Wierd Q to take a derivative of

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The discussion focuses on finding the derivative of the function y=(x)/((x+2)(x+3)(x+4)). Participants recommend using the Quotient Rule and the Leibniz Product Rule for differentiation. A suggested approach is to rewrite the function as y=(x) * (x+2)(^-1) * (x+3)(^-1) * (x+4)(^-1) or to expand the denominator to y=x/(x^3 + 9x^2 + 26x + 24) before applying the Product Rule. The derivative can then be calculated using the formula for the derivative of a reciprocal function.

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the4thcafeavenue
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weird Q to take a derivative of...

y=(x)/((x+2)(x+3)(x+4)).
how to do u take the derivative? HELP! :confused:
 
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So is it

y(x)=\frac{x}{(x+2)(x+3)(x+4)}

Do u know to apply the

1.Quotient's derivative rule.
2.Leibniz product rule...?

Daniel.
 
AARGH!

DO NOT DOUBLE, TRIPLE, OR QUADRUPLE POST IN THE FUTURE!
 
The easiest way would be to rewrite it like this first:

y=(x) * (x+2)(^-1) * (x+3)(^-1) * (x+4)(^-1)
 
haha, i guess i'd get mroe help dat way, but, oops hehe
 
the easiest way is actually probably to just multiply out the denominator:

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

and then just use the product rule and the fact that

\frac{d}{dx}\frac{1}{f(x)} = -\frac{f^\prime (x)}{\left[f(x)\right]^2}
 
the4thcafeavenue said:
haha, i guess i'd get mroe help dat way, but, oops hehe
At least, now you've learned your lesson, right? :wink:
 

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