OpenStudy (anonymous):

Find the second and first derivatives of the function: y= ((x+3)(x^2-3x+9))/(x^3) I got -3/x^2 and 6/x^3 Is that correct? If not how do you get the answer? Thanks.

4 years ago
zepdrix (zepdrix):

$\Large y\quad=\quad \frac{(x+3)(x^2-3x+9)}{x^3}$Oh boy this one is a doozy +_+ Hmm

4 years ago
zepdrix (zepdrix):

So I guess we would start by applying the Quotient Rule,$\large \color{royalblue}{y'}\quad=\quad \frac{\color{royalblue}{[(x+3)(x^2-3x+9)]'}x^3-(x-3)(x^2-3x+9)\color{royalblue}{(x^3)'}}{(x^3)^2}$ Hmm yah this is no good, we should simply first I guess huh? XD

4 years ago
OpenStudy (anonymous):

Yeah I would go with : $y=\frac{(x+2)(x^2-3x+9)}{x^3}=\frac{x^3-x^2+3x+18}{x^3}=1-(x)^{-1}+3(x)^{-2}+18(x)^{-3}$ So then you can apply the power rule instead!: $y'=x^{-2}-6x^{-3}-54x^{-4}$ And: $y''=-2x^{-3}+18x^{-4}+216x^{-5}$

4 years ago
OpenStudy (anonymous):

$(x+3)(2x-3)+(x ^{2}-3x+9)(1) = 2x^2-3x+6x-9+x^2-3x+9= 3x^2 = (3x^2)/x^3$

4 years ago
OpenStudy (anonymous):

So what would the first derivative be?

4 years ago
OpenStudy (anonymous):

@KL2013 We notate the first derivative of y as: $\frac{dy}{dx}=\frac{d}{dx}y=y'$ And the second derivative as: $\frac{d^2y}{dx^2}=\frac{d^2}{dx^2}y=y''$

4 years ago
OpenStudy (anonymous):

I included both, $$y'\phantom{.}and\phantom{.}y''$$

4 years ago
OpenStudy (anonymous):

Okay thank you!

4 years ago
OpenStudy (anonymous):

Anytime :)

4 years ago