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anonymous

  • 5 years ago

Finding the derivative...

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  1. anonymous
    • 5 years ago
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    h(x)=\[\int\limits_{sinx}^{1}\ln(t^2)dt\]

  2. anonymous
    • 5 years ago
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    oh hey girl

  3. anonymous
    • 5 years ago
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    help!!!

  4. anonymous
    • 5 years ago
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    I need your help Sapph! :)

  5. anonymous
    • 5 years ago
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    you want the derivative of the integral or the integral itself?

  6. anonymous
    • 5 years ago
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    Umm.. I'm not sure.. What's the difference?

  7. anonymous
    • 5 years ago
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    we just need the derivative i believe

  8. anonymous
    • 5 years ago
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    I think it's asking about the derivative

  9. anonymous
    • 5 years ago
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    Yes, it's the derivative it's asking for.

  10. anonymous
    • 5 years ago
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    using the Fundamental theorem of calculus: \[d/dt \int\limits\limits_{\sin x}^{1} \ln (t^2) dt = - \cos x . \ln(\sin^2 x)\]

  11. anonymous
    • 5 years ago
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    Can you sort of explain this...?

  12. anonymous
    • 5 years ago
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    ok here is a general formula: \[d/dx \int\limits_{g(x)}^{h(x)}f(t) dt = h'(x) f(h(x)) - g'(x) f(g(x))\] I hope that does not complicate it more

  13. anonymous
    • 5 years ago
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    does that make sense?

  14. anonymous
    • 5 years ago
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    I suck at explaining things :(

  15. anonymous
    • 5 years ago
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    Hm.. So it's sort of like the product rule?

  16. anonymous
    • 5 years ago
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    hmm not really.. you just take the derivative of the upper border multiplied by the function after substituting in t minus the same thing with lower border

  17. anonymous
    • 5 years ago
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    Alright well that makes sense :) Thank you :)

  18. anonymous
    • 5 years ago
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    I'm trying to work this out myself.. And well.. I can't seem to get the answer. For the first part I get 0 times f(1)... minus cosx.. How do you get ln?

  19. anonymous
    • 5 years ago
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    it's the same ln in the integral, I just put sinx in the place of t

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