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onegirl

  • one year ago

Find the derivative f'(x)

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

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

  3. onegirl
    • one year ago
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    ohhhh okay

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

  5. terenzreignz
    • one year ago
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    What he said ^

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

  7. onegirl
    • one year ago
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    yea but its t

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

  9. onegirl
    • one year ago
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    okay

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

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

  13. onegirl
    • one year ago
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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

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

  15. agent0smith
    • one year ago
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    @terenzreignz

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

  17. agent0smith
    • one year ago
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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.

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

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

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

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

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

  23. terenzreignz
    • one year ago
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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 :)

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

  25. onegirl
    • one year ago
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    lol

  26. onegirl
    • one year ago
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    thanks guys!

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

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

  29. agent0smith
    • one year ago
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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)

  30. onegirl
    • one year ago
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    okay thanks for the explanation

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

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