Where does the square come from?

**astuart** · August 7th 2012, 05:16 PM

I need to find the derivative of the following..

g(t) = ln(t²e^{-t^2})

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

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

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

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

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

Where is the t²coming from?

**Prove It** · August 7th 2012, 05:40 PM

Originally Posted by astuart

I need to find the derivative of the following..

g(t) = ln(t²e^{-t^2})

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

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

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

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

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

Where is the t²coming from?

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

**astuart** · August 7th 2012, 05:55 PM

Originally Posted by Prove It

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

Ah. Is it from this??

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

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?

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