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krypton

  • 2 years ago

y=d/dx integralcosx to 0 t/(1+t)dt

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  1. krypton
    • 2 years ago
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    \[d/dx \int\limits_{0}^{cosx} \frac{ t }{ 1+t } dt\]

  2. Outkast3r09
    • 2 years ago
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    \[\frac{cosx}{1+cosx}\]

  3. krypton
    • 2 years ago
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    how did u do it?

  4. Zekarias
    • 2 years ago
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    @Outkast3r09 I think u made a miss take in your first step (While u made the integral)

  5. Outkast3r09
    • 2 years ago
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    the derivative of an integral is simply the integral itself with the variables in other words ... integrals and derivatives cancel eachother out

  6. Outkast3r09
    • 2 years ago
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    \[\frac{cosx}{1+cosx} \]is your answer

  7. Zekarias
    • 2 years ago
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    It should be\[\cos(x)-\ln(1+\cos(x))\]

  8. Outkast3r09
    • 2 years ago
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    |dw:1349804491261:dw|

  9. Zekarias
    • 2 years ago
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    Now take the derivative

  10. Outkast3r09
    • 2 years ago
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    it shouldn't be anything @ zekarias

  11. Outkast3r09
    • 2 years ago
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    @zekarias ... you don't need to do anything

  12. Zekarias
    • 2 years ago
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    What u get finally?

  13. krypton
    • 2 years ago
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    so the answer is just cosx/1+cosx ?

  14. Outkast3r09
    • 2 years ago
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    yes

  15. krypton
    • 2 years ago
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    thanks,wolfram gives me the same answer,just wanted to make sure how it was done

  16. Outkast3r09
    • 2 years ago
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    it's just simply knowing how an integral work... there is a lot of work to do it zekarias way but if you just know that the the derivative on an integral will bring you back to the f(x) within the integral... that's all you need to know(This works anytime the lower limit is some real number) because if you think about it.... |dw:1349804918350:dw| if you take the derivative of f(n)... you'll always get zero

  17. krypton
    • 2 years ago
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    oh thanks alot sweedy :)

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