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?

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:

$\displaystyle \frac{(3(x+h)^2+4(x+h)-6)-(3x^2+4x-6)}{h}=\frac{6xh+3h^2+4h}{h}$

Re: Deriving from first principals

Hello, Swiftspice!

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

Quote:

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

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

I break this up into four steps:

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

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

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

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

Here we go!

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

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

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

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

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

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

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

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