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Derivatives and integrals help

  1. Dec 2, 2011 #1
    derivative of integral over e^t to t^5 (sqrt(8+x^4)) dx

    I know I need to use the chain rule and I can take the derivative of the integral without respect to e^t and t^5. If you know the answer, can you answer and tell me how to do it?! Calculus final on Monday...
     
  2. jcsd
  3. Dec 2, 2011 #2

    Ray Vickson

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    Have you never seen the formula [tex] \frac{d}{dt}\int_{A(t)}^{B(t)} f(x,t) dx =
    \frac{dB(t)}{dt} \left. f(x,t)\right|_{x=B(t)}
    - \frac{dA(t)}{dt} \left. f(x,t)\right|_{x=A(t)}
    + \int_{A(t)}^{B(t)} \frac{\partial f(x,t)}{\partial t} \, dx ? [/tex]

    RGV
     
  4. Dec 2, 2011 #3
    Umm no I have not :(
    How do I put the pieces from my integral into that?
     
  5. Dec 2, 2011 #4

    SammyS

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    Hi polulech. Welcome to PF.

    Suppose F(x) is the anti-derivative of [itex]\displaystyle \sqrt{8+x^4\,}\,.[/itex]

    Then by the fundamental theorem of calculus, [itex]\displaystyle \int_a^b\sqrt{8+x^4\,}\,dx=F(b)-F(a)\,.[/itex]

    In the case of your integral you have: [itex]\displaystyle \int_{e^t}^{\,t^5}\sqrt{8+x^4\,} \,dx=F(t^5)-F(e^t)\,.[/itex]

    You know that [itex]\displaystyle \frac{d}{dx}F(x)=\sqrt{8+x^4\,}\,. [/itex] Combine this result with the chain rule to find the derivative of your integral.
     
  6. Dec 2, 2011 #5
    how would I do that
     
  7. Dec 2, 2011 #6

    SammyS

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    Can you find [itex]\displaystyle \frac{d}{dt}F(t^5)\,,[/itex] if you know that F'(x)=√(8 + x4) ?
     
  8. Dec 2, 2011 #7
    would i plug t^5 into the √8+x^4 and then calculate the derivative?
     
  9. Dec 3, 2011 #8

    SammyS

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    No.

    F(x) is a function whose derivative is √(8+x4).

    The chain rule says that [itex]\displaystyle\frac{d}{dt}F(t^5)=F\,'(t^5)\cdot \frac{d}{dt}(t^5)\,.[/itex]

    Furthermore, [itex]\displaystyle F\,'(t^5)=\sqrt{8+t^{20}}[/itex]
     
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