so, did you even think about the hint?
2x/(1-x^2) = 2x/((1-x)(1+x)
= -[1/(x+1) + 1/(x-1)]
The nth derivative of 1/(x+1) and 1/(x-1) is easy to figure, right?
Find the nth derivative of y=2x/(1-x^2)
(Hint: Consider -[(1/(x-1))+(1/(x+1))].)
3 answers
Or how do I find the formula for this?
huh? Let's take f=1/u. (Let u=x+1 or u=x-1)
f = 1/u = u^-1
f' = -1 u-2
f" = (-1)(-2)u-3
...
fn = (-1)nn! u-(n+1)
f = 1/u = u^-1
f' = -1 u-2
f" = (-1)(-2)u-3
...
fn = (-1)nn! u-(n+1)