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Given that $f(x) = \dfrac{x}{x+ (2/x)}$

The derivative function is given by $f'(x) = \dfrac{Ax^2 + Bx + C}{(x^2+D)^2}$

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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    @ 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.2011-01-31

1 Answers 1

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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 $\frac{-2x}{(x^2+a)^2}$.

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    @Finzz: The derivative of $\frac{1}{x^2+a^2}$ is $\frac{-2x}{(x^2+a)^2}$. So for the actual problem i.e. the derivative of $1 - \frac{2}{x^2+2}$, it is $-2 \times \frac{-2x}{(x^2+a)^2} = \frac{4x}{(x^2+a)^2}$.2011-02-08