Question

how do i use macluerins expansion on
(e^2x -1)/(e^2x +1)

Answer 1

just to clarify

\[e ^{2x}-1 \over e ^{2x}+1 \]

Answer 2

i need the terms to include x to power 5

Answer 3

now i know \[e^{2x}=1+ 2x + {(2x)^2 \over2!} + ...\]

Answer 4

but i'm not sure how that helps with this

Answer 5

you are on the right track

Answer 6

problem is i'll end up with the terms of the expansion in numerator and denominator

Answer 7

plus i've tried to simplify to
\[1- {2\over e^{2x} + 1}\]

Answer 8

i've also tried directly differentiating the whole expression but it rapidly gets complex after 2 levels of differentiation

Answer 9

\[\frac{1+ 2x + {(2x)^2 \over2!} + ... -1}{1+ 2x + {(2x)^2 \over2!} + ... +1}\]
\[\frac{ 2x + {(2x)^2 \over2!} + ... }{2+ 2x + {(2x)^2 \over2!} + ... }\]

Answer 10

apparently the answer is \[\approx x - {x^3\over 3} +{ 2x^5 \over15}\]

Answer 11

i got as far as you imran but nowhere near that answer