If u = (8x2 + y2) (log x - log y), then \(\left( {{\rm{x}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{x}}}}\,{\rm{ + }}\,{\rm{y}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{y}}}}} \right)\) equals

1
u/2
2
2u
3
u
4
-2u

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