- #1
mateomy
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Second derivative...
Okay, this is a rough one for me. It was a question I got on my test, and (obviouslly) didnt get right. I am studying all my old exams for my final in 2 days and this is the last of the problems that I can't wrap my head around...any help would be greatly appreciated.
[tex]
\frac{d^2}{dx^2} \int_0^{x}\,(\Big\,\int_1^{sint}\sqrt{1+u^2}\,du)\Big\,dt
[/tex]
I know, through the fundamental theorem of calculus that I can just replace (so to speak) the 't' in 'sint' to a 'sinx'. and then replace the [itex]u^2[/itex] with 'sinx'. I think that's it for the first part, then I think to find the second derivative I just derive it again? I dunno...Im seriously lost; all those parenthesis and variable changes are throwing me off. I am not necessarily looking for an answer (although that would definitely help), more so just looking for some direction. Thanks in advance.
Homework Statement
Okay, this is a rough one for me. It was a question I got on my test, and (obviouslly) didnt get right. I am studying all my old exams for my final in 2 days and this is the last of the problems that I can't wrap my head around...any help would be greatly appreciated.
[tex]
\frac{d^2}{dx^2} \int_0^{x}\,(\Big\,\int_1^{sint}\sqrt{1+u^2}\,du)\Big\,dt
[/tex]
I know, through the fundamental theorem of calculus that I can just replace (so to speak) the 't' in 'sint' to a 'sinx'. and then replace the [itex]u^2[/itex] with 'sinx'. I think that's it for the first part, then I think to find the second derivative I just derive it again? I dunno...Im seriously lost; all those parenthesis and variable changes are throwing me off. I am not necessarily looking for an answer (although that would definitely help), more so just looking for some direction. Thanks in advance.