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Math Help - Deriving from first principals

  1. #1
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    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?
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  2. #2
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    Re: Deriving from first principals

    Quote Originally Posted by Swiftspice View Post
    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}
    Last edited by Plato; August 22nd 2013 at 03:04 AM.
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  3. #3
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    Re: Deriving from first principals

    Hello, Swiftspice!

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


    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
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