Question

For the brave at heart...

I need help checking the derivative(s) of the following function:

"y = x/(x^2 - 4)"

The following are what I got for the first & second derivatives, respectively:
"y' = (-x^2 - 4)/(x^2 - 4)^2" ,
"y" = [(2(x^2 - 4))(x(x^2 - 4) - x(-x^2 - 4))]/(x^2 - 4)^2

Yeah... Disgusting, right? That's why I'm hoping to have some burial land soul go through this one and check my work, because I know even a tiny mistake can ruin a full-page problem.

(To be continued... )

Answer 1

*Note:
1. I used the Quotient Rule to find the derivative I did.
2. No, what I have written is not the initial second derivative- I did as much combining like terms/ factoring as I knew how to. The 'original' second derivative, without moving, adding, or changing anything from the format of the Q.R. equation is: "[-2x((x^2 - 4)^2) - (-x^2 - 4)(2(x^2 - 4))(2x)]/(x^2 - 4)^4"

Answer 2

Any and all help is greatly appreciated! :)

Answer 3

y = x/(x^2 -4) let x =u let (x^2-4)= v
dy/dx = (v*du/dx-u*dv/dx) / v^2
du/dx = 1 dv/dx = 2x v^2 = (x^4 - 8x^2+ 16)
y' = (x^2-4)(1)-x(2x) all divided by (x^4-8x^2+16)
(x^2 -4 -2x^2)/(x^4-8x^2+16)
(-3x^2 -4)/(x^4 -8x^2+16)

Answer 4

?? So... Is my work right? :) Oh, and I think that it should be -1x^2 in the last step you did. Right? 1x^2 - 2x^2 = -1x^2 ?

Answer 5

Use wolframalpha to check your answers
http://www.wolframalpha.com/input/?i=++derivative+x%2F%28x^2+-+4%29

Answer 6

Ohhhh!! Awesome! This website is amazing!! :) Thanks @eliassaab !

Answer 7

Yw

Answer 8

You can also practice calculus on my interactive site
http://www.saab.org/calculus.cgi

Answer 9

@wio any thoughts? :)

Answer 10

http://www.wolframalpha.com/input/?i=d%2Fdx++x%2F(x%5E2+-+4)
This is what I think.

Answer 11

Touché. ;) Alright, guess I'll go that route.