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onegirl

Find the derivative f'(x)

  • 11 months ago
  • 11 months ago

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  1. onegirl
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    I got - integral sign ln (t^2 + 1) is that correct?

    • 11 months ago
  2. agent0smith
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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 \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]

    • 11 months ago
  3. onegirl
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    ohhhh okay

    • 11 months ago
  4. agent0smith
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    Wait, is this finding f'(x)? That t limit... is that meant to be x? This would just be zero.

    • 11 months ago
  5. terenzreignz
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    What he said ^

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

    • 11 months ago
  7. onegirl
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    yea but its t

    • 11 months ago
  8. agent0smith
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    \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)\]

    • 11 months ago
  9. onegirl
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    okay

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

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

    • 11 months ago
  12. onegirl
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    okay and that is what my original problem had a t but okay thanks!

    • 11 months ago
  13. onegirl
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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

    • 11 months ago
  14. agent0smith
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    Unless it's \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =\] \[ \Large = -\frac{ dt }{ dx }\ln (t^2+1)\]

    • 11 months ago
  15. agent0smith
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    @terenzreignz

    • 11 months ago
  16. terenzreignz
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    implicit? It's tempting, but I'm more inclined to believe a typo.

    • 11 months ago
  17. agent0smith
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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.

    • 11 months ago
  18. terenzreignz
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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... \[\Large \frac{ d }{ dx } -\int\limits\limits\limits_{-1}^{x}\ln (t^2 +1) dt =\]

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

    • 11 months ago
  20. terenzreignz
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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... \[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\] \[\Large -\ln(t^2+1)\]

    • 11 months ago
  21. onegirl
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    Hmm..i'll check on with my teacher on this one..

    • 11 months ago
  22. agent0smith
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    @terenzreignz shouldn't it be an x^2+1 in your last step?

    • 11 months ago
  23. terenzreignz
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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 :)

    • 11 months ago
  24. agent0smith
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    Yeah i made a few above, with all the damn x's and t's...

    • 11 months ago
  25. onegirl
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    lol

    • 11 months ago
  26. onegirl
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    thanks guys!

    • 11 months ago
  27. terenzreignz
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    I only did go online for a few giggles... lol ----------------------------------------- Terence out

    • 11 months ago
  28. onegirl
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    so @agent0smith you think its a typo? and that it should be with an x and not a t?

    • 11 months ago
  29. agent0smith
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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)

    • 11 months ago
  30. onegirl
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    okay thanks for the explanation

    • 11 months ago
  31. agent0smith
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    No prob. But also @terenzreignz i guess it makes sense to switch limits and change the sign too.

    • 11 months ago
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