The first fundamental theorem of calculus if the upper limit is a function.

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Just curious.



If my function is
$$g(x,y) = int_0^x+y/4f(t),dt$$
What is the partial derivative of the function w.r.t $x$?



Is it $f(x+y/4)$? What's the partial derivative of $y$ then?




calculus




share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    Just curious.



    If my function is
    $$g(x,y) = int_0^x+y/4f(t),dt$$
    What is the partial derivative of the function w.r.t $x$?



    Is it $f(x+y/4)$? What's the partial derivative of $y$ then?




    calculus




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Just curious.



      If my function is
      $$g(x,y) = int_0^x+y/4f(t),dt$$
      What is the partial derivative of the function w.r.t $x$?



      Is it $f(x+y/4)$? What's the partial derivative of $y$ then?




      calculus




      share|cite|improve this question













      Just curious.



      If my function is
      $$g(x,y) = int_0^x+y/4f(t),dt$$
      What is the partial derivative of the function w.r.t $x$?



      Is it $f(x+y/4)$? What's the partial derivative of $y$ then?





      calculus





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Michael Hardy

      204k23185460




      204k23185460









      asked 2 days ago









      Zhidong Li

      204




      204




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Your answer to the first question is correct and as for the second question the answer is



          $$fracpartialpartial y g (x, y) = frac 1 4 f(x + y/4)$$



          Take a look at Leibniz formula!






          share|cite|improve this answer




























            up vote
            3
            down vote













            For partial derivatives of $$ g(x,y) = int_0^x+y/4f(t)dt$$



            You may apply the chain rule.



            For example $$frac partial partial yint_0^x+y/4f(t)dt= frac 14 f(x+y/4)$$






            share|cite|improve this answer




























              up vote
              2
              down vote













              Viewing the chain rule as differentiation by substitution often helps clarify:
              beginalign
              g & = int_0^uf(t),dt quadtextand u = x + frac y 4 \[10pt]
              fracdgdu & = f(u) quad textand fracpartial upartial y = frac 1 4 \[10pt]
              f(u) cdot frac 1 4 & = fleft( x + frac y 4 right) cdot frac 1 4.
              endalign






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                1
                down vote



                accepted










                Your answer to the first question is correct and as for the second question the answer is



                $$fracpartialpartial y g (x, y) = frac 1 4 f(x + y/4)$$



                Take a look at Leibniz formula!






                share|cite|improve this answer

























                  up vote
                  1
                  down vote



                  accepted










                  Your answer to the first question is correct and as for the second question the answer is



                  $$fracpartialpartial y g (x, y) = frac 1 4 f(x + y/4)$$



                  Take a look at Leibniz formula!






                  share|cite|improve this answer























                    up vote
                    1
                    down vote



                    accepted







                    up vote
                    1
                    down vote



                    accepted






                    Your answer to the first question is correct and as for the second question the answer is



                    $$fracpartialpartial y g (x, y) = frac 1 4 f(x + y/4)$$



                    Take a look at Leibniz formula!






                    share|cite|improve this answer













                    Your answer to the first question is correct and as for the second question the answer is



                    $$fracpartialpartial y g (x, y) = frac 1 4 f(x + y/4)$$



                    Take a look at Leibniz formula!







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered 2 days ago









                    pointguard0

                    512315




                    512315




















                        up vote
                        3
                        down vote













                        For partial derivatives of $$ g(x,y) = int_0^x+y/4f(t)dt$$



                        You may apply the chain rule.



                        For example $$frac partial partial yint_0^x+y/4f(t)dt= frac 14 f(x+y/4)$$






                        share|cite|improve this answer

























                          up vote
                          3
                          down vote













                          For partial derivatives of $$ g(x,y) = int_0^x+y/4f(t)dt$$



                          You may apply the chain rule.



                          For example $$frac partial partial yint_0^x+y/4f(t)dt= frac 14 f(x+y/4)$$






                          share|cite|improve this answer























                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            For partial derivatives of $$ g(x,y) = int_0^x+y/4f(t)dt$$



                            You may apply the chain rule.



                            For example $$frac partial partial yint_0^x+y/4f(t)dt= frac 14 f(x+y/4)$$






                            share|cite|improve this answer













                            For partial derivatives of $$ g(x,y) = int_0^x+y/4f(t)dt$$



                            You may apply the chain rule.



                            For example $$frac partial partial yint_0^x+y/4f(t)dt= frac 14 f(x+y/4)$$







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered 2 days ago









                            Mohammad Riazi-Kermani

                            27k41850




                            27k41850




















                                up vote
                                2
                                down vote













                                Viewing the chain rule as differentiation by substitution often helps clarify:
                                beginalign
                                g & = int_0^uf(t),dt quadtextand u = x + frac y 4 \[10pt]
                                fracdgdu & = f(u) quad textand fracpartial upartial y = frac 1 4 \[10pt]
                                f(u) cdot frac 1 4 & = fleft( x + frac y 4 right) cdot frac 1 4.
                                endalign






                                share|cite|improve this answer

























                                  up vote
                                  2
                                  down vote













                                  Viewing the chain rule as differentiation by substitution often helps clarify:
                                  beginalign
                                  g & = int_0^uf(t),dt quadtextand u = x + frac y 4 \[10pt]
                                  fracdgdu & = f(u) quad textand fracpartial upartial y = frac 1 4 \[10pt]
                                  f(u) cdot frac 1 4 & = fleft( x + frac y 4 right) cdot frac 1 4.
                                  endalign






                                  share|cite|improve this answer























                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    Viewing the chain rule as differentiation by substitution often helps clarify:
                                    beginalign
                                    g & = int_0^uf(t),dt quadtextand u = x + frac y 4 \[10pt]
                                    fracdgdu & = f(u) quad textand fracpartial upartial y = frac 1 4 \[10pt]
                                    f(u) cdot frac 1 4 & = fleft( x + frac y 4 right) cdot frac 1 4.
                                    endalign






                                    share|cite|improve this answer













                                    Viewing the chain rule as differentiation by substitution often helps clarify:
                                    beginalign
                                    g & = int_0^uf(t),dt quadtextand u = x + frac y 4 \[10pt]
                                    fracdgdu & = f(u) quad textand fracpartial upartial y = frac 1 4 \[10pt]
                                    f(u) cdot frac 1 4 & = fleft( x + frac y 4 right) cdot frac 1 4.
                                    endalign







                                    share|cite|improve this answer













                                    share|cite|improve this answer



                                    share|cite|improve this answer











                                    answered 2 days ago









                                    Michael Hardy

                                    204k23185460




                                    204k23185460






















                                         

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