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1 solution
Take partial derivative with respect to x 1 :
h ′ ( x 1 x 2 ) x 2 = h ′ ( x 1 ) .
This should be true for all x 1 .
Plug x 1 = 0 . 5 :
h ′ ( 0 . 5 x 2 ) x 2 = h ′ ( 0 . 5 ) = − 2 .
Denoting x = 0 . 5 x 2 , we can rewrite it as: h ′ ( x ) = − 1 / x .
By taking integral, we get h ( x ) = − l n ( x ) + C .
Using h ( 1 ) = 0 , we find C = 0 and h ( x ) = − l n ( x ) .