g)=d(f,x)/d(g,x)   
   >    
   > reference math handbook    
   > www.mathhandbook.com   
      
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