Here's the question you clicked on:
onegirl
Find the derivative f'(x)
I got - integral sign ln (t^2 + 1) is that correct?
^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 \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]
Wait, is this finding f'(x)? That t limit... is that meant to be x? This would just be zero.
Much more interesting if it were an x instead... here... \[\Large \frac{ d }{ dx } \int\limits\limits_{\color{red}x}^{-1}\ln (t^2 +1) dt =\]
\[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)\]
perhaps something like this? \[\Large \frac{ d }{ d\color{red}t } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\]
@onegirl that one with t will just be zero. Since you're basically taking the derivative of a constant... zero.
okay and that is what my original problem had a t but okay thanks!
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
Unless it's \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =\] \[ \Large = -\frac{ dt }{ dx }\ln (t^2+1)\]
implicit? It's tempting, but I'm more inclined to believe a typo.
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.
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... \[\Large \frac{ d }{ dx } -\int\limits\limits\limits_{-1}^{x}\ln (t^2 +1) dt =\]
If it is an x then \[\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (x^2+1)\] \[\Large = - \ln (x^2+1)\]
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... \[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\] \[\Large -\ln(t^2+1)\]
Hmm..i'll check on with my teacher on this one..
@terenzreignz shouldn't it be an x^2+1 in your last step?
cr*p typos are contagious :D should really stop relying on copy-paste XD \[\Large -\ln(x^2+1)\] Thanks for pointing that out :)
Yeah i made a few above, with all the damn x's and t's...
I only did go online for a few giggles... lol ----------------------------------------- Terence out
so @agent0smith you think its a typo? and that it should be with an x and not a t?
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)
okay thanks for the explanation
No prob. But also @terenzreignz i guess it makes sense to switch limits and change the sign too.