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Mathematics 7 Online
OpenStudy (anonymous):

find the derivative of y with respect to x:

OpenStudy (anonymous):

\[y=1/x(x+1)(x+2)\]

OpenStudy (amorfide):

|dw:1350454366994:dw| this?

OpenStudy (anonymous):

sorry...1 is the numerator...everything else in the den.

OpenStudy (amorfide):

do you know the chain rule?

OpenStudy (anonymous):

y=1/(x(x+1)(x+2))

OpenStudy (anonymous):

I pretty much know nothing for my calc class. I just need to get through this assignment that's due tomorrow and then I start my sessions with my tutor...any help is greatly appreciated

OpenStudy (mimi_x3):

\[\large \frac{1}{x(x+1)(x+2)} \]

OpenStudy (mimi_x3):

well you can use the product rule or the quotient rule

OpenStudy (mimi_x3):

have you tried expanding it

OpenStudy (anonymous):

So...1\[\frac{ 1 }{ x ^{3}+3x ^{2}+2x}?\]

OpenStudy (amorfide):

you could bring it to the top, so all the powers change signs to a negative

OpenStudy (anonymous):

\[y'=\frac{ -(3x ^{2}+6x+2) }{ x ^{2}(x ^{2}+3x+2)^{2} }..???

OpenStudy (anonymous):

sorry...that's so ugly. I'll retype it.

OpenStudy (anonymous):

\[y'=\frac{ -(3x ^{2}+6x+2) }{ x ^{2}(x ^{2}+3x+2)^{2} }\]

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