Question

Simplify the expression. (Do not expand the expression in the denominator.)
((e^(7x) + e^(-7x) (e^(7x) + e^(-7x) - (e^(7x) - e^(-7x)(e^(7x) - e^(-7x)) / ((e^(7x) + e^(-7x))^2

Answer 1

Can you type it in latex? You are missing a few parenthesis :P

Answer 2

latex?

Answer 3

(e^(7x) + e^(-7x))^2 - (e^(7x) - e^(-7x))^2 is the numerator
((e^(7x) + e^(-7x))^2 is the denominator

Answer 4

Okay so you have:
\[\frac{(e^{7x}+e^{-7x})^2-(e^{7x}-e^{-7x})^2}{(e^{7x}+e^{-7x})^2}\]
?

Answer 5

Yes

Answer 6

Okay. Start off my multiplying the top out completely.
\[(e^{7x}+e^{-7x})^2=e^{7x}*e^{7x}+e^{-7x}*e^{-7x}+e^{-7x}*e^{7x}+e^{-7x}*e^{7x}=e^{14x}+e^{-14x}+2\]

Answer 7

The last one should end as:
\[e^{14x}+e^{-14x}+2\]
Then the second part is the exact same except with a negative sign. so:
\[(e^{7x}-e^{-7x})^2=-e^{7x}*e^{-7x}+e^{7x}*e^{7x}+e^{-7x}*e^{-7x}-e^{7x}*e^{-7x}\]
Which equals:
\[e^{14x}+e^{-14x}-2\]

Answer 8

Distribute the negative on the second term giving:
\[e^{14x}+e^{-14x}+2-e^{14x}-e^{-14x}+2=4\]

Answer 9

So now you have:
\[\frac{4}{(e^{7x}+e^{-7x})^2}\]

Answer 10

Ahhhh coolies, thanks a heap!!!

Answer 11

If the writing looks bad you can refresh the page.

Answer 12

Nope, looks good XD

Answer 13

No problem :P Sorry it took a bit to type up. Latex is the way I was writing the math btw :P It just looks nice instead of all the text parenthesis and what not.

Answer 14

Yeah it does :)