Thread: Finding the derivative of a fraction

1. 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?

2. Re: Finding the derivative of a fraction

Hello,

Try to use the following rule:
$\displaystyle (\frac{f}{g})'=\frac{f'g-g'f}{g^2}$

3. Re: Finding the derivative of a fraction

I have but Im still getting it wrong

4. 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.

5. Re: Finding the derivative of a fraction

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

Finding the derivative should now be a piece of cake

6. 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

7. Re: Finding the derivative of a fraction

Not sure how, this is my first time on this forum

8. Re: Finding the derivative of a fraction

$\displaystyle y= \frac{4x+x^3-x^4}{x^2}=\frac{x(4+x^2-x^3)}{x^2}=\frac{4+x^2-x^3}{x}\Rightarrow \frac{dy}{dx}=$ can you get it from here using the quotient rule?

9. Re: Finding the derivative of a fraction

Ohhhhhhh, yeah I got it thanks so much Joe

10. Re: Finding the derivative of a fraction

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:

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

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

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

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

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

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

- Hollywood

11. Re: Finding the derivative of a fraction

Oh yeah your right, I didn't see that. Thanks Hollywood

12. Re: Finding the derivative of a fraction

私に大きな助けを共有していただきありがとうございます。私の家へようこそ！中国ドラマ宮廷女官ジャクギ

nice job!

14. Re: Finding the derivative of a fraction

no problem JellyOnion good luck