If $F(x,y)=int_{f(x,y)}^{g(x,y)}h(x,y,t), dt$, then find $partial F/partial x$.












2















Let $f,g:mathbb{R}^2to mathbb{R}$ and $h:mathbb{R}^3tomathbb{R}$ be $C^1$, define $F(x,y)=int_{f(x,y)}^{g(x,y)}h(x,y,t), dt$. Find $frac{partial F}{partial x}$.




So I split $$F(x,y)=int_0^{g(x,y)}h(x,y,t), dt-int_0^{f(x,y)}h(x,y,t), dt.$$



And then using FTC I believe I have $$F(x,y)=H(x,y,g(x,y))-H(x,y,0)-H(x,y,f(x,y))+H(x,y,0)=H(x,y,g(x,y))-H(x,y,f(x,y))$$



Then differentiating this new thing with respect to x.



$$frac{partial F}{partial x}=frac{partial H}{partial x}+frac{partial H}{partial y}frac{partial y}{partial x}+frac{partial H}{partial g(x,y)}frac{partial g(x,y)}{partial x}-frac{partial H}{partial x}-frac{partial H}{partial y}frac{partial y}{partial x}-frac{partial H}{partial f(x,y)}frac{partial f(x,y)}{partial x}$$



Is this correct? From here I would simplify the expression and expand out the partials. But I'm not sure that the FTC works in this way for $H$, since it's a function of 3 variables.






multivariable-calculus proof-verification







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  • As $y$ is a free variable, the 2nd and 5th terms are zero.
    – Rafa Budría
    Nov 28 '18 at 16:40
















2















Let $f,g:mathbb{R}^2to mathbb{R}$ and $h:mathbb{R}^3tomathbb{R}$ be $C^1$, define $F(x,y)=int_{f(x,y)}^{g(x,y)}h(x,y,t), dt$. Find $frac{partial F}{partial x}$.




So I split $$F(x,y)=int_0^{g(x,y)}h(x,y,t), dt-int_0^{f(x,y)}h(x,y,t), dt.$$



And then using FTC I believe I have $$F(x,y)=H(x,y,g(x,y))-H(x,y,0)-H(x,y,f(x,y))+H(x,y,0)=H(x,y,g(x,y))-H(x,y,f(x,y))$$



Then differentiating this new thing with respect to x.



$$frac{partial F}{partial x}=frac{partial H}{partial x}+frac{partial H}{partial y}frac{partial y}{partial x}+frac{partial H}{partial g(x,y)}frac{partial g(x,y)}{partial x}-frac{partial H}{partial x}-frac{partial H}{partial y}frac{partial y}{partial x}-frac{partial H}{partial f(x,y)}frac{partial f(x,y)}{partial x}$$



Is this correct? From here I would simplify the expression and expand out the partials. But I'm not sure that the FTC works in this way for $H$, since it's a function of 3 variables.






multivariable-calculus proof-verification







share|cite|improve this question
























  • As $y$ is a free variable, the 2nd and 5th terms are zero.
    – Rafa Budría
    Nov 28 '18 at 16:40














2












2








2








Let $f,g:mathbb{R}^2to mathbb{R}$ and $h:mathbb{R}^3tomathbb{R}$ be $C^1$, define $F(x,y)=int_{f(x,y)}^{g(x,y)}h(x,y,t), dt$. Find $frac{partial F}{partial x}$.




So I split $$F(x,y)=int_0^{g(x,y)}h(x,y,t), dt-int_0^{f(x,y)}h(x,y,t), dt.$$



And then using FTC I believe I have $$F(x,y)=H(x,y,g(x,y))-H(x,y,0)-H(x,y,f(x,y))+H(x,y,0)=H(x,y,g(x,y))-H(x,y,f(x,y))$$



Then differentiating this new thing with respect to x.



$$frac{partial F}{partial x}=frac{partial H}{partial x}+frac{partial H}{partial y}frac{partial y}{partial x}+frac{partial H}{partial g(x,y)}frac{partial g(x,y)}{partial x}-frac{partial H}{partial x}-frac{partial H}{partial y}frac{partial y}{partial x}-frac{partial H}{partial f(x,y)}frac{partial f(x,y)}{partial x}$$



Is this correct? From here I would simplify the expression and expand out the partials. But I'm not sure that the FTC works in this way for $H$, since it's a function of 3 variables.






multivariable-calculus proof-verification







share|cite|improve this question
















Let $f,g:mathbb{R}^2to mathbb{R}$ and $h:mathbb{R}^3tomathbb{R}$ be $C^1$, define $F(x,y)=int_{f(x,y)}^{g(x,y)}h(x,y,t), dt$. Find $frac{partial F}{partial x}$.




So I split $$F(x,y)=int_0^{g(x,y)}h(x,y,t), dt-int_0^{f(x,y)}h(x,y,t), dt.$$



And then using FTC I believe I have $$F(x,y)=H(x,y,g(x,y))-H(x,y,0)-H(x,y,f(x,y))+H(x,y,0)=H(x,y,g(x,y))-H(x,y,f(x,y))$$



Then differentiating this new thing with respect to x.



$$frac{partial F}{partial x}=frac{partial H}{partial x}+frac{partial H}{partial y}frac{partial y}{partial x}+frac{partial H}{partial g(x,y)}frac{partial g(x,y)}{partial x}-frac{partial H}{partial x}-frac{partial H}{partial y}frac{partial y}{partial x}-frac{partial H}{partial f(x,y)}frac{partial f(x,y)}{partial x}$$



Is this correct? From here I would simplify the expression and expand out the partials. But I'm not sure that the FTC works in this way for $H$, since it's a function of 3 variables.






multivariable-calculus proof-verification




multivariable-calculus proof-verification






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edited Nov 28 '18 at 19:08









Brahadeesh

6,13042361




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asked Nov 28 '18 at 14:53









AColoredReptile

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  • As $y$ is a free variable, the 2nd and 5th terms are zero.
    – Rafa Budría
    Nov 28 '18 at 16:40


















  • As $y$ is a free variable, the 2nd and 5th terms are zero.
    – Rafa Budría
    Nov 28 '18 at 16:40
















As $y$ is a free variable, the 2nd and 5th terms are zero.
– Rafa Budría
Nov 28 '18 at 16:40




As $y$ is a free variable, the 2nd and 5th terms are zero.
– Rafa Budría
Nov 28 '18 at 16:40










1 Answer
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Your approach is correct.



Since $h : mathbb{R}^3 to mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y in mathbb{R}$ the function $t mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
$$
int_{f(x,y)}^{g(x,y)} h(x,y,t), dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
$$



The chain rule you have used for computing the partial derivative $partial H / partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $partial y / partial x$ can be removed as $partial y / partial x = 0$. So your final answer should be
$$
frac{partial F}{partial x} = frac{partial H}{partial g} frac{partial g}{partial x} - frac{partial H}{partial f} frac{partial f}{partial x}.
$$



Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $partial H / partial t$ in place of $partial H / partial f$ and $partial H / partial g$ would be an improvement.






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    Your approach is correct.



    Since $h : mathbb{R}^3 to mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y in mathbb{R}$ the function $t mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
    $$
    int_{f(x,y)}^{g(x,y)} h(x,y,t), dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
    $$



    The chain rule you have used for computing the partial derivative $partial H / partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $partial y / partial x$ can be removed as $partial y / partial x = 0$. So your final answer should be
    $$
    frac{partial F}{partial x} = frac{partial H}{partial g} frac{partial g}{partial x} - frac{partial H}{partial f} frac{partial f}{partial x}.
    $$



    Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $partial H / partial t$ in place of $partial H / partial f$ and $partial H / partial g$ would be an improvement.






    share|cite|improve this answer


























      1














      Your approach is correct.



      Since $h : mathbb{R}^3 to mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y in mathbb{R}$ the function $t mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
      $$
      int_{f(x,y)}^{g(x,y)} h(x,y,t), dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
      $$



      The chain rule you have used for computing the partial derivative $partial H / partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $partial y / partial x$ can be removed as $partial y / partial x = 0$. So your final answer should be
      $$
      frac{partial F}{partial x} = frac{partial H}{partial g} frac{partial g}{partial x} - frac{partial H}{partial f} frac{partial f}{partial x}.
      $$



      Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $partial H / partial t$ in place of $partial H / partial f$ and $partial H / partial g$ would be an improvement.






      share|cite|improve this answer
























        1












        1








        1






        Your approach is correct.



        Since $h : mathbb{R}^3 to mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y in mathbb{R}$ the function $t mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
        $$
        int_{f(x,y)}^{g(x,y)} h(x,y,t), dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
        $$



        The chain rule you have used for computing the partial derivative $partial H / partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $partial y / partial x$ can be removed as $partial y / partial x = 0$. So your final answer should be
        $$
        frac{partial F}{partial x} = frac{partial H}{partial g} frac{partial g}{partial x} - frac{partial H}{partial f} frac{partial f}{partial x}.
        $$



        Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $partial H / partial t$ in place of $partial H / partial f$ and $partial H / partial g$ would be an improvement.






        share|cite|improve this answer












        Your approach is correct.



        Since $h : mathbb{R}^3 to mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y in mathbb{R}$ the function $t mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
        $$
        int_{f(x,y)}^{g(x,y)} h(x,y,t), dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
        $$



        The chain rule you have used for computing the partial derivative $partial H / partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $partial y / partial x$ can be removed as $partial y / partial x = 0$. So your final answer should be
        $$
        frac{partial F}{partial x} = frac{partial H}{partial g} frac{partial g}{partial x} - frac{partial H}{partial f} frac{partial f}{partial x}.
        $$



        Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $partial H / partial t$ in place of $partial H / partial f$ and $partial H / partial g$ would be an improvement.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 19:05









        Brahadeesh

        6,13042361




        6,13042361






























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