Partial Differential












0












$begingroup$


If f is a function of x. Some times people write:



enter image description here



Is this a rule? does it have a proof? need help.



Thanks.






derivatives







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$endgroup$

















    0












    $begingroup$


    If f is a function of x. Some times people write:



    enter image description here



    Is this a rule? does it have a proof? need help.



    Thanks.






    derivatives







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If f is a function of x. Some times people write:



      enter image description here



      Is this a rule? does it have a proof? need help.



      Thanks.






      derivatives







      share|cite|improve this question









      $endgroup$




      If f is a function of x. Some times people write:



      enter image description here



      Is this a rule? does it have a proof? need help.



      Thanks.






      derivatives




      derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 9:22









      DiamondDiamond

      143




      143






















          2 Answers
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          0












          $begingroup$

          begin{align*}
          x_i &= x_i(t) text{ and}\
          F &= f(x_1, x_2, ..., x_n)\
          frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
          dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
          end{align*}



          In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
          begin{align*}
          dF &=frac{partial f}{partial x_1} dx_1 \
          &= frac{df}{dx_1} dx_1
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OK but why capital F?
            $endgroup$
            – Diamond
            Dec 13 '18 at 17:40










          • $begingroup$
            That's just an arbitrary symbol, notice that F=f anyway.
            $endgroup$
            – TED
            Dec 14 '18 at 1:29



















          0












          $begingroup$

          $df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.



          In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so you mean although they wrote it as partial derivative they mean common derivative.
            $endgroup$
            – Diamond
            Dec 13 '18 at 15:32










          • $begingroup$
            "common" : "usual" or "ordinary" (not quite sure of the correct translation).
            $endgroup$
            – JJacquelin
            Dec 13 '18 at 15:51












          Your Answer





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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

          votes









          0












          $begingroup$

          begin{align*}
          x_i &= x_i(t) text{ and}\
          F &= f(x_1, x_2, ..., x_n)\
          frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
          dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
          end{align*}



          In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
          begin{align*}
          dF &=frac{partial f}{partial x_1} dx_1 \
          &= frac{df}{dx_1} dx_1
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OK but why capital F?
            $endgroup$
            – Diamond
            Dec 13 '18 at 17:40










          • $begingroup$
            That's just an arbitrary symbol, notice that F=f anyway.
            $endgroup$
            – TED
            Dec 14 '18 at 1:29
















          0












          $begingroup$

          begin{align*}
          x_i &= x_i(t) text{ and}\
          F &= f(x_1, x_2, ..., x_n)\
          frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
          dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
          end{align*}



          In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
          begin{align*}
          dF &=frac{partial f}{partial x_1} dx_1 \
          &= frac{df}{dx_1} dx_1
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            OK but why capital F?
            $endgroup$
            – Diamond
            Dec 13 '18 at 17:40










          • $begingroup$
            That's just an arbitrary symbol, notice that F=f anyway.
            $endgroup$
            – TED
            Dec 14 '18 at 1:29














          0












          0








          0





          $begingroup$

          begin{align*}
          x_i &= x_i(t) text{ and}\
          F &= f(x_1, x_2, ..., x_n)\
          frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
          dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
          end{align*}



          In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
          begin{align*}
          dF &=frac{partial f}{partial x_1} dx_1 \
          &= frac{df}{dx_1} dx_1
          end{align*}






          share|cite|improve this answer











          $endgroup$



          begin{align*}
          x_i &= x_i(t) text{ and}\
          F &= f(x_1, x_2, ..., x_n)\
          frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
          dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
          end{align*}



          In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
          begin{align*}
          dF &=frac{partial f}{partial x_1} dx_1 \
          &= frac{df}{dx_1} dx_1
          end{align*}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 9:46

























          answered Dec 13 '18 at 9:41









          TEDTED

          34




          34












          • $begingroup$
            OK but why capital F?
            $endgroup$
            – Diamond
            Dec 13 '18 at 17:40










          • $begingroup$
            That's just an arbitrary symbol, notice that F=f anyway.
            $endgroup$
            – TED
            Dec 14 '18 at 1:29


















          • $begingroup$
            OK but why capital F?
            $endgroup$
            – Diamond
            Dec 13 '18 at 17:40










          • $begingroup$
            That's just an arbitrary symbol, notice that F=f anyway.
            $endgroup$
            – TED
            Dec 14 '18 at 1:29
















          $begingroup$
          OK but why capital F?
          $endgroup$
          – Diamond
          Dec 13 '18 at 17:40




          $begingroup$
          OK but why capital F?
          $endgroup$
          – Diamond
          Dec 13 '18 at 17:40












          $begingroup$
          That's just an arbitrary symbol, notice that F=f anyway.
          $endgroup$
          – TED
          Dec 14 '18 at 1:29




          $begingroup$
          That's just an arbitrary symbol, notice that F=f anyway.
          $endgroup$
          – TED
          Dec 14 '18 at 1:29











          0












          $begingroup$

          $df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.



          In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so you mean although they wrote it as partial derivative they mean common derivative.
            $endgroup$
            – Diamond
            Dec 13 '18 at 15:32










          • $begingroup$
            "common" : "usual" or "ordinary" (not quite sure of the correct translation).
            $endgroup$
            – JJacquelin
            Dec 13 '18 at 15:51
















          0












          $begingroup$

          $df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.



          In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so you mean although they wrote it as partial derivative they mean common derivative.
            $endgroup$
            – Diamond
            Dec 13 '18 at 15:32










          • $begingroup$
            "common" : "usual" or "ordinary" (not quite sure of the correct translation).
            $endgroup$
            – JJacquelin
            Dec 13 '18 at 15:51














          0












          0








          0





          $begingroup$

          $df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.



          In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.






          share|cite|improve this answer











          $endgroup$



          $df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.



          In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 15:46

























          answered Dec 13 '18 at 9:45









          JJacquelinJJacquelin

          45.4k21856




          45.4k21856












          • $begingroup$
            so you mean although they wrote it as partial derivative they mean common derivative.
            $endgroup$
            – Diamond
            Dec 13 '18 at 15:32










          • $begingroup$
            "common" : "usual" or "ordinary" (not quite sure of the correct translation).
            $endgroup$
            – JJacquelin
            Dec 13 '18 at 15:51


















          • $begingroup$
            so you mean although they wrote it as partial derivative they mean common derivative.
            $endgroup$
            – Diamond
            Dec 13 '18 at 15:32










          • $begingroup$
            "common" : "usual" or "ordinary" (not quite sure of the correct translation).
            $endgroup$
            – JJacquelin
            Dec 13 '18 at 15:51
















          $begingroup$
          so you mean although they wrote it as partial derivative they mean common derivative.
          $endgroup$
          – Diamond
          Dec 13 '18 at 15:32




          $begingroup$
          so you mean although they wrote it as partial derivative they mean common derivative.
          $endgroup$
          – Diamond
          Dec 13 '18 at 15:32












          $begingroup$
          "common" : "usual" or "ordinary" (not quite sure of the correct translation).
          $endgroup$
          – JJacquelin
          Dec 13 '18 at 15:51




          $begingroup$
          "common" : "usual" or "ordinary" (not quite sure of the correct translation).
          $endgroup$
          – JJacquelin
          Dec 13 '18 at 15:51


















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