Function problem

**Jools** · February 9th 2010, 07:58 AM

Hey there. I am working on a function problem with which I am having a hard time getting started on.

Differentiate the following function:

f(x) = 3x/x^2 + 4

Once I can get it to polynomial form I'll be good to go, but I'm a little stuck as it stands now. Could somebody give me a nudge to get me started? I think it has something to do with x^-n = 1/x^n.... Thanks.

**Nyrox** · February 9th 2010, 08:11 AM

The derivative of a quotient of functions is given by the following rule:

$\left[\frac{f}{g}\right]'=\frac{f'\cdot g-f\cdot g'}{g^2}$ .

I think you can manage from there

**Jools** · February 9th 2010, 09:37 AM

I have never seen this rule before.... So my f(x) = 3x/x^2 + 4, but what detremines my g(x)? Thanks again.

**Plato** · February 9th 2010, 09:42 AM

Originally Posted by Jools

I have never seen this rule before.... So my f(x) = 3x/x^2 + 4, but what detremines my g(x)? Thanks again.

What course are you doing?
Why were you asked to find the derivative of a quotient if you have never seen the quotient rule?

**Jools** · February 9th 2010, 09:47 AM

It's a calculus and vectors pre-university course. We were show the power, constant, sum and difference, and product rules, as well as the chain rule for powers of polynomials...

**danielomalmsteen** · February 9th 2010, 09:52 AM

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

using te quotient rule

$f`(x) = \frac{3(x^2+4) - 6x^2}{(x^2+4)^2}$

**Henryt999** · February 9th 2010, 09:53 AM

Why dont you solve it this way then,
$f(x) \frac{3x}{x^2}+4$

You can simply to $f(x) \frac{3}{x} +4$

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

And this you probably seen before?
Or did I miss someting?... is that 4 suppose to be in the denominator? That "hint" you gave in your first post made me think that 4 is not in the term $\frac{3x}{x^2}$

**Plato** · February 9th 2010, 10:01 AM

Originally Posted by Jools

It's a calculus and vectors pre-university course. We were show the power, constant, sum and difference, and product rules, as well as the chain rule for powers of polynomials...

$\left[ {\frac{f}<br /> {g}} \right]^\prime = \left[ {fg^{ - 1} } \right]^\prime = f'g^{ - 1} + f\left( { - g^{ - 2} g'} \right) = \frac{{f'g - fg'}}<br /> {{g^2 }}$

**Jools** · February 9th 2010, 10:21 AM

Alright thanks guys. I think I've got it now. I was never shown the quotient rule. My only question is how come the derivative for each of the two last methods is so different?

**Jools** · February 9th 2010, 05:32 PM

K... Here's how I did it. It's a little different approach...

f(x) = 3x / x^2 + 4
= (3X)(X^2 + 4)^-1

Then used a combination of the product rule and chain rule to come up with:

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

Look right?

**Nyrox** · February 10th 2010, 01:39 AM

You're missing a square. It's -6x^2

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