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Math2400

  • one year ago

can someone explain this derivative ? i have the answer but the steps are unclear to me

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  1. Math2400
    • one year ago
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    \[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\]

  2. amistre64
    • one year ago
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    what is the fundamental thrm of calculus for integration?

  3. Math2400
    • one year ago
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    \[ \int\limits_a^b f(x)dx=F(b)-F(a)\]

  4. amistre64
    • one year ago
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    good, but lets use different notation, f= f', and F = f this helps to see the relationship between F and f in a more normal fashion. \[\int_{a}^{b}f'(x)~dx=f(b)-f(a)\] same setup, just more familiar notations what happens when we take the derivative of both sides?

  5. Math2400
    • one year ago
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    okay got that, and um you get the derivative..? lol

  6. amistre64
    • one year ago
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    whats it look like tho, we are working a generalization

  7. Math2400
    • one year ago
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    don't u have to undo the antiderivative first on the Right side?

  8. Math2400
    • one year ago
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    *left

  9. amistre64
    • one year ago
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    we already have, its the right side

  10. amistre64
    • one year ago
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    the right side is the 'undoing' of the derivative of the left side. f' comes from f

  11. Math2400
    • one year ago
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    oh okay so if i take the derivative of f(b)-f(a)?

  12. amistre64
    • one year ago
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    yep

  13. amistre64
    • one year ago
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    almost, a and b need not be constants, they can be functions, so we apply the chain rule \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=\frac d{dx}f(b)-\frac d{dx}f(a)\] \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=b'~f'(b)-a'~f'(a)\]

  14. amistre64
    • one year ago
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    notice here, that we already know what f' is ... we know a and b, therefore we know a' and b'

  15. Math2400
    • one year ago
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    ok so u u just need to get the derivatives of a and b and multiy by f'

  16. amistre64
    • one year ago
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    \[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\] f'(a) = cos(3(0)+1) f'(b) = cos(3x^5+1) a(x) = 0, a' = 0 b(x) = x^5, b' = 5x^4

  17. amistre64
    • one year ago
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    you have all the parts required to determine the results, without havig to intgrate first, and then take the derivative ... its a shortcut if you will

  18. Math2400
    • one year ago
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    so when i re-write it into the equ. i use f'(a) and f'(b) instead of putting the original....so there should be no "t" variable

  19. amistre64
    • one year ago
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    correct, the a and b are in the domain of t in this case.

  20. Math2400
    • one year ago
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    i got it actually!

  21. Math2400
    • one year ago
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    thank u so much :)

  22. amistre64
    • one year ago
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    youre welcome

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