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## Main Question or Discussion Point

**wierd Q to take a derivative of....**

y=(x)/((x+2)(x+3)(x+4)).

how to do u take the derivative??? HELP!!!!!!

- Thread starter the4thcafeavenue
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- #1

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y=(x)/((x+2)(x+3)(x+4)).

how to do u take the derivative??? HELP!!!!!!

- #2

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[tex] y(x)=\frac{x}{(x+2)(x+3)(x+4)} [/tex]

Do u know to apply the

1.Quotient's derivative rule.

2.Leibniz product rule...?

Daniel.

- #3

arildno

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DO NOT DOUBLE, TRIPLE, OR QUADRUPLE POST IN THE FUTURE!!!!!!!!!!!!!!!!!!!!!

:grumpy:

- #4

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The easiest way would be to rewrite it like this first:

y=(x) * (x+2)(^-1) * (x+3)(^-1) * (x+4)(^-1)

y=(x) * (x+2)(^-1) * (x+3)(^-1) * (x+4)(^-1)

- #5

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haha, i guess i'd get mroe help dat way, but, oops hehe

- #6

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[tex]\frac{x}{(x+2)(x+3)(x+4)} = \frac{x}{x^3 + 9x^2 + 26x + 24}[/tex]

and then just use the product rule and the fact that

[tex]\frac{d}{dx}\frac{1}{f(x)} = -\frac{f^\prime (x)}{\left[f(x)\right]^2}[/tex]

- #7

arildno

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At least, now you've learnt your lesson, right?the4thcafeavenue said:haha, i guess i'd get mroe help dat way, but, oops hehe

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