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differentiate. h(t)=t/1-t^2
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did u mean t/(1-t^2) ?
yes that is what i meant, sorry for the confusion
okay so use quotient rule: lodehi take hidelo draw a line and square the lo.. lol
[(1-t^2)*1 - [t(-2t)]] / (1-t^2)^2
[ 1-t^2 + 2t^2 ] / (1-t^2)^2 = = (t^2+1)/(1-t^2)^2
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the minus 2t in the problem is the derivative of 1-t^2, right?
yes.
the question i have is what happens to the 2t^2
2t^2 -t^2 it becomes t^2..
okay that makes sense... thanks
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\[\int\limits_{ }^{}t/1-t^2dt=u=1-t^2 du=-2tdt\] \[-1/2\int\limits_{}^{}du/u = -1/2\ln(1-t^2)+c\] looks like on a good day ln(1-t^2)^-1/2+C
i dont know what you guys are doing but this problem is SCEAMING ln
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