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Mathematics 18 Online
OpenStudy (anonymous):

differentiate. h(t)=t/1-t^2

OpenStudy (bahrom7893):

did u mean t/(1-t^2) ?

OpenStudy (anonymous):

yes that is what i meant, sorry for the confusion

OpenStudy (bahrom7893):

okay so use quotient rule: lodehi take hidelo draw a line and square the lo.. lol

OpenStudy (bahrom7893):

[(1-t^2)*1 - [t(-2t)]] / (1-t^2)^2

OpenStudy (bahrom7893):

[ 1-t^2 + 2t^2 ] / (1-t^2)^2 = = (t^2+1)/(1-t^2)^2

OpenStudy (anonymous):

the minus 2t in the problem is the derivative of 1-t^2, right?

OpenStudy (bahrom7893):

yes.

OpenStudy (anonymous):

the question i have is what happens to the 2t^2

OpenStudy (bahrom7893):

2t^2 -t^2 it becomes t^2..

OpenStudy (anonymous):

okay that makes sense... thanks

OpenStudy (anonymous):

\[\int\limits_{ }^{}t/1-t^2dt=u=1-t^2 du=-2tdt\] \[-1/2\int\limits_{}^{}du/u = -1/2\ln(1-t^2)+c\] looks like on a good day ln(1-t^2)^-1/2+C

OpenStudy (anonymous):

i dont know what you guys are doing but this problem is SCEAMING ln

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