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Thread: [SOLVED] Derivatives of integrals

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Derivatives of integrals

    I "need" you, MHF.
    I would like you to verify the answers I got (I can't trust Mathematica) and help me with the 2 last problems.
    Find the derivatives of the following integrals :
    1)$\displaystyle F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}$. I got $\displaystyle F'(x)=-\frac{1}{1+x^2+sin^2x}$.
    2)$\displaystyle F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}$. I got $\displaystyle F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3(x)dt)+(\int_1^x sin^3(x)dt)^2}$.
    3)$\displaystyle F(x)=\int_a^x xf(t)dt$. I don't know how to start this one.
    4)$\displaystyle F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy)$. I am not able to do this one. But I have started something : I wrote $\displaystyle F$ as a composition of 3 functions, that is $\displaystyle f \circ g \circ h$.
    $\displaystyle f(g)=sin(g)$.
    $\displaystyle g(h)=\int_0^x sin(h)dy$.
    $\displaystyle h(y)=\int_0^y sin^3(t)dt$.
    $\displaystyle f'(g)=cos(g)$
    $\displaystyle g'(h)=$... I couldn't find it. I guess I'm already wrong writing $\displaystyle g(h)$. All those variables confuse me a lot!
    $\displaystyle h'(y)=sin^3(y)$.
    I wanted to use the chain rule for this one. $\displaystyle F'(x)$ would be $\displaystyle (f' \circ g *g' \circ h * h')(x)$. I'm not even sure if this is right.
    Also, I may post one more derivative of an integral.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi arbolis

    Quote Originally Posted by arbolis View Post
    Find the derivatives of the following integrals :
    1)$\displaystyle F(x)=\int_x^b \frac{dt}{1+t^2+sin^2(t)}$. I got $\displaystyle F'(x)=-\frac{1}{1+x^2+sin^2x}$.
    OK
    2)$\displaystyle F(x)=\int_3^{\int_1^x sin^3(t)dt} \frac{du}{1+sin^6(u)+u^2}$. I got $\displaystyle F'(x)=\frac{sin^3x}{1+sin^6(\int_1^x sin^3({\color{red}t})dt)+(\int_1^x sin^3({\color{red}t})dt)^2}$.
    OK

    (By the way, $\displaystyle \int_1^x\sin^3t\,\mathrm{d}t$ can be computed using $\displaystyle \sin^3x=\sin ^2x \sin x=\sin x -\cos ^2x \sin x$)

    3)$\displaystyle F(x)=\int_a^x xf(t)dt$. I don't know how to start this one.
    The integration is done with respect to $\displaystyle t$ so one can write $\displaystyle F(x)=x\int_a^xf(t)\,\mathrm{d}t$ and differentiate using the product rule. ($\displaystyle u(x)=x,\,v(x)=\int_a^xf(t)\,\mathrm{d}t$)

    4)$\displaystyle F(x)=sin(\int_0^x sin(\int_0^y sin^3(t)dt)dy)$. I am not able to do this one. But I have started something : I wrote $\displaystyle F$ as a composition of 3 functions, that is $\displaystyle f \circ g \circ h$.
    $\displaystyle f(g)=sin(g)$.
    $\displaystyle g(h)=\int_0^x sin(h)dy$.
    $\displaystyle h(y)=\int_0^y sin^3(t)dt$.
    $\displaystyle f'(g)=cos(g)$
    $\displaystyle g'(h)=$... I couldn't find it. I guess I'm already wrong writing $\displaystyle g(h)$. All those variables confuse me a lot!
    $\displaystyle h'(y)=sin^3(y)$.
    I wanted to use the chain rule for this one. $\displaystyle F'(x)$ would be $\displaystyle (f' \circ g *g' \circ h * h')(x)$. I'm not even sure if this is right.
    $\displaystyle F(x)=\sin\left(\int_0^x \sin\left[\int_0^y \sin^3 t\, \mathrm{d}t\right]\,\mathrm{d}y\right)$

    Using composite functions is the right idea but you can't define $\displaystyle g(h)=\int_0^x \sin h \,\mathrm{d}y$ since the integration is done with respect to $\displaystyle y$ and $\displaystyle h$ depends on $\displaystyle y$. You'll have to use $\displaystyle g(x)= \int_0^x \sin\left[\int_0^y \sin^3(t)\,\mathrm{d}t\right]\,\mathrm{d}y $. This gives $\displaystyle F(x)=f\circ g(x)$ hence $\displaystyle F'(x)=g'(x)\cdot f'\circ g(x)$. (chain rule)

    You've already found $\displaystyle f'$ but what is $\displaystyle g'$ ? Remember that $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left[ \int_0^x u(y)\,\mathrm{d}y\right]=u(x)$. What does $\displaystyle u(y)$ correspond to here ?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Hello Écureuil volant!
    Oops, sorry for the 2), I mistyped, so I got the right answer (on my sheet).
    For the 3),
    The integration is done with respect to so one can write and differentiate using the product rule. ()
    Thanks! I was very confused. I understand better now. So $\displaystyle F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t)$.
    For the 4),
    What does correspond to here ?
    It corresponds to $\displaystyle sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big )$.
    So $\displaystyle F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big) $? Or I am confused again about the variables?
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  4. #4
    MHF Contributor arbolis's Avatar
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    And finally the last one :$\displaystyle F(x)=\int_a^b \frac{x}{1+t^2+sin^2t} dt$. I considered x as a constant when under the integral since it's a dt one, as you just teached me. I got $\displaystyle F'(x)=\int_a^b \frac{dt}{1+t^2+sin^2t}$. Hope it's right.
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by arbolis View Post
    Hello Écureuil volant!

    For the 3), Thanks! I was very confused. I understand better now. So $\displaystyle F'(x)=\int_a^xf(t)\,\mathrm{d}t+x^2f(t)$.
    You meant $\displaystyle F'(x)=\int_a^x f(t)\,\mathrm{d}t+{\color{red}xf(x)}$ didn't you ?
    For the 4), It corresponds to $\displaystyle sin \big (\int_0^y \sin^3 t\, \mathrm{d}t\big )$.
    So $\displaystyle F'(x)=cos \left( \int_0^x \sin\left[\int_0^y \sin^3 (t) dt\right] dy\right) \cdot \sin \big( \int_0^x \sin^3(t) dt \big) $?
    That's it

    EDIT :
    Quote Originally Posted by arbolis View Post
    And finally the last one [...] Hope it's right.
    It is.
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  6. #6
    MHF Contributor arbolis's Avatar
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    You meant didn't you ?
    Yes sorry about that, I did it mentally and probably too fast as we can see. Thanks a lot for your precious help/time.
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