How do I take the derivative of $int g(x - y) mathrm dx$ with respect to $y$? [closed]

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How do I take the derivative of $displaystyleint g(x - y),mathrm dx$ with respect to $y$? Is it just $-g(x - y),mathrm dx$?



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integration derivatives




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closed as off-topic by choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos Aug 21 at 14:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
    – TenaliRaman
    Aug 18 at 6:12











  • Sounds nice to me!
    – Mostafa Ayaz
    Aug 18 at 6:59














up vote
1
down vote

favorite












How do I take the derivative of $displaystyleint g(x - y),mathrm dx$ with respect to $y$? Is it just $-g(x - y),mathrm dx$?



Thanks.




integration derivatives




share|cite|improve this question














closed as off-topic by choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos Aug 21 at 14:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
    – TenaliRaman
    Aug 18 at 6:12











  • Sounds nice to me!
    – Mostafa Ayaz
    Aug 18 at 6:59












up vote
1
down vote

favorite









up vote
1
down vote

favorite











How do I take the derivative of $displaystyleint g(x - y),mathrm dx$ with respect to $y$? Is it just $-g(x - y),mathrm dx$?



Thanks.




integration derivatives




share|cite|improve this question














How do I take the derivative of $displaystyleint g(x - y),mathrm dx$ with respect to $y$? Is it just $-g(x - y),mathrm dx$?



Thanks.





integration derivatives





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 6:19









an4s

2,0632417




2,0632417










asked Aug 18 at 6:06









Riya

413




413




closed as off-topic by choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos Aug 21 at 14:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos Aug 21 at 14:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Xander Henderson, Jyrki Lahtonen, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
    – TenaliRaman
    Aug 18 at 6:12











  • Sounds nice to me!
    – Mostafa Ayaz
    Aug 18 at 6:59












  • 1




    The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
    – TenaliRaman
    Aug 18 at 6:12











  • Sounds nice to me!
    – Mostafa Ayaz
    Aug 18 at 6:59







1




1




The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
– TenaliRaman
Aug 18 at 6:12





The integral should still be there. en.wikipedia.org/wiki/Leibniz_integral_rule
– TenaliRaman
Aug 18 at 6:12













Sounds nice to me!
– Mostafa Ayaz
Aug 18 at 6:59




Sounds nice to me!
– Mostafa Ayaz
Aug 18 at 6:59










3 Answers
3






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Let $$G(x)=int g(x)dx$$therefore $$G(x-y)=int g(x-y)dx$$so we have $$dfracddyint g(x-y)dx=-g(x-y)$$






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    Differentiating under the integral sign, $$fracddyint g(x-y)dx=intpartial_y g(x-y)dx=-int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.






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    • Might be better written as $C(y)$.
      – Deepak
      Aug 18 at 6:36










    • @Deepak Thanks; fixed.
      – J.G.
      Aug 18 at 6:38

















    up vote
    1
    down vote













    Yes. Consider $x$ a constant parameter.






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      Let $$G(x)=int g(x)dx$$therefore $$G(x-y)=int g(x-y)dx$$so we have $$dfracddyint g(x-y)dx=-g(x-y)$$






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted










        Let $$G(x)=int g(x)dx$$therefore $$G(x-y)=int g(x-y)dx$$so we have $$dfracddyint g(x-y)dx=-g(x-y)$$






        share|cite|improve this answer






















          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Let $$G(x)=int g(x)dx$$therefore $$G(x-y)=int g(x-y)dx$$so we have $$dfracddyint g(x-y)dx=-g(x-y)$$






          share|cite|improve this answer












          Let $$G(x)=int g(x)dx$$therefore $$G(x-y)=int g(x-y)dx$$so we have $$dfracddyint g(x-y)dx=-g(x-y)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 18 at 6:59









          Mostafa Ayaz

          9,6633730




          9,6633730




















              up vote
              2
              down vote













              Differentiating under the integral sign, $$fracddyint g(x-y)dx=intpartial_y g(x-y)dx=-int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.






              share|cite|improve this answer






















              • Might be better written as $C(y)$.
                – Deepak
                Aug 18 at 6:36










              • @Deepak Thanks; fixed.
                – J.G.
                Aug 18 at 6:38














              up vote
              2
              down vote













              Differentiating under the integral sign, $$fracddyint g(x-y)dx=intpartial_y g(x-y)dx=-int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.






              share|cite|improve this answer






















              • Might be better written as $C(y)$.
                – Deepak
                Aug 18 at 6:36










              • @Deepak Thanks; fixed.
                – J.G.
                Aug 18 at 6:38












              up vote
              2
              down vote










              up vote
              2
              down vote









              Differentiating under the integral sign, $$fracddyint g(x-y)dx=intpartial_y g(x-y)dx=-int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.






              share|cite|improve this answer














              Differentiating under the integral sign, $$fracddyint g(x-y)dx=intpartial_y g(x-y)dx=-int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Aug 18 at 6:37

























              answered Aug 18 at 6:32









              J.G.

              13.8k11424




              13.8k11424











              • Might be better written as $C(y)$.
                – Deepak
                Aug 18 at 6:36










              • @Deepak Thanks; fixed.
                – J.G.
                Aug 18 at 6:38
















              • Might be better written as $C(y)$.
                – Deepak
                Aug 18 at 6:36










              • @Deepak Thanks; fixed.
                – J.G.
                Aug 18 at 6:38















              Might be better written as $C(y)$.
              – Deepak
              Aug 18 at 6:36




              Might be better written as $C(y)$.
              – Deepak
              Aug 18 at 6:36












              @Deepak Thanks; fixed.
              – J.G.
              Aug 18 at 6:38




              @Deepak Thanks; fixed.
              – J.G.
              Aug 18 at 6:38










              up vote
              1
              down vote













              Yes. Consider $x$ a constant parameter.






              share|cite|improve this answer
























                up vote
                1
                down vote













                Yes. Consider $x$ a constant parameter.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Yes. Consider $x$ a constant parameter.






                  share|cite|improve this answer












                  Yes. Consider $x$ a constant parameter.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 18 at 6:11









                  A. Pongrácz

                  3,887625




                  3,887625












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