# Finding the derivative of a fraction

• Apr 10th 2013, 11:24 PM
JellyOnion
Finding the derivative of a fraction
How do I find the derivative of (4x+(x^3)-(x^4))/x^2, where x does not equal 0

I tried using the quotient rule but I can't seem to get the right answer. Can some please work this out step by step so I can see where I'm going wrong?
• Apr 10th 2013, 11:39 PM
Also sprach Zarathustra
Re: Finding the derivative of a fraction
Hello,

Try to use the following rule:
$(\frac{f}{g})'=\frac{f'g-g'f}{g^2}$
• Apr 10th 2013, 11:58 PM
JellyOnion
Re: Finding the derivative of a fraction
I have but Im still getting it wrong
• Apr 11th 2013, 12:07 AM
Educated
Re: Finding the derivative of a fraction
It would help if you showed us your working... we're not going to give you the full answer with working if we see no effort on your part.
• Apr 11th 2013, 12:09 AM
Prove It
Re: Finding the derivative of a fraction
Quote:

Originally Posted by JellyOnion
How do I find the derivative of (4x+(x^3)-(x^4))/x^2, where x does not equal 0

I tried using the quotient rule but I can't seem to get the right answer. Can some please work this out step by step so I can see where I'm going wrong?

$\displaystyle \frac{4x + x^3 - x^4}{x^2} = \frac{4x}{x^2} + \frac{x^3}{x^2} - \frac{x^4}{x^2} = \frac{4}{x} + x - x^2 = 4x^{-1} + x - x^2$

Finding the derivative should now be a piece of cake :)
• Apr 11th 2013, 12:31 AM
JellyOnion
Re: Finding the derivative of a fraction
But then wouldn't you get (-4x^-2)+1-2x?

My textbook says the answer is (-2x^3+x^2-4)/x^2
• Apr 11th 2013, 12:32 AM
JellyOnion
Re: Finding the derivative of a fraction
Not sure how, this is my first time on this forum
• Apr 11th 2013, 02:01 AM
joeDIT
Re: Finding the derivative of a fraction
$y= \frac{4x+x^3-x^4}{x^2}=\frac{x(4+x^2-x^3)}{x^2}=\frac{4+x^2-x^3}{x}\Rightarrow \frac{dy}{dx}=$ can you get it from here using the quotient rule?
• Apr 11th 2013, 04:04 AM
JellyOnion
Re: Finding the derivative of a fraction
Ohhhhhhh, yeah I got it thanks so much Joe :)
• Apr 11th 2013, 06:30 PM
hollywood
Re: Finding the derivative of a fraction
Quote:

Originally Posted by JellyOnion
But then wouldn't you get (-4x^-2)+1-2x?

My textbook says the answer is (-2x^3+x^2-4)/x^2

Aren't those the same thing?

If you want to use the quotient rule directly, it's more work, but here's what it looks like:

$\left(\frac{4x+x^3-x^4}{x^2}\right)' =$

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

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

$\frac{4x^2+3x^4-4x^5-8x^2-2x^4+2x^5}{x^4}=$

$\frac{-4x^2+x^4-2x^5}{x^4}=$

$-\frac{4}{x^2}+1-2x$

- Hollywood
• Apr 11th 2013, 09:22 PM
JellyOnion
Re: Finding the derivative of a fraction
Oh yeah your right, I didn't see that. Thanks Hollywood
• Apr 11th 2013, 11:01 PM