Question

can someone walk me through how to simplify this?

Answer 1

yes

Answer 2

what needs simplifying?

it's the quotient rule, if you like, but that thing you linked does all the work.

Answer 3

..... no I mean how did they get from the left side part to the right.... . Like how did they get (1-2e^x)

Answer 4

they factored it

and they got \(1 - 2 e^{\color{red}{2}x} = 0\)

it's a mess but there's nothing more to it than that.

[out of interest, did the original DE come from some Calculation of Variations application? it just looks like it should.]

Answer 5

but would did they factor out...
like could you please show me? I'm honestly confused and tried factoring but I'm not getting the same answer

Answer 6

It is
\(\bigl((1+e^{2x})^{3/2} e^x - e^x (3/2) (1+e^{2x})^{1/2} e^{2x}2\bigr)(1+e^{2x})^{-3}\)
\(\iff\)
\(\bigl(e^{x}(1+e^{2x})^{1/2} ((1+e^{3x})^1 - 3e^{2x}\bigr) (1+e^{2x})^{-3} \)

and \(1+e^{3x} - 3e^{2x} = 1 - 2e^{2x}\).
Then the argument is that the fraction is equal to zero only if the numerator is equal to zero, and since \(e^x\) and \(1+e^{2x}\) are always strictly positive, the only term which might be equal to zero is \(1-2e^{2x}\).