# differentiation from first principals

• May 9th 2006, 01:54 AM
differentiation from first principals
Hi Math help forum!

I'm finding it difficult to differentiation from first principals and wanted to know if there is a simple method I can follow. Could you please give an example aswell thanks.

regards

• May 9th 2006, 07:40 AM
TD!
"from first principles" would be using the (limit) definition?
• May 9th 2006, 09:33 AM
Thanks for the reply

Yes any method as long as it easy to follow and you get the right answer! :D
• May 9th 2006, 09:36 AM
TD!
Well there is no standard 'way' of doing them, since evaluating limits can go very different, depending on what limit it is. For what kind of functions do you have to be able to get its derivative through the definition?
• May 9th 2006, 11:28 AM
Jameson
Just so I'm clear, is this what you are talking about?

$f'(x)=\lim_{\Delta{x}\rightarrow{0}}\frac{f({x}{+} \Delta{x})-f(x)}{\Delta{x}}$?
• May 9th 2006, 02:03 PM
ThePerfectHacker
I realized that you used
Code:

\rightarrow
While you can have used,
Code:

\to
• May 9th 2006, 02:46 PM
yeah jameson when the limit turns zero.
• May 9th 2006, 05:50 PM
Jameson
Quote:

Originally Posted by dadon
yeah jameson when the limit turns zero.

That's what I wrote. Hence the $\Delta{x}\to{0}$. What seems to be the problem? These problems usually require some basic algebra manipulation. Try finding the derivatives of $f(x)=x^2$ and $f(x)=\frac{1}{x}$ using the limit definition. Or is there a specific one we can help you with?
• May 9th 2006, 05:51 PM
Jameson
PerfectHacker:

Thanks. A small time saver. :)
• May 9th 2006, 06:40 PM
ThePerfectHacker
Quote:

Originally Posted by Jameson
PerfectHacker:

Thanks. A small time saver. :)

I also made the same thing, then I was curious to see what Code TD! used and a saw he used a simpler one used it ever since.
• May 10th 2006, 01:02 AM
re:
So say I had to find from first principals $\frac{dy}{dx}$ of the following:

$y = 16x + \frac{1}{x^2}$

cheers guys
• May 10th 2006, 09:16 AM
Jameson
Set up your limit.

$\frac{dy}{dx}=\lim_{\Delta{x}\to{0}}\frac{16(x{+}{ \Delta}x)+\frac{1}{(x{+}{\Delta}{x})^2}-16x-\frac{1}{x^2}}{\Delta{x}}$

Is this where you're having trouble?
• May 10th 2006, 09:17 AM