Okay guise i need your utmost help i have a test tomorrow on this and on how to graph it, I am having trouble finding the derivative of -2x/((x^2-1)^2) and then on how to graph it :( help!

Question

dumbcow · Answer

looks like quotient rule:
= (f'g - fg')/g^2

f = -2x
f' = -2

g = (x^2-1)^2
For g' use chain rule
u = x^2 -1
du = 2x
g' = d/du (u^2) *du
g' = 4x(x^2 -1) = 4x^3-4x

(f/g)' = (-2(x^2-1)^2 - (-2x)(4x^3-4x))/ (x^2-1)^4
then expand and simplify

= (6x^4-4x^2-2)/(x^2-1)^4

dumbcow · Answer

as far as graphing it.

do you need to graph the original function or the derivative?
either way find the vertical and horizontal asymptotes, then plug in points

anonymous · Answer

well i found the first derivative from the original equation and got my asymptotes, So my question now is do I plug in points to the original equation?

dumbcow · Answer

it depends on which function you are graphing

anonymous · Answer

it tells me to sketch the graph of f , meaning the original given equation so that means i plug in points there?

dumbcow · Answer

ok then yes, pick x-values on either side of vertical asymptotes and plug them into original equation

anonymous · Answer

okay thank you! this will help me alot on my test tomorrow :)