# Find derivative of: (3 - x^2)/(4 + x^2); Quotient Rule vs. Product Rule

1. Jan 10, 2013

### Lo.Lee.Ta.

Hi, I got the right answer when I used the Quotient Rule but not when I used the Product Rule...
I think it might be an algebra mistake...

Product Rule Method:

f'(x) = (3 - x^2)*(4 + x^2)^-1

= (3 - x^2)[(-1(4 + x^2)^-2)*2x] + [(4 + x^2)^-1](-2x)

= [(3 - x^2)(-2x)]/[(4 + x^2)^2] + [(-2x)/(4 + x^2)]

To get the same denominator here, I thought I might square the factor on the right.

= (-6x + 2x^3 + 4x^2)/[(4 + x^2)^2] <----- This is not right.

THE RIGHT ANSWER IS: -14x/[(4 + x^2)^2]
I found that by using the Quotient Rule.
...But where am I going wrong with the Product Rule...?

P.S. Oh, and a side note:
If I wrote out the Product Rule answer like this: (-6x + 2x^3 + 4x^2)/(x^4 + 8x^2 + 16)

Would the 4x^2 and the 8x^2 be able to cancel somewhat?
Sometimes I get confused about what can cancel.
Could it turn into:
(-6x + 2x^3)/(x^4 + 2 + 16) = (-6x + 2x^3)/(x^4 + 18)?

I know that's not the right answer, but is this how it would cancel?

Thank you very much!

2. Jan 10, 2013

### Staff: Mentor

given a fraction like (a+b / (a + c) you can't cancel the a terms out to get b / c

but if you have a * b / a * c then you can cancel the a factors out to get b / c

with that in mind try reviewing your work to see where you went wrong, the product and quotient rules should give the same answers.

3. Jan 10, 2013

### epenguin

rewritten a bit more correctly what you surely meant

As far as I understand what you are saying, there is just no rationale for your operation of squaring. You have a sum, represent it X + Y . You are saying if you square a bit of it -> X + Y2 that will be the same thing?

You have something of form A/B2 + C/B .

In general how do you combine that into one?

4. Jan 10, 2013

### HallsofIvy

Staff Emeritus
Be careful to write what you mean! This is f, not f'. You have not yet differentiated.
$f(x)= (3- x^2)(4+ x^2)^{-1}$

Okay, this is correct.

Yes, it is not right. If you have 2/3 and want to make the denominator 9, you multiply both numerator and denominator to get 6/9 which, of course, represents the same number as 2/3. If you square, you get 4/9 which is NOT the same.

Going back to
$$-\frac{2x(3- x^2)}{(4+x^2)^2}- \frac{2x}{4+ x^2}$$
if you multipy both numerator and denominator of the second fraction by $4+ x^2$ you get
$$-\frac{6x- 2x^3}{(4+ x^2)^2}- \frac{8x+ 2x^3}{(4+ x^2)^2}$$
$$= \frac{-6x+ 2x^3- 8x- 2x^3}{(4+ x^2)^2}= \frac{-14x}{(4+x^2)^2}$$

Last edited: Jan 10, 2013
5. Jan 10, 2013

### Lo.Lee.Ta.

WOW! HallsofIvy! You gave me the BEST answer I have ever gotten in my experience on Physics Forums!

You wrote out EXACTLY where I went wrong, and explained WHY it was wrong!

I get it now! My squaring method was off!

I worked the problem by multiplying instead by (4 + x^2) and got the right answer!

Thank you SO much for your help!
With the other answers, I didn't understand what was wrong with my method!

If this were YahooAnswers, you'd get voted "Best Answer!" :)

6. Jan 10, 2013

### mtayab1994

Just apply what you've learned: $$\frac{(3-x^{2})'(4+x^{2})-(4+x^{2})'(3-x^{2})}{(4+x^{2})^{2}}=\frac{-14x}{(x^{2}+4)^{2}}$$

7. Jan 11, 2013

### epenguin

You heard it here first. :tongue: