anonymous
  • anonymous
Find the derivative f'(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
http://assets.openstudy.com/updates/attachments/51858a78e4b0bc9e008a60b4-onegirl-1367706243068-untitled10.jpg
anonymous
  • anonymous
I got - integral sign ln (t^2 + 1) is that correct?
agent0smith
  • agent0smith
^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 \[\Large \frac{ d }{ dx } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =\] \[\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1)\]

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

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