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onegirl

  • 2 years ago

Find the derivative f'(x)

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

  2. agent0smith
    • 2 years 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
    • 2 years ago
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    ohhhh okay

  4. agent0smith
    • 2 years 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
    • 2 years ago
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    What he said ^

  6. terenzreignz
    • 2 years 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
    • 2 years ago
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    yea but its t

  8. agent0smith
    • 2 years 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
    • 2 years ago
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    okay

  10. terenzreignz
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay and that is what my original problem had a t but okay thanks!

  13. onegirl
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @terenzreignz

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

  17. agent0smith
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Hmm..i'll check on with my teacher on this one..

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

  23. terenzreignz
    • 2 years 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
    • 2 years ago
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    Yeah i made a few above, with all the damn x's and t's...

  25. onegirl
    • 2 years ago
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    lol

  26. onegirl
    • 2 years ago
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    thanks guys!

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

  28. onegirl
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay thanks for the explanation

  31. agent0smith
    • 2 years 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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