can someone check my answer Id check it myself but its a even problem $\displaystyle \frac {1}{t^2+3t-1} $ $\displaystyle (t^2+3t-1)^{-1} $ $\displaystyle -1(t^2+3t-1)^{-2}(2t+3) $ $\displaystyle - \frac {2t+3}{(t^2+3t-1)^2} $
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Looks correct
Try doing the same problem using the 1st principle
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