# How would i rewrite this problem

• Feb 20th 2010, 10:52 AM
x5pyd3rx
How would i rewrite this problem
Sorry im redoing it accidently submitted the wrong thing
• Feb 20th 2010, 11:06 AM
x5pyd3rx
Here was my attempt it was actuly a derivative problem but i decided not to post in the calculus section.

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

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

$= \frac{1}{(2-2e^{6x})^{\frac{2}{3}}} - e^{6x}$
• Feb 20th 2010, 11:11 AM
e^(i*pi)
Quote:

Originally Posted by x5pyd3rx
Here was my attempt it was actuly a derivative problem but i decided not to post in the calculus section.

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

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

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

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.

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

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

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 AM
icemanfan
I'm assuming you are taking the derivative of f(x):

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

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 $(1 - e^{6x})^{-\frac{2}{3}}$ term.

(third step) You should be multiplying by $-e^{6x}$, not subtracting $e^{6x}$.
• Feb 20th 2010, 11:14 AM
x5pyd3rx
Ahhh ok gotcha.. (no need to change the post btw)
• Feb 20th 2010, 11:16 AM
icemanfan
Quote:

Originally Posted by e^(i*pi)
$\frac{1}{3}(1-e^{6x})^{2/3} \cdot -6e^{6x}$

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

A slight correction:

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

$= -2e^{6x}(1-e^{6x})^{-2/3}$
• Feb 20th 2010, 11:16 AM
x5pyd3rx
Quote:

Originally Posted by icemanfan
I'm assuming you are taking the derivative of f(x):

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

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 $(1 - e^{6x})^{-\frac{2}{3}}$ term.

(third step) You should be multiplying by $-e^{6x}$, not subtracting $e^{6x}$.

Your right... I did i just wrote it on here wrong. thanks for the help both of you
• Feb 20th 2010, 11:22 AM
x5pyd3rx
So in that case i can rewrite it as

$

= \frac{-2e^{6x}}{(1-e^{6x})^{2/3}}
$
• Feb 20th 2010, 11:38 AM
icemanfan
Quote:

Originally Posted by x5pyd3rx
So in that case i can rewrite it as

$

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

Correct.