Having trouble simplifying

frank789

Junior Member
Joined
Sep 16, 2017
Messages
58
Hi all, having trouble with this one simplification I found inside of a derivative.

(ex + e-x)2 - (ex - e-x)2
--------------------------
(ex + e-x)2

so I seperated and cancelled first too (thats one minus and then the fraction)

(ex - e-x)2
1 - -----------
(ex + e-x)2

when I expand and collect the binomials i got

(ex - e-x)2 = e2x - e0 - e0 + e-2x

and

(ex + e-x)2​ = e2x + e0 + e0 + e-2x

I guess im not sure where to go from here or how to keep simplyfing.

a nudge in the right direction would be much appreciated :) (also how do you post properly formatted math?)
 
Hi all, having trouble with this one simplification I found inside of a derivative.

(ex + e-x)2 - (ex - e-x)2
--------------------------
(ex + e-x)2

so I seperated and cancelled first too (thats one minus and then the fraction)

(ex - e-x)2
1 - -----------
(ex + e-x)2

when I expand and collect the binomials i got

(ex - e-x)2 = e2x - e0 - e0 + e-2x

\(\displaystyle = e^{2x} - 2 + \dfrac{1}{e^{2x}} = \dfrac{e^{4x} - 2e^{2x} + 1}{e^{2x}}.\) SEE WHAT I DID HERE?

and

(ex + e-x)2​ = e2x + e0 + e0 + e-2x

\(\displaystyle = e^2x + 2 + \dfrac{1}{e^{2x}} = \dfrac{e^{4x} + 2e^{2x} + 1}{e^{2x}}.\) SEE WHAT I DID HERE?

I guess im not sure where to go from here or how to keep simplyfing.

a nudge in the right direction would be much appreciated :) (also how do you post properly formatted math?)
Answering your last question first, there is a fussy formatting language available here, but I recommend that students ignore it and concentrate on math. Just write things out in line using grouping symbols and ^ to indicate superscripts and _ to indicate subscripts.

So you can write ((e^x + e^(-x))^2 - (e^x - e^(-x))^2) / ((e^x + e^(-x))^2).

So continuing on

\(\displaystyle 1 - \dfrac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}= 1 - \dfrac{ \dfrac{e^{4x} - 2e^{2x} + 1}{e^{2x}}}{\dfrac{e^{4x} + 2e^{2x} + 1}{e^{2x}}} = 1 - \dfrac{e^{4x} - 2e^{2x} + 1}{e^{4x} + 2e^{2x} + 1} = \)

\(\displaystyle \dfrac{e^{4x} + 2e^{2x} + 1 - (e^{4x} - 2e^{2x} + 1)}{e^{4x} + 2e^{2x} + 1} = \dfrac{4e^{2x}}{(e^{2x} + 1)^2}.\)

Does that answer your questions?
 
I now see a much simpler solution based on these two lemmas.

\(\displaystyle (a + b)^2 - (a - b)^2 = a^2 + 2ab + b^2 - (a^2 - 2ab + b^2) = 4ab.\)

\(\displaystyle (c + c^{-1})^2 = c^2 + 2c^{(1-1)} + c^{-2} = c^2 + 2 + \dfrac{1}{c^2} =\)

\(\displaystyle \dfrac{c^4 + 2c^2 + 1}{c^2} = \dfrac{(c^2 + 1)^2}{c^2}.\)

THUS

\(\displaystyle \dfrac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2} = \dfrac{4 * e^x * e^{-x}}{\dfrac{\{(e^x)^2 + 1\}^2}{(e^x)^2}} = \dfrac{4}{\dfrac{(e^{2x} + 1)^2}{e^{2x}}} = \dfrac{4e^{2x}}{(e^{2x} + 1)^2}.\)
 
thanks jeff much appreciated. the higher exponent comes from the common denominator.
 
and your expansion formula is awesome as well, your are right that is probably the better way to explain it. thanks again
 
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