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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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    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.2011-01-31
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    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 $Ax^2+Bx+C$.2011-01-31
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    Why is it you "can't find B"?2011-01-31
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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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    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}$2011-01-31
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    @Sivaram: From there I know that D = 2 and that C is most likely 0 as well as A because we can't use x^2 and don't need a constant of C, so that means B has to be -2?2011-01-31
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    @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.2011-01-31
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    @Arturo: Right, thanks. that's exactly how I did it, but the system I enter my answers into accepts everything except B = -2 so I will email my instructor.2011-01-31
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    He said I went wrong with my quotient rule. :S2011-01-31
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    Apparently the answer for B is 4. Could anyone help find that?2011-02-08
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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