Question

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.

Answer 1

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

Answer 2

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

Answer 3

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:
\[y''=-2x^{-3}+18x^{-4}+216x^{-5}\]

Answer 4

\[(x+3)(2x-3)+(x ^{2}-3x+9)(1) = 2x^2-3x+6x-9+x^2-3x+9= 3x^2 = (3x^2)/x^3\]

Answer 5

So what would the first derivative be?

Answer 6

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

Answer 7

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

Answer 8

Okay thank you!

Answer 9

Anytime :)