Help on a calc question? Posted in comments

Question

sleepyjess · Answer

$$f(x) = \frac{ 6x^4 - 9x^2 }{ 3x^3 }$$

sleepyjess · Answer

My work so far using quotient rule:

$f'(x) = \dfrac{f'g - g'f}{g^2}$
f = $6x^4 - 9x^2$
g = $3x^3$

f' = $24x^3 - 18x$
g' = $9x^2$

$f'(x) = \dfrac{(24x^3-18x^2)(3x^3) - (9x^2)(6x^4 - 9x^2)}{3x{^3}^2}$

sleepyjess · Answer

After multiplying everything out:

$f'(x) = \dfrac{72x^6 - 54x^4 -56x^6 + 81x^4}{3x^5}$

Simplified: 

$f'(x) = \dfrac{18x^6 + 27x^4}{3x^5}$

The answer sheet says the answer should be $f'(x) = 2 + \dfrac3{x^2}$, so did I do something wrong?

sleepyjess · Answer

@zepdrix @phi

yoooww · Answer

solid mensuration ?  can you help me ? :)

phi · Answer

do you have to use the quotient rule ?
because I would first simplify the problem by dividing by 3x^3 first

sleepyjess · Answer

No, I was thinking that would be the easiest was to go about the problem. How would I divide the 3x^3?

phi · Answer

as you know
a/b  + c/b  is (a+c)/b
but you can go "backwards"

mww · Answer

Recall $$\frac{ a+b }{ c } = \frac{ a }{ c }+\frac{ b }{ c }$$

sleepyjess · Answer

Ah, so $\dfrac{6x^4}{3x^3} - \dfrac{9x^2}{3x^3}$ 

$\dfrac{6x^4}{3x^3}$ divides to 2x, and $\dfrac{9x^2}{3x^3}$ divides to 3x^-1 or 3/x

phi · Answer

btw, using the quotient rule should work, but it's messy
the bottom should be (3x^3)^2  which becomes 9 x^6

now if you simplify (divide both terms) I think it simplifies to the book answer

sleepyjess · Answer

2x - $\dfrac3x$, but how does that go to $2 + \dfrac3{x^2}$?

phi · Answer

now take the derivative

phi · Answer

$$ f(x) = \frac{ 6x^4 - 9x^2 }{ 3x^3 }= 2x - 3 x^{-1} $$

sleepyjess · Answer

Gotcha, that makes it 2 + 3/x^2. So everything up to the 2x - 3/x was just simplifying the original?

phi · Answer

yes.  notice your original also simplifies to this result.
but simplify first then derivative is a lot less work than derivative, then simplify