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anonymous

  • one year ago

find F'(x) when F(x)= integral from 4 to x^2 (2 sqrt(1+t^3) dt)

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  1. freckles
    • one year ago
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    \[F(x)=\int\limits_4^{x^2} g(t) dt=G(t)|_4^{x^2}=G(x^2)-G(4) \\ \text{so we have } F(x)=G(x^2)-G(4) \\ \text{ differentiate both sides }\]

  2. freckles
    • one year ago
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    \[F'(x)=(x^2)'g(x^2)-0 \text{ by chain rule and constant rule }\]

  3. freckles
    • one year ago
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    @aawowaa are you there?

  4. IrishBoy123
    • one year ago
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    +

  5. anonymous
    • one year ago
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    yes, I'm going over it.

  6. freckles
    • one year ago
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    you understand that g(t) is 2sqrt(1+t^3)

  7. freckles
    • one year ago
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    And that I was using G(t) to represent the antiderivative of g(t) that is G'=g

  8. anonymous
    • one year ago
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    yes, I understand

  9. anonymous
    • one year ago
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    so now I plug in 2sqrt(1+t^3) back?

  10. freckles
    • one year ago
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    g(t)=2sqrt(1+t^3) so g(x^2)=?"

  11. anonymous
    • one year ago
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    =2sqrt(1+x^6)

  12. freckles
    • one year ago
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    ok so that is what you replace g(x^2) with

  13. freckles
    • one year ago
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    don't forget to find (x^2)' which is next to g(x^2)

  14. anonymous
    • one year ago
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    thank you I got it

  15. freckles
    • one year ago
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    np

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