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onegirl
 one year ago
Best ResponseYou've already chosen the best response.0I got  integral sign ln (t^2 + 1) is that correct?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2^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 \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Wait, is this finding f'(x)? That t limit... is that meant to be x? This would just be zero.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Much more interesting if it were an x instead... here... \[\Large \frac{ d }{ dx } \int\limits\limits_{\color{red}x}^{1}\ln (t^2 +1) dt =\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1perhaps something like this? \[\Large \frac{ d }{ d\color{red}t } \int\limits\limits_{t}^{1}\ln (t^2 +1) dt =\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2@onegirl that one with t will just be zero. Since you're basically taking the derivative of a constant... zero.

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0okay and that is what my original problem had a t but okay thanks!

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0thats 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

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Unless it's \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =\] \[ \Large = \frac{ dt }{ dx }\ln (t^2+1)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1implicit? It's tempting, but I'm more inclined to believe a typo.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1How 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... \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{1}^{x}\ln (t^2 +1) dt =\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2If it is an x then \[\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{x}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (x) \right] \ln (x^2+1)\] \[\Large =  \ln (x^2+1)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1Of 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... \[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\] \[\Large \ln(t^2+1)\]

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0Hmm..i'll check on with my teacher on this one..

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2@terenzreignz shouldn't it be an x^2+1 in your last step?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1cr*p typos are contagious :D should really stop relying on copypaste XD \[\Large \ln(x^2+1)\] Thanks for pointing that out :)

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Yeah i made a few above, with all the damn x's and t's...

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.1I only did go online for a few giggles... lol  Terence out

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0so @agent0smith you think its a typo? and that it should be with an x and not a t?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2Probably, 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)

onegirl
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks for the explanation

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.2No prob. But also @terenzreignz i guess it makes sense to switch limits and change the sign too.
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