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?

Thanks in advance

Re: Finding the derivative of a fraction

Hello,

Try to use the following rule:

$\displaystyle (\frac{f}{g})'=\frac{f'g-g'f}{g^2}$

Re: Finding the derivative of a fraction

I have but Im still getting it wrong

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.

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?

Thanks in advance

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

Finding the derivative should now be a piece of cake :)

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

Re: Finding the derivative of a fraction

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

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?

Re: Finding the derivative of a fraction

Ohhhhhhh, yeah I got it thanks so much Joe :)

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:

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

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

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

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

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

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

- Hollywood

Re: Finding the derivative of a fraction

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

Re: Finding the derivative of a fraction

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Re: Finding the derivative of a fraction

Re: Finding the derivative of a fraction

no problem JellyOnion good luck