I need to find the derivative of the following..

g(t) = ln(t^{2}e^{-t^2})

$\displaystyle g(t) = ln t^2 + ln e^(-t^2)$

= $\displaystyle g(t) = 2 ln t + -t^2$ (ln e = 1, right?)

g '(t)= $\displaystyle 2 ( 1 / t ) + (-2t)$

g '(t)= $\displaystyle 2 - 2t / t$

$\displaystyle g '(t) = 2(1 - t) / t$

According to my textbook though, I should have gotten this...

$\displaystyle g '(t) = 2(1-t^2) / t$

Where is the t^{2 }coming from?