Where does the square come from?

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?

Re: Where does the square come from?

Quote:

Originally Posted by

**astuart** 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?

$\displaystyle \displaystyle \begin{align*} \frac{2}{t} - 2t &= \frac{2}{t} - \frac{2t^2}{t} \\ &= \frac{2 - 2t^2}{t} \end{align*}$

Re: Where does the square come from?

Quote:

Originally Posted by

**Prove It** $\displaystyle \displaystyle \begin{align*} \frac{2}{t} - 2t &= \frac{2}{t} - \frac{2t^2}{t} \\ &= \frac{2 - 2t^2}{t} \end{align*}$

Ah. Is it from this??

$\displaystyle 2t / t - 2t / 1 (t) = 2t / t - 2t^2 - t$

Basically, to subtract fractions they must have the same denominator - therefore to get 2t to be 2t / t, it has to be multiplied by t?