f(x)+= (x^2 +1)
...........(x^2 -1)
x squared plus 1 over x squared minus 1
2007-03-08 15:55:39 · 6 answers · asked by Jeff W 1 in Science & Mathematics ➔ Mathematics
use u/v rule
u=x^2+1
du/dx= 2x
v=x^2-1
dv/dx=2x
f'(x)= [v.du/dx - u. dv/dx]/v^2
= [(x^2-1)2x - (x^2+1)2x]/(x^2-1)^2
= [2x^3-2x-2x^3-2x]/(x^2-1)^2
= - 4x/(x^2-1)^2
2007-03-08 16:02:44 · answer #1 · answered by Maths Rocks 4 · 0⤊ 0⤋
Use the quotient rule. The derivative is:
2x / (x^2 - 1) - [2(x^2 + 1) * x / (x^2 - 1)^2]
2007-03-09 00:00:32 · answer #2 · answered by Anonymous · 0⤊ 0⤋
((x^2 -1)*2x) -(x^2 +1)*2x / (x^2 - 1)^2
use hodhi-hidho/hoho
where ho = bottom, hi = top, d = derivative
2007-03-09 00:02:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
it,s right answer is
f(x)=(x^2+1)
=1(x^2+1)^1-1
=(x^2+)^0
f(x)=1 Answer!
2007-03-09 00:10:15 · answer #4 · answered by stranger_eyes2007 1 · 0⤊ 1⤋
simplify, then
-2/(x^2-1)
then will be most simple
2007-03-09 00:13:37 · answer #5 · answered by Suiram 2 · 0⤊ 1⤋
you mean (x^2+1)/(x^2-1) ?
if so
you have
[(x^2+1)'*(x^2-1)-(x^2+1)*(x^2-1)']/(x^2-1)^2
[2x*(x^2-1)-(x^2+1)*2x]/(x^2-1)^2
2x(x^2-1-x^2-1)/(x^2-1)^2
-4x/(x^2-1)^2
2007-03-09 00:06:10 · answer #6 · answered by djin 2 · 0⤊ 0⤋