Switching Orders of Differentiation and Integration in case of Function of 2 Variables












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I encountered the following equation in my calculus lecture notes:
$$
frac{d}{dx}int_0^xf(x,y)dy = f(x,x)+int_0^xfrac{partial f}{partial x}(x,y)dy
$$

Could you show me how to prove the above equality?










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    $begingroup$


    I encountered the following equation in my calculus lecture notes:
    $$
    frac{d}{dx}int_0^xf(x,y)dy = f(x,x)+int_0^xfrac{partial f}{partial x}(x,y)dy
    $$

    Could you show me how to prove the above equality?










    share|cite|improve this question









    $endgroup$















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      1








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      1



      $begingroup$


      I encountered the following equation in my calculus lecture notes:
      $$
      frac{d}{dx}int_0^xf(x,y)dy = f(x,x)+int_0^xfrac{partial f}{partial x}(x,y)dy
      $$

      Could you show me how to prove the above equality?










      share|cite|improve this question









      $endgroup$




      I encountered the following equation in my calculus lecture notes:
      $$
      frac{d}{dx}int_0^xf(x,y)dy = f(x,x)+int_0^xfrac{partial f}{partial x}(x,y)dy
      $$

      Could you show me how to prove the above equality?







      multivariable-calculus






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      asked Jan 25 at 17:39









      A Slow LearnerA Slow Learner

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          This is a particular version of what is known as the Leibniz Integral Rule. The wikipedia link provides a nice proof of the general case(s), and it is not too hard to see how this particular form follows






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            $begingroup$

            This is a particular version of what is known as the Leibniz Integral Rule. The wikipedia link provides a nice proof of the general case(s), and it is not too hard to see how this particular form follows






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is a particular version of what is known as the Leibniz Integral Rule. The wikipedia link provides a nice proof of the general case(s), and it is not too hard to see how this particular form follows






              share|cite|improve this answer









              $endgroup$
















                0












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                0





                $begingroup$

                This is a particular version of what is known as the Leibniz Integral Rule. The wikipedia link provides a nice proof of the general case(s), and it is not too hard to see how this particular form follows






                share|cite|improve this answer









                $endgroup$



                This is a particular version of what is known as the Leibniz Integral Rule. The wikipedia link provides a nice proof of the general case(s), and it is not too hard to see how this particular form follows







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 18:53









                Marat AlievMarat Aliev

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