2nd derivative trouble

[ande1717]

1. The problem statement, all variables and given/known data

f(x)=x/x^2+1
Find horiz. asymptotes and use them w/ concavity and intervals to sketch the curve

2. Relevant equations



3. The attempt at a solutionf'(x)= 1-x^2/(x^2+1)^2
But then I can't seem to work through taking the 2nd derivative, perhaps I am not using the chain rule right.
I get -4x^5-2x^3-2x/(x^2+1)^4
But thats not right... please help!

[CompuChip]

Assuming you meant f(x)=x/(x^2+1) and not f(x)=(x/x^2)+1, the first derivative is correct. So again using the quotient rule, the second derivative will be
\frac{ (1 - x^2)' (x^2 + 1)^2 - (1 - x^2) ((x^2 + 1)^2 )' }{(x^2+1)^4}
Can you work out the two derivatives that are there separately?

[ande1717]

So I get that f''(x)= ((-2x)((x^2+1)^2)-1-x^2)((2)(x^2+1))(2x))/(x^2+1)^4
I think thats right but not sure about distributing through.

I get (2x^5-2x^3-4x)/(x^2+1)^4

But the posted answer is (2x^5-4x^3-6x)/(x^2+1)^4

[CompuChip]

The first line is correct, so you probably did something wrong in the expansion.
Try working out
((-2x)((x^2+1)^2)
and
-(1-x^2)((2)(x^2+1))(2x))
separately, and only then adding them.
Forgetting about the denominator (x^2 + 1)^4 for a while, you should get 2 x^5 - 4x^3 - 6x.

And look at the bright side: probably it's some stupid writing error, at least you know you can differentiate :smile:

[ande1717]

Thanks for all the help, most of my errors are in the algebra...
on the left side I get (-2x^5-2x) and the right -((1-x^2)(4x^3+4x)) which goes to -(4x^3+4x-4x^5-4x^3).... Left - Right I get 2x^5-6x!! still missing the -4x^3 I don't know where i am going wrong.

[CompuChip]

on the left side I get (-2x^5-2x)
You do? I don't.
What does (x^2 + 1)^2 expand to? It's not x^4 + 1!
If you want, write it out: (a + b)^2 = (a + b)(a + b) = ....
Then remember that formula (or at least, remember to remember that something is going on whenever you see it) forever :smile:

[ande1717]

now I understand....I was convinced my mistake was on the right not the left. Thanks a ton.