Sorry im redoing it accidently submitted the wrong thing

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- Feb 20th 2010, 10:52 AMx5pyd3rxHow would i rewrite this problem
Sorry im redoing it accidently submitted the wrong thing

- Feb 20th 2010, 11:06 AMx5pyd3rx
Here was my attempt it was actuly a derivative problem but i decided not to post in the calculus section.

$\displaystyle \frac {1}{3} (1-e^{6x})^{\frac{-2}{3}} * (- e^{6x} * 6)$

$\displaystyle = (2 - 2e^{6x}) ^{\frac{-2}{3}} * -e^{6x}$

$\displaystyle = \frac{1}{(2-2e^{6x})^{\frac{2}{3}}} - e^{6x}$ - Feb 20th 2010, 11:11 AMe^(i*pi)
You can't distribute the two to the binomial expression because that expression is an infinite sum and the 2 is not raised to the same power.

$\displaystyle \frac{1}{3}(1-e^{6x})^{2/3} \cdot -6e^{6x}$

$\displaystyle = 2e^{6x}(1-e^{6x})^{2/3}$

You cannot simplify that further.

**EDIT**: If you're planning on finding the derivative use the product rule and the chain rule :) - Feb 20th 2010, 11:13 AMicemanfan
I'm assuming you are taking the derivative of f(x):

$\displaystyle f(x) = (1 - e^{6x})^{\frac{1}{3}}$

In that case, you made errors in both the second and third steps.

(second step) You can't put the multiple of 2 inside the $\displaystyle (1 - e^{6x})^{-\frac{2}{3}}$ term.

(third step) You should be multiplying by $\displaystyle -e^{6x}$, not subtracting $\displaystyle e^{6x}$. - Feb 20th 2010, 11:14 AMx5pyd3rx
Ahhh ok gotcha.. (no need to change the post btw)

- Feb 20th 2010, 11:16 AMicemanfan
- Feb 20th 2010, 11:16 AMx5pyd3rx
- Feb 20th 2010, 11:22 AMx5pyd3rx
So in that case i can rewrite it as

$\displaystyle

= \frac{-2e^{6x}}{(1-e^{6x})^{2/3}}

$ - Feb 20th 2010, 11:38 AMicemanfan