Answer
$y'=\frac{1-3t^2}{1+t-t^3}$
Work Step by Step
$y=\ln|1+t-t^3|\\
y'=\frac{1}{1+t-t^3}\times(0+1-3t^2)\\
y'=\frac{1-3t^2}{1+t-t^3}$
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