Simple Differentiation Question

**Rudey** · January 23rd 2010, 09:49 PM

Hi,

I've hit abit of a roadblock. I was wondering, how would I solve this (using the differntiation rules, not by first principles.) I can't see how to approach this. Could someone show working (such as expanding / factorising) in order to be able to use those rules to solve:

Differentiate with respect to x: $\frac{4x^2+2x}{x^2}$

Thanks!

**VonNemo19** · January 23rd 2010, 09:53 PM

Originally Posted by Rudey

Hi,

I've hit abit of a roadblock. I was wondering, how would I solve this (using the differntiation rules, not by first principles.) I can't see how to approach this. Could someone show working (such as expanding / factorising) in order to be able to use those rules to solve:

Differentiate with respect to x: $\frac{4x^2+2x}{x^2}$

Thanks!

Note that $\frac{4x^2+2x}{x^2}=\frac{1}{x^2}{(4x^2+2x)}$ .

So, we can distribute the $\frac{1}{x^2}$ like so:

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

So, can you find $\frac{d}{dx}[4+2x^{-1}]$ ?

**Rudey** · January 23rd 2010, 10:17 PM

Hi. Thanks for reply. I can find the derivative of $4 + 2x^-1$ = -2x^x-2

I'm still confused as what happens with distributing the 1/x^2. I don't see how you go that? Could you explain?

Sorry about that

Thanks!

**VonNemo19** · January 23rd 2010, 10:37 PM

Originally Posted by Rudey

Hi. Thanks for reply. I can find the derivative of $4 + 2x^-1$ = -2x^-2

I'm still confused as what happens with distributing the 1/x^2. I don't see how you go that? Could you explain?

Sorry about that

Thanks!

Yeah, no problem

So, take a look at $\frac{a+b}{c}$ .

It should be clear that this is equivalant to $\frac{a}{c}+\frac{b}{c}$

And that this is equivalent to $\frac{1}{c}a+\frac{1}{c}b$ .

Now, since $\frac{1}{c}$ is common to both terms, I can factor it out giving me

$\frac{1}{c}(a+b)$ .

Can you see how this applies to your problem? Do you remember this from algebra. It is calle the Distributive Law. It's a biggie. And you're gonna need it.

**Rudey** · January 23rd 2010, 10:43 PM

Hi. Ah yeah I remember that.

So when differentiating this do you differentiate $\frac{1}{x^2}$ and multiple it by the derivative of $4x^2+2x$ ?

**Prove It** · January 23rd 2010, 10:48 PM

Originally Posted by Rudey

Hi. Ah yeah I remember that.

So when differentiating this do you differentiate $\frac{1}{x^2}$ and multiple it by the derivative of $4x^2+2x$ ?

No.

$\frac{4x^2 + 2x}{x^2} = 4 + \frac{2}{x}$ .

So you just differentiate $4 + \frac{2}{x}$ .

If you didn't want to simplify at the beginning, you would have to use the Quotient Rule, but that's a lot messier...

**VonNemo19** · January 23rd 2010, 10:54 PM

Originally Posted by Rudey

Hi. Ah yeah I remember that.

So when differentiating this do you differentiate $\frac{1}{x^2}$ and multiple it by the derivative of $4x^2+2x$ ?

No. All of the algebra was done as to simplify the expression so that taking the derivative would be easy.

Note that there are a whole bunch of ways to differentiate something:

The Quotient Rule is one way we coulda went:

$\frac{dy}{dx}=\frac{x^2\cdot(8x+2)-(4x^2+2x)\cdot2x}{(x^2)^2}=\frac{8x^3+2x^2-8x^3-4x^2}{x^4}=-2x^{-2}$

But why go through all that trouble if we coulda simplified the expression first, and the used the power rule, which is so much easier.

**Rudey** · January 23rd 2010, 11:08 PM

Oh. I get it!!! Thank you both so much!!!

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