Question

What is the second derivative of f(x) = x/ (x^2-1)

Answer 1

I got the first derivative as -x^2-1 / (x^2-1)^2
but i dont know how to do the 2nd derivative

Answer 2

Did you try a sub-y :)

Answer 3

I need to do it the quotient rule way

Answer 4

sorry mis-read

Answer 5

So Just apply quotient rule again

Answer 6

i got it till -2x(x^2-1)^2 - 4x^3 - 4x / (x^2-1)^4

it looks wrong to me..so i dont know

Answer 7

\[f'=\frac{-x^2-1}{(x^2-1)^2}\]
\[f''=\frac{(-2x)(x^2-1)^2-(-x^2-1)(2 \cdot (x^2-1) \cdot (2x)}{(x^2-1)^4} \]

Answer 8

Did you get this far correct?

Answer 9

Final answer should be
\[\frac{d^2}{dx^2}\left(\frac{x}{x^2-1}\right)=\frac{2 x \left(x^2+3\right)}{\left(x^2-1\right)^3}\]

Answer 10

why is there 2 (-x^2 -1)? there should be only 1 of them on the numerator isnt it

Answer 11

what...i got like -2x(x^4 +3) / (x^2 -1 )^4...

Answer 12

Hey cmkc I left the top the same and used chain rule for the bottom

Answer 13

ohhh i know what i did wrong!! thank you both of u!!

Answer 14

For that part of the quotient rule anyways

Answer 15

yea, forgot to put the initial ..haha thanks so much!!

Answer 16

I still got it wrong
can u check?

Answer 17

wait

Answer 18

Ok thanks! I will be waiting :)

Answer 19

\[y=\frac{x}{x^2-1}\]
\[y'=\frac{(x^2-1)-x(2x)}{(x^2-1)^2}\]
\[y'=\frac{-x^2-1}{(x^2-1)^2}\]
\[y''=\frac{(x^2-1)^2(-2x)-(-x^2-1)(2)(x^2-1)(2x)}{(x^2-1)^4}\]
\[y''=\frac{-2x(x^2-1)^2-(-2x^3-2x)(2)(x^2-1)}{(x^2-1)^4}\]
\[y''=\frac{(x^2-1)[2x(x^2-1)+4x^3+4x]}{(x^2-1)^4}\]
\[y''=\frac{[-2x^3+2x+4x^3+4x]}{(x^2-1)^3}\]
\[y''=\frac{2x^3+6x}{(x^2-1)^3}\]
\[y''=\frac{2x(x^2+3)}{(x^2-1)^3}\]

Answer 20

thank u thank u

Answer 21

no prob