Finding a f'(x) in a fraction

**hemi** · May 29th 2009, 04:19 PM

I don't know how I should go about solving down on this problem for $f'(x)$ :

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

If someone could show me in a step-by-step fashion that would be great.

**Random Variable** · May 29th 2009, 04:23 PM

You could use the quotient rule or you could just divide each term in the numerator by x and then take the derivative.

**Spec** · May 29th 2009, 04:23 PM

Quotient rule - Wikipedia, the free encyclopedia

In this case, it's easier to just split it up.

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

**hemi** · May 29th 2009, 07:45 PM

I'm still a bit confused on the quotient rule, can you go a bit further with the math explaination?

Thanks a lot!

**Random Variable** · May 29th 2009, 08:12 PM

Originally Posted by hemi

I'm still a bit confused on the quotient rule, can you go a bit further with the math explaination?

Thanks a lot!

For this problem there is no reason to use the quotient rule. But here's what it would look like:

if $f(x) = \frac {g(x)}{h(x)}$ then $f'(x) = \frac {h(x)g'(x)-g(x)h'(x)}{[h(x)]^{2}}$

so $f'(x) = \frac {x \frac {d}{dx} (-6x^{3}+3x^{2}-2x+1) - (-6x^3+3x^{2}-2x+1) \frac {d}{dx}x} {x^{2}}$

$= \frac {x(-18x^{2}+6x-2) - (-6x^{3}+3x^{2}-2x+1)}{x^{2}}$

$= \frac {-12x^{3} +3x^{2}-1}{x^{2}}$

$= -12x + 3 - \frac {1}{x^{2}}$

The quotient rule is just an extension of the product rule.

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