1. Which derivative rules?

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

What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?

2. you may want to consider writing as 3 expressions

1-3/x +1/x^3 and using the power rule on each

otherwise all you need is the quotient rule

3. Originally Posted by anon_404
$\frac{x^3-3x^2+1}{x^3}$

What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?
You could use the quotient rule and then the chain rule on the numerator but it is much simpler to divide each part by x^3:

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

the RHS is much easier to differentiate because each one can be differentiated separately

4. Originally Posted by anon_404
$\frac{x^3-3x^2+1}{x^3}$

What derivate rules would be needed to find the derivative of this function? I am guessing product rule and quotient. Would i first use the product rule on the numerator, then use the quotient rule on the resulting fraction?
Work from the outside in! That is, the main function is a quotient, so apply the quotient rule to it:
$\left(\frac{x^3-3x^2+ 1}{x^3}\right)'= \frac{(x^3- 3x^2+ 1)'(x^3)- (x^3- 3x^2+ 1)(x^3)'}{x^6}$

Then differentiate numerator and denominator. While " $x^3$" and " $3x^2$" are "products", you don't really use the "product rule": use the "power law", $(x^n)'= nx^{n-1}$ and the fact that, for c a constant, $(cf(x))'= c f'(x)$.

5. Using the quotient rule i now have $\frac{3x^2}{x^6}$ but using a derivative calculator gives $3x^2-6x$ how could i arrive at this answer instead?

6. Neither one is correct

quotient rule gives

[(3x^2-6x)x^3 -3x^2(x^3 -3x^2+1)]\x^6

= {3x^5-6x^4 -3x^5+9x^4 -3x^2}/x^6

= (3x^4-3x^2)/x^6

= 3x^2( x^2-1)/x^6

= 3(x^2-1)/x^4

7. Your calculator must be broken. That's the derivative of the numerator. =p

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

$(1 - 3x^{-1} + x^{-3})' = -3(-1)x^{-2} + -3x^{-4}$

8. Originally Posted by Calculus26

= 3x^2( x^2-1)/x^6

= 3(x^2-1)/x^4
Can you explain how you get these 2 lines? sorry i am having trouble with the algebra.

9. 1. factor out 3x^2 from (3x^4-3x^2)

2. divide x^2/x^6

10. I generally hate usin the quotient rule because it's so long winded. I usually just rewrite quotients like this one like this

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

Now it becomes easier to differentiate because you can do the power rule in your head!

$\frac{d}{dx}(1-3x^{-1}+x^{-3})=3x^{-2}+^-3x^{-4}$