ln ⁡ ( 1 + x y ) = x yWhen I try to implicitly differentiate this I get 1 1 + x y ( y + x y ′ ) = (y + xy')At which point the ( y + x y ′ ) terms cancel out, leaving no y ′ to solve for.However, the answer to this is − y x ... How do you get this?

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ln  ⁡  (  1  +  x  y  )  =  x  yWhen I try to implicitly differentiate this I get      1          1      +      x      y        (  y  +  x      y    ′    ) = (y + xy&#039;)At which point the   (  y  +  x      y    ′    ) terms cancel out, leaving no       y    ′   to solve for.However, the answer to this is   −      y    x  ... How do you get this?

tkaljegs · Accepted Answer

You can only cancel the   y  +  x      y    ′   terms if that is not equal to zero, otherwise you&#039;re dividing by zero. Assuming that is not equal to zero gets you      1          1      +      x      y        =  1so   1  +  x  y  =  1, or   x  y  =  0, which gives you   log  ⁡  1  =  0 in the original equation. So the solution is based on the notion that for the equation to be non-trivial,   y  +  x      y    ′    =  0.Incidentally, based on the hint in abel&#039;s post, I tried solving the differential equation, and I got                    d        y            y        =  −                    d        x            x      which yields  ln  ⁡  y  =  −  ln  ⁡  x  +  c  ⇒  y  =  k      /    x  .

sveiparnu · Accepted Answer

the only real solution for   ln  ⁡  (  1  +  u  )  =  u is   u  =  0. so your equation implies   x  y  =  0  , which in turn gives   y  =  0 identically. hence             d      y              d      x        =  0.

Implicit differentiation: ln(1+xy)=xy

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