Simplify Derivative with Substitution











up vote
3
down vote

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I try to evaluate:



$$ frac{partial}{partial x} log{u(x, y, z)}$$



Mathematica gives:



$$ frac{1}{x+y+z}$$



I want to simplify the expression with my function:



$$ frac{1}{u(x, y, z)}$$



How to do that?



Thanks.



u[x_, y_, z_] = x + y + z

Simplify[D[Log[u[x, y, z]], x]]





calculus-and-analysis simplifying-expressions







share|improve this question


























    up vote
    3
    down vote

    favorite












    I try to evaluate:



    $$ frac{partial}{partial x} log{u(x, y, z)}$$



    Mathematica gives:



    $$ frac{1}{x+y+z}$$



    I want to simplify the expression with my function:



    $$ frac{1}{u(x, y, z)}$$



    How to do that?



    Thanks.



    u[x_, y_, z_] = x + y + z
    
Simplify[D[Log[u[x, y, z]], x]]





    calculus-and-analysis simplifying-expressions







    share|improve this question
























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I try to evaluate:



      $$ frac{partial}{partial x} log{u(x, y, z)}$$



      Mathematica gives:



      $$ frac{1}{x+y+z}$$



      I want to simplify the expression with my function:



      $$ frac{1}{u(x, y, z)}$$



      How to do that?



      Thanks.



      u[x_, y_, z_] = x + y + z
      
Simplify[D[Log[u[x, y, z]], x]]





      calculus-and-analysis simplifying-expressions







      share|improve this question













      I try to evaluate:



      $$ frac{partial}{partial x} log{u(x, y, z)}$$



      Mathematica gives:



      $$ frac{1}{x+y+z}$$



      I want to simplify the expression with my function:



      $$ frac{1}{u(x, y, z)}$$



      How to do that?



      Thanks.



      u[x_, y_, z_] = x + y + z
      
Simplify[D[Log[u[x, y, z]], x]]





      calculus-and-analysis simplifying-expressions




      calculus-and-analysis simplifying-expressions






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 22 at 17:33









      R zu

      1797




      1797






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          7
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            Nov 22 at 18:04










          • @Rzu, good point.
            – kglr
            Nov 22 at 18:05


















          up vote
          4
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          
Derivative[0, 1, 0][u] ^:= 1&
          
Derivative[0, 0, 1][u] ^:= 1&
          

          
D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            Nov 22 at 19:29












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            Nov 22 at 19:39













          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          7
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            Nov 22 at 18:04










          • @Rzu, good point.
            – kglr
            Nov 22 at 18:05















          up vote
          7
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            Nov 22 at 18:04










          • @Rzu, good point.
            – kglr
            Nov 22 at 18:05













          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer














          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 22 at 18:05

























          answered Nov 22 at 17:36









          kglr

          175k9197402




          175k9197402












          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            Nov 22 at 18:04










          • @Rzu, good point.
            – kglr
            Nov 22 at 18:05


















          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            Nov 22 at 18:04










          • @Rzu, good point.
            – kglr
            Nov 22 at 18:05
















          A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
          – R zu
          Nov 22 at 18:04




          A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
          – R zu
          Nov 22 at 18:04












          @Rzu, good point.
          – kglr
          Nov 22 at 18:05




          @Rzu, good point.
          – kglr
          Nov 22 at 18:05










          up vote
          4
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          
Derivative[0, 1, 0][u] ^:= 1&
          
Derivative[0, 0, 1][u] ^:= 1&
          

          
D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            Nov 22 at 19:29












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            Nov 22 at 19:39

















          up vote
          4
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          
Derivative[0, 1, 0][u] ^:= 1&
          
Derivative[0, 0, 1][u] ^:= 1&
          

          
D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            Nov 22 at 19:29












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            Nov 22 at 19:39















          up vote
          4
          down vote










          up vote
          4
          down vote









          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          
Derivative[0, 1, 0][u] ^:= 1&
          
Derivative[0, 0, 1][u] ^:= 1&
          

          
D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer












          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          
Derivative[0, 1, 0][u] ^:= 1&
          
Derivative[0, 0, 1][u] ^:= 1&
          

          
D[Log[u[x, y, z]], x]



          1/u[x, y, z]








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 22 at 19:26









          Carl Woll

          66.7k385174




          66.7k385174












          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            Nov 22 at 19:29












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            Nov 22 at 19:39




















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            Nov 22 at 19:29












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            Nov 22 at 19:39


















          What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
          – R zu
          Nov 22 at 19:29






          What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
          – R zu
          Nov 22 at 19:29














          @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
          – Carl Woll
          Nov 22 at 19:39






          @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
          – Carl Woll
          Nov 22 at 19:39




















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