# Where does the square come from?

• Aug 7th 2012, 05:16 PM
astuart
Where does the square come from?
I need to find the derivative of the following..

g(t) = ln(t2e-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 t2 coming from?
• Aug 7th 2012, 05:40 PM
Prove It
Re: Where does the square come from?
Quote:

Originally Posted by astuart
I need to find the derivative of the following..

g(t) = ln(t2e-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 t2 coming from?

\displaystyle \displaystyle \begin{align*} \frac{2}{t} - 2t &= \frac{2}{t} - \frac{2t^2}{t} \\ &= \frac{2 - 2t^2}{t} \end{align*}
• Aug 7th 2012, 05:55 PM
astuart
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?