# Help rearranging a derivative

Given that f(x) = x / (x+ (2/x))

The derivative function is given by f'(x) = (Ax^2 + Bx + C) / (x^2+D)^2

where

A=

B=

C=

D=

From rearranging the original equation, I can get D=2 C=0 and A=0, but I cant seem to find B. Maybe someone could help?

EDIT: I found the answer to be 4, but don't see how this makes sense, since I can only clearly see how -2 would work.

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Perhaps start by rewriting the denominator of your given function $f(x)$ as a single fraction. Then rewrite the whole thing as a single fraction. – Jesse Madnick Jan 31 '11 at 5:49
Your parenthesis in the formula for the derivative are unbalanced, so it is not clear what you have. It is also unclear if $x^2+D$ is to divide just $C$, or the entire $Ax^2+Bx+C$. – Arturo Magidin Jan 31 '11 at 6:02
Thanks I'll fix it – Finzz Jan 31 '11 at 7:16
Why is it you "can't find B"? – Arturo Magidin Jan 31 '11 at 19:06
@ Arturo The system I enter my answers into accepts everything except B = -2 so I figured I was doing something wrong. I guess I will just email my instructor. – Finzz Jan 31 '11 at 19:53

HINT: Rearrange $f(x) = \frac{x^2}{x^2 + 2} = 1 - \frac{2}{x^2 + 2}$. Use that the derivative of a constant is $0$ and the derivative of $\frac{1}{x^2 + a}$ is $\frac{-2x}{(x^2+a)^2}$.
Do you mean rearrange the entire equality? Could you prove quick how $f(x) = \frac{x^2}{x^2 + 2} = 1 - \frac{2}{x^2 + 2}$? I know that you rearrange the original f(x) to equal $f(x) = \frac{x^2}{x^2 + 2}$ – Finzz Jan 31 '11 at 7:22
@Finzz: to get the equality of $f(x)$, just add a "clever $0$" to the numerator and break up the sum: $$\frac{x^2}{x^2+2} = \frac{x^2+(2-2)}{x^2+2} = \frac{(x^2+2)-2}{x^2+2} = \frac{x^2+2}{x^2+2} - \frac{2}{x^2+2} = 1 - \frac{2}{x^2+2}.$$And, yes, for $\frac{Ax^2+Bx+C}{(x^2+D)^2}$ to be equal to $$\frac{-2x}{(x^2+2)^2}=\frac{0x^2 -2x +0}{(x^2+2)^2},$$you need the numerator and denominators to be identical. – Arturo Magidin Jan 31 '11 at 16:55