# Simple Differentiation Question

• Jan 23rd 2010, 09:49 PM
Rudey
Simple Differentiation Question
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: $\displaystyle \frac{4x^2+2x}{x^2}$

Thanks!
• Jan 23rd 2010, 09:53 PM
VonNemo19
Quote:

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: $\displaystyle \frac{4x^2+2x}{x^2}$

Thanks!

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

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

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

So, can you find $\displaystyle \frac{d}{dx}[4+2x^{-1}]$ ?
• Jan 23rd 2010, 10:17 PM
Rudey
Hi. Thanks for reply. I can find the derivative of $\displaystyle 4 + 2x^-1$ = -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?

Thanks!
• Jan 23rd 2010, 10:37 PM
VonNemo19
Quote:

Originally Posted by Rudey
Hi. Thanks for reply. I can find the derivative of $\displaystyle 4 + 2x^-1$ = -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?

Thanks!

Yeah, no problem

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

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

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

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

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

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.
• Jan 23rd 2010, 10:43 PM
Rudey
Hi. Ah yeah I remember that.

So when differentiating this do you differentiate $\displaystyle \frac{1}{x^2}$ and multiple it by the derivative of $\displaystyle 4x^2+2x$?
• Jan 23rd 2010, 10:48 PM
Prove It
Quote:

Originally Posted by Rudey
Hi. Ah yeah I remember that.

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

No.

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

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

If you didn't want to simplify at the beginning, you would have to use the Quotient Rule, but that's a lot messier...
• Jan 23rd 2010, 10:54 PM
VonNemo19
Quote:

Originally Posted by Rudey
Hi. Ah yeah I remember that.

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

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:

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

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.
• Jan 23rd 2010, 11:08 PM
Rudey
Oh. I get it!!! Thank you both so much!!!