Question

Find the derivative f'(x)

Answer 1

http://assets.openstudy.com/updates/attachments/51858a78e4b0bc9e008a60b4-onegirl-1367706243068-untitled10.jpg

Answer 2

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

Answer 3

^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

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

Answer 4

ohhhh okay

Answer 5

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

This would just be zero.

Answer 6

What he said ^

Answer 7

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

Answer 8

yea but its t

Answer 9

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

Answer 10

okay

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

@terenzreignz

Answer 17

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

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

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...
\[\Large \color{green}{\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)}\]

\[\Large -\ln(t^2+1)\]

Answer 22

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

Answer 23

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

Answer 24

cr*p
typos are contagious :D
should really stop relying on copy-paste XD
\[\Large -\ln(x^2+1)\]

Thanks for pointing that out :)

Answer 25

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

Answer 26

lol

Answer 27

thanks guys!

Answer 28

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

Answer 29

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

Answer 30

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)

Answer 31

okay thanks for the explanation

Answer 32

No prob.

But also @terenzreignz i guess it makes sense to switch limits and change the sign too.