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onegirl Group Title

Find the derivative f'(x)

  • one year ago
  • one year ago

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  1. onegirl Group Title
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    http://assets.openstudy.com/updates/attachments/51858a78e4b0bc9e008a60b4-onegirl-1367706243068-untitled10.jpg

    • one year ago
  2. onegirl Group Title
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    I got - integral sign ln (t^2 + 1) is that correct?

    • one year ago
  3. agent0smith Group Title
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    ^Close Similar to the last one, except with two limits... similar process though. The derivative cancels out the integral, then you just basically plug in the limits and differentiate the limits \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]

    • one year ago
  4. onegirl Group Title
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    ohhhh okay

    • one year ago
  5. agent0smith Group Title
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    Wait, is this finding f'(x)? That t limit... is that meant to be x? This would just be zero.

    • one year ago
  6. terenzreignz Group Title
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    What he said ^

    • one year ago
  7. terenzreignz Group Title
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    Much more interesting if it were an x instead... here... \[\Large \frac{ d }{ dx } \int\limits\limits_{\color{red}x}^{-1}\ln (t^2 +1) dt =\]

    • one year ago
  8. onegirl Group Title
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    yea but its t

    • one year ago
  9. agent0smith Group Title
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    \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)\]

    • one year ago
  10. onegirl Group Title
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    okay

    • one year ago
  11. terenzreignz Group Title
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    perhaps something like this? \[\Large \frac{ d }{ d\color{red}t } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\]

    • one year ago
  12. agent0smith Group Title
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    @onegirl that one with t will just be zero. Since you're basically taking the derivative of a constant... zero.

    • one year ago
  13. onegirl Group Title
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    okay and that is what my original problem had a t but okay thanks!

    • one year ago
  14. onegirl Group Title
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    thats what confused me when i was looking at this video he was substituting x and i had a t so i didn't know what to do http://www.youtube.com/watch?v=PGmVvIglZx8

    • one year ago
  15. agent0smith Group Title
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    Unless it's \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =\] \[ \Large = -\frac{ dt }{ dx }\ln (t^2+1)\]

    • one year ago
  16. agent0smith Group Title
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    @terenzreignz

    • one year ago
  17. terenzreignz Group Title
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    implicit? It's tempting, but I'm more inclined to believe a typo.

    • one year ago
  18. agent0smith Group Title
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    Yeah, i've only ever seen these types when one of the limits is an x, and it's dt. Not a t when it's dt.

    • one year ago
  19. terenzreignz Group Title
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    How much simpler this would be were it an x as the lower limit instead \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] Switch the limits, and put a negative sign... \[\Large \frac{ d }{ dx } -\int\limits\limits\limits_{-1}^{x}\ln (t^2 +1) dt =\]

    • one year ago
  20. agent0smith Group Title
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    If it is an x then \[\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (x^2+1)\] \[\Large = - \ln (x^2+1)\]

    • one year ago
  21. terenzreignz Group Title
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    Of course, the negative sign may cross the dy/dx, the derivative of a negative is the negative of the derivative \[\Large -\frac{ d }{ dx } \int\limits\limits\limits_{-1}^{x}\ln (t^2 +1) dt =\] And here, you can just go ahead and use the first fundamental theorem of Calculus... \[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\] \[\Large -\ln(t^2+1)\]

    • one year ago
  22. onegirl Group Title
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    Hmm..i'll check on with my teacher on this one..

    • one year ago
  23. agent0smith Group Title
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    @terenzreignz shouldn't it be an x^2+1 in your last step?

    • one year ago
  24. terenzreignz Group Title
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    cr*p typos are contagious :D should really stop relying on copy-paste XD \[\Large -\ln(x^2+1)\] Thanks for pointing that out :)

    • one year ago
  25. agent0smith Group Title
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    Yeah i made a few above, with all the damn x's and t's...

    • one year ago
  26. onegirl Group Title
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    lol

    • one year ago
  27. onegirl Group Title
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    thanks guys!

    • one year ago
  28. terenzreignz Group Title
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    I only did go online for a few giggles... lol ----------------------------------------- Terence out

    • one year ago
  29. onegirl Group Title
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    so @agent0smith you think its a typo? and that it should be with an x and not a t?

    • one year ago
  30. agent0smith Group Title
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    Probably, cos it just doesn't look right with a t. But @terenzreignz you don't need to have the x as a lower limit and pull out negatives etc... the way i showed gets the same result (and you can shortcut it since you know the derivative of the upper limit will be zero, leaving 0 - ln(x^2+1) since x was the lower limit)

    • one year ago
  31. onegirl Group Title
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    okay thanks for the explanation

    • one year ago
  32. agent0smith Group Title
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    No prob. But also @terenzreignz i guess it makes sense to switch limits and change the sign too.

    • one year ago
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