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## onegirl Group Title Find the derivative f'(x) one year ago one year ago

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1. onegirl Group Title
2. onegirl Group Title

I got - integral sign ln (t^2 + 1) is that correct?

3. agent0smith Group Title

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

4. onegirl Group Title

ohhhh okay

5. agent0smith Group Title

Wait, is this finding f'(x)? That t limit... is that meant to be x? This would just be zero.

6. terenzreignz Group Title

What he said ^

7. terenzreignz Group Title

Much more interesting if it were an x instead... here... $\Large \frac{ d }{ dx } \int\limits\limits_{\color{red}x}^{-1}\ln (t^2 +1) dt =$

8. onegirl Group Title

yea but its t

9. agent0smith Group Title

$\Large \frac{ d }{ dx } \int\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =$ $\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (t^2+1)$

10. onegirl Group Title

okay

11. terenzreignz Group Title

perhaps something like this? $\Large \frac{ d }{ d\color{red}t } \int\limits\limits_{t}^{-1}\ln (t^2 +1) dt =$

12. agent0smith Group Title

@onegirl that one with t will just be zero. Since you're basically taking the derivative of a constant... zero.

13. onegirl Group Title

okay and that is what my original problem had a t but okay thanks!

14. onegirl Group Title

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

15. agent0smith Group Title

Unless it's $\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (t) \right] \ln (t^2+1) =$ $\Large = -\frac{ dt }{ dx }\ln (t^2+1)$

16. agent0smith Group Title

@terenzreignz

17. terenzreignz Group Title

implicit? It's tempting, but I'm more inclined to believe a typo.

18. agent0smith Group Title

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.

19. terenzreignz Group Title

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... $\Large \frac{ d }{ dx } -\int\limits\limits\limits_{-1}^{x}\ln (t^2 +1) dt =$

20. agent0smith Group Title

If it is an x then $\Large \frac{ d }{ dx } \int\limits\limits\limits\limits_{x}^{-1}\ln (t^2 +1) dt =$ $\Large = \left[ \frac{ d }{ dx }(-1) \right] \ln( (-1)^2 + 1) -\left[ \frac{ d }{ dx } (x) \right] \ln (x^2+1)$ $\Large = - \ln (x^2+1)$

21. terenzreignz Group Title

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... $\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}$ $\Large -\ln(t^2+1)$

22. onegirl Group Title

Hmm..i'll check on with my teacher on this one..

23. agent0smith Group Title

@terenzreignz shouldn't it be an x^2+1 in your last step?

24. terenzreignz Group Title

cr*p typos are contagious :D should really stop relying on copy-paste XD $\Large -\ln(x^2+1)$ Thanks for pointing that out :)

25. agent0smith Group Title

Yeah i made a few above, with all the damn x's and t's...

26. onegirl Group Title

lol

27. onegirl Group Title

thanks guys!

28. terenzreignz Group Title

I only did go online for a few giggles... lol ----------------------------------------- Terence out

29. onegirl Group Title

so @agent0smith you think its a typo? and that it should be with an x and not a t?

30. agent0smith Group Title

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)

31. onegirl Group Title

okay thanks for the explanation

32. agent0smith Group Title

No prob. But also @terenzreignz i guess it makes sense to switch limits and change the sign too.