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How to apply the fundamental theorem to partial derivatives?

  1. Jul 21, 2016 #1
    1. The problem statement, all variables and given/known data
    "Under mild continuity restrictions, it is true that if ##F(x)=\int_a^b g(t,x)dt##,

    then ##F'(x)=\int_a^b g_x(t,x)dt##.

    Using this fact and the Chain Rule, we can find the derivative of

    ##F(x)=\int_{a}^{f(x)} g(t,x)dt##

    by letting

    ##G(u,x)=\int_a^u g(t,x)dt##,

    where ##u=f(x)##."

    Given:

    ##F(x)=\int_{0}^{x^2} \sqrt{t^4+x^3}dt##,

    find ##\frac{dF}{dx}##.

    2. Relevant equations
    ##\frac{dF}{dx}=\frac{dF}{dt}\frac{dt}{dx}+\frac{dF}{du}\frac{du}{dx}##

    3. The attempt at a solution
    ##\frac{dF}{du}=\frac{d}{du} \int_{0}^{x^2}\sqrt{t^4+x^3}dt=\sqrt{(x^2)^4+x^3}=\sqrt{x^8+x^3}##
    ##\frac{du}{dx}=\frac{d}{dx}(x^2)=2x## by letting ##u=x^2##

    ##\frac{dF}{dt}=F'(t)=\int_{0}^{x^2} g_x(t^4+x^3)dt=\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}##

    I know that: ##\frac{dF}{dx}=2x\sqrt{x^8+x^3}+\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}\frac{dt}{dx}##

    But, I'm at a loss at how to calculate: ##\frac{dt}{dx}##, because I don't know how to define ##t##, so I don't know how to differentiate it with respect to ##x##.
     
  2. jcsd
  3. Jul 22, 2016 #2

    Delta²

    User Avatar
    Gold Member

    in short the way you doing it is t=x but you cant replace that inside the integral. In short the result is
    $$2x\sqrt{x^8+x^3}+\int_{0}^{x^2}\frac{3x^2}{2\sqrt{t^4+x3}}dt$$.

    However in order to be more accurate, we have to define ##F(x)=G(u(x),x)## with ##G(u,v)=\int_0^{u}\sqrt{t^4+v^3}dt## therefore

    $$\frac{dF}{dx}=\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}$$.

    with ##u(x)=x^2## ##v(x)=x##.

    Calculating the above expression you ll get the result.
     
  4. Jul 22, 2016 #3
    So ##\frac{∂G}{∂v}\frac{∂v}{dx}## corresponds to ##\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}## and ##\frac{∂G}{∂u}\frac{∂u}{∂x}## corresponds to ##2x\sqrt{x^8+x^3}##? And ##v'(x)=1##, so it's just ##\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}+2x\sqrt{x^8+x^3}##. I guess that makes sense; I mean, I just need to perform the proper substitutions and such. Anyway, thank you for your help, Delta.
     
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