# Deriving from first principals

• Aug 21st 2013, 11:55 PM
Swiftspice
Deriving from first principals
I don't quite understand how to derive from first principals.

I understand that it is a point x+h - x for delta x and how h is delta y but i still have trouble working through them. i have tried googleing a few examples but the formatting hurts my brain.

Would anyone mind going through an example for me of 3x^2+4x-6

I believe it starts at (3(x+h)^2+4(x+h)-6)-(3x^2+4x-6) all over h lim as h -> 0 but would someone mind taking me step by step through it?
• Aug 22nd 2013, 04:01 AM
Plato
Re: Deriving from first principals
Quote:

Originally Posted by Swiftspice
I understand that it is a point x+h - x for delta x and how h is delta y but i still have trouble working through them. i have tried googleing a few examples but the formatting hurts my brain.
Would anyone mind going through an example for me of 3x^2+4x-6

You must be able to do very simple algebra to do this:
$\frac{(3(x+h)^2+4(x+h)-6)-(3x^2+4x-6)}{h}=\frac{6xh+3h^2+4h}{h}$
• Aug 22nd 2013, 09:20 AM
Soroban
Re: Deriving from first principals
Hello, Swiftspice!

I'll show you the way I taught it to my Calculus classes.

Quote:

Differentiate $f(x) \:=\:3x^2+4x-6$ from first principles.

Definition: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

I break this up into four steps:

[1] Find $f(x+h)$. . . . Replace $x$ with $x+h$ . . . and simplify.

[2] Subtract $f(x)$ . . . Subtract the original function . . . and simplify.

[3] Divide by $h$ . . . . Factor and reduce.

[4] Let $h\to0$ . . . . . .Let $h = 0$ . . . and simplify.

Here we go!

$[1]\;f(x+h)\;=\;3(x+h)^2 + 4(x+h) - 6 \;=\; 3x^2 + 6xh + 3h^2 + 4x + 4h - 6$

$[2]\;f(x+h) - f(x) \;=\;(3x^2 + 6xh + 3h^2 + 4x + 4h - 6) - (3x^2 + 4x - 6)$

. . . . . . . . . . . . . . $=\;3x^2 + 6xh + 3h^2 + 4x + 4h - 6 - 3x^2 - 4x + 6$

. . . . . . . . . . . . . . $=\; 6xh + 3h^2 + 4h$

$[3]\;\frac{f(x+h)-f(x)}{h} \;=\;\frac{6xh + 3h^2 + 4h}{h} \;=\;\frac{h(6x+3h+4)}{h}$

. . . . . . . . . . . . . . $=\;6x+3h+4$

$[4]\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}(6x+3h+4) \;=\;6x + 0 + 4 \;=\;6x+4$

Therefore: . $f'(x) \;=\;6x+4$