- #1

rock.freak667

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## Homework Statement

Part of a question here...

Show that

[tex]\frac{d}{dx} \left[ \frac{x}{(4-x^2)^n} \right ] = \frac{1-2n}{(4-x^2)^n} - \frac{8n}{(4-x^2)^{n+1}}[/tex]

## Homework Equations

[tex]\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} [/tex]

## The Attempt at a Solution

So, using the quotient law I get

[tex] \frac{(4-x^2)^n \times 1 -(x) \times n(4-x^2)^{n-1} \times -2x}{(4-x^2)^{2n}}[/tex]

[tex]=\frac{1-2nx^2(4-x^2)^{-1}}{(4-x^2)^n}[/tex]

[tex] =\frac{1}{(4-x^2)^n} - \frac{2nx^2}{(4-x^2)^{n+1}}[/tex]

I seem to have the denominators correct but not the numerators. Did I do it wrong or are there more ways to simplify?

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