I don't know how I should go about solving down on this problem for $\displaystyle f'(x)$:
$\displaystyle f(x)=\frac{-6x^3+3x^2-2x+1}{x}$
If someone could show me in a step-by-step fashion that would be great.
Quotient rule - Wikipedia, the free encyclopedia
In this case, it's easier to just split it up.
$\displaystyle f(x)=\frac{-6x^3+3x^2-2x+1}{x}=-6x^2+3x-2+\frac{1}{x}$
For this problem there is no reason to use the quotient rule. But here's what it would look like:
if $\displaystyle f(x) = \frac {g(x)}{h(x)} $ then $\displaystyle f'(x) = \frac {h(x)g'(x)-g(x)h'(x)}{[h(x)]^{2}} $
so $\displaystyle f'(x) = \frac {x \frac {d}{dx} (-6x^{3}+3x^{2}-2x+1) - (-6x^3+3x^{2}-2x+1) \frac {d}{dx}x} {x^{2}} $
$\displaystyle = \frac {x(-18x^{2}+6x-2) - (-6x^{3}+3x^{2}-2x+1)}{x^{2}} $
$\displaystyle = \frac {-12x^{3} +3x^{2}-1}{x^{2}} $
$\displaystyle = -12x + 3 - \frac {1}{x^{2}} $
The quotient rule is just an extension of the product rule.