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

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

onegirl Group TitleBest ResponseYou've already chosen the best response.0
ohhhh okay
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
What he said ^
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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

onegirl Group TitleBest ResponseYou've already chosen the best response.0
yea but its t
 one year ago

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

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

agent0smith Group TitleBest 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.
 one year ago

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

onegirl Group TitleBest ResponseYou've already chosen the best response.0
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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
@terenzreignz
 one year ago

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
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

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

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

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

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

onegirl Group TitleBest ResponseYou've already chosen the best response.0
thanks guys!
 one year ago

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

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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

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

agent0smith Group TitleBest ResponseYou've already chosen the best response.2
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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