Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

onegirlBest ResponseYou've already chosen the best response.0
I got  integral sign ln (t^2 + 1) is that correct?
 11 months ago

agent0smithBest 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 \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]
 11 months ago

agent0smithBest 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.
 11 months ago

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

agent0smithBest ResponseYou've already chosen the best response.2
\[\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)\]
 11 months ago

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

agent0smithBest 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.
 11 months ago

onegirlBest ResponseYou've already chosen the best response.0
okay and that is what my original problem had a t but okay thanks!
 11 months ago

onegirlBest 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
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.2
Unless it's \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =\] \[ \Large = \frac{ dt }{ dx }\ln (t^2+1)\]
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
implicit? It's tempting, but I'm more inclined to believe a typo.
 11 months ago

agent0smithBest 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.
 11 months ago

terenzreignzBest 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... \[\Large \frac{ d }{ dx } \int\limits\limits\limits_{1}^{x}\ln (t^2 +1) dt =\]
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.2
If it is an x then \[\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{x}^{1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(1) \right] \ln( (1)^2 + 1) \left[ \frac{ d }{ dx } (x) \right] \ln (x^2+1)\] \[\Large =  \ln (x^2+1)\]
 11 months ago

terenzreignzBest 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... \[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\] \[\Large \ln(t^2+1)\]
 11 months ago

onegirlBest ResponseYou've already chosen the best response.0
Hmm..i'll check on with my teacher on this one..
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.2
@terenzreignz shouldn't it be an x^2+1 in your last step?
 11 months ago

terenzreignzBest 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 :)
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.2
Yeah i made a few above, with all the damn x's and t's...
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
I only did go online for a few giggles... lol  Terence out
 11 months ago

onegirlBest 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?
 11 months ago

agent0smithBest 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)
 11 months ago

onegirlBest ResponseYou've already chosen the best response.0
okay thanks for the explanation
 11 months ago

agent0smithBest 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.
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.