Wierd Q to take a derivative of

[the4thcafeavenue]

wierd Q to take a derivative of...

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

 

[dextercioby]

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.

 

[arildno]

AARGH!

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

 

[Vega]

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

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

 

[the4thcafeavenue]

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

 

[Data]

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}

 

[arildno]

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

At least, now you've learned your lesson, right?