## onegirl Find the derivative f'(x) 11 months ago 11 months ago

1. onegirl
2. onegirl

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

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

4. onegirl

ohhhh okay

5. agent0smith

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

6. terenzreignz

What he said ^

7. terenzreignz

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

8. onegirl

yea but its t

9. agent0smith

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

10. onegirl

okay

11. terenzreignz

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

12. agent0smith

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

13. onegirl

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

14. onegirl

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

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

@terenzreignz

17. terenzreignz

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

18. 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.

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

20. agent0smith

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

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

22. onegirl

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

23. agent0smith

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

24. 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 :)

25. agent0smith

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

26. onegirl

lol

27. onegirl

thanks guys!

28. terenzreignz

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

29. onegirl

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

30. 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)

31. onegirl

okay thanks for the explanation

32. agent0smith

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