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Math Help - derivative of an integral

  1. #1
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    derivative of an integral

    d/dx of the integral from 0 --> x^2 of sin^2t * dt

    can you guys understand that. i don't know how to do that fancy synthax.

    and the question i have about this was whether there was some kind of trick to this problem and is there a specific terminology for this type of problem so i can study up further.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    Let F(x)=\int_0^x\sin^2t\,\mathrm{d}t. Then, you know \frac{\mathrm{d}F}{\mathrm{d}x} and you're looking for \frac{\mathrm{d}F(x^2)}{\mathrm{d}x} which simply is the derivative of a composite function.

    is there a specific terminology for this type of problem
    I don't know...
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  3. #3
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    so is the answer:

    2xcos^2(x^2)

    thanks
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  4. #4
    Super Member flyingsquirrel's Avatar
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    No, it's not a cosine.
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    Quote Originally Posted by DINOCALC09 View Post
    so is the answer:

    2xcos^2(x^2)

    thanks
    First tell us what is \frac{dF}{dx}? I think you are missing this itself!
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  6. #6
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    sin^2(x^2)
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  7. #7
    Super Member flyingsquirrel's Avatar
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    No, \frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\,  \mathrm{d}t\right)=f(x) so there shouldn't have any x^2 in the expression of \frac{\mathrm{d}F}{\mathrm{d}x}
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  8. #8
    GAMMA Mathematics
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    Quote Originally Posted by flyingsquirrel View Post
    No, \frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\,  \mathrm{d}t\right)=f(x) so there shouldn't have any x^2 in the expression of \frac{\mathrm{d}F}{\mathrm{d}x}
    Exactly, think about this...

    \frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\,  \mathrm{d}t\right)

    \frac{\mathrm{d}}{\mathrm{d}x}(F(x)-F(0))

    F(0) is a constant, so the derivative is zero.

    F'(x) - c'

    f(x)
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    No, \frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\,  \mathrm{d}t\right)=f(x) so there shouldn't have any x^2 in the expression of \frac{\mathrm{d}F}{\mathrm{d}x}
    Look at the original: the upper limit is x^2.

    So the answer will be f(x^2) \cdot 2x by the chain rule.

    -Dan
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  10. #10
    Behold, the power of SARDINES!
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    Maybe this will help...

    let

    g(u)=\int_{0}^{u}h(t)dt

    by the fundemental theorem of Calculus we get

    \frac{dg}{du}=h(u)

    so for example

    f(x)=\int_{0}^{x}\tan(t)dt
    then
    \frac{df}{dx}=\tan(x)

    now back to your function

    let

    f(x)=\int_{0}^{x^2}\sin^2(t)dt

    the problem is your upper limit of integration is not linear so we need to fix it up the the FTC applies

    let g(u)=\int_{0}^{u}\sin^2(t)dt \mbox{ and } u(x)=x^2

    then f(x)=g(u)=\int_0^{x^2}\sin^2(t)dt

    now using the chain rule and the FTC we get

    \frac{df}{dx}=\frac{dg}{du}\frac{du}{dx}=(\sin^2(u  ))(2x)

    now we need to sub the value of u back in and we get

    \frac{df}{dx}=2x\sin^2(x^2)
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