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 Pre-Calculus and Calculus
 First Principles
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/02/2007 :  23:40:37  Show Profile
Q: Provide the First Principles definition of dy/dx.

I've been looking through my text and cannot find a clear definition of what First Principles is. However, it does show how to differentiate using first principles, as a method (of which my wrist has suffered countless uses).

The answer I've come up with is.

Let A (x, f(x)) be any point on the curve f(x).
Let P be any other point, different from A by Δx, defined as P (x + Δx, f(x + Δx)).

Therefore, the slope of the secant AP

= Δy/Δx = [f(x + Δx) - f(x)] / Δx

And the derivative (dy/dx) can be defined as

dy/dx = Δy/Δx lim(Δx->0) [f(x + Δx) - f(x)] / Δx

Am I in the ballpark?
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 10/03/2007 :  08:31:46  Show Profile
"First Principles" appears in many fields of study and practice. Generally, it means only the most fundamental methodology possible. It is not always obvious what that means. I think your limit definition of the derivative is a good shot at it.
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/03/2007 :  13:08:04  Show Profile
Thanks!!
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/04/2007 :  14:53:51  Show Profile
I am further confounded by this as a later question asks:

For any function f(x), express dy/dx using limit notation. Give a clear explanation for your given answer.

I've revised my First Principles definition to:

Let A (x, f(x)) be the point on the curve f(x) where we want to find the slope.
Let P (x + Δx, f(x + Δx) be another point, different from x of A by Δx.

The slope (m) of the secant AP is given by

Δy/Δx = [f(x + Δx) - f(x)] / [(x + Δx) - x] = [f(x + Δx) - f(x)] / Δx

As P approaches A (and Δx approaches zero), the slope (m) of the secant AP approaches the slope of the tangent to the curve f(x) at A.

lim(P->A) Slope of AP = lim(Δx->0) Δy/Δx

The derivative of f(x) is the instantaneous rate of change of y with respect to x (Δy/Δx) at any particular point on the curve f(x). Therefore the derivative (dy/dx) can be defined as

lim(Δx->0) Δy/Δx

or

lim(Δx->0) [f(x + Δx) - f(x)] / Δx

I can't seem to express this definition without limit notation so I am at a standstill.

Anyway, there's a teacher chat in 2:30 that I hope I can make. In the mean time, I'll read through the text again... and try the Google.

I'd appreciate any input.
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/04/2007 :  18:47:44  Show Profile
Yes David, I do. And I have used it many times. I appreciate your query.

I think all this may require is

y' = lim(Δx->0) [f(x + Δx) - f(x)] / Δx

or

f'(x) = lim(Δx->0) [f(x + Δx) - f(x)] / Δx

Where the "y'" and "f'(x)" are the limit notation asked for.

Edited by - uberclay on 10/04/2007 18:51:27
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/04/2007 :  19:19:34  Show Profile
Much respect for those chat teachers. They have a lot of students eager for answers.

I'm all set. What they are looking for is

dy/dx = f'(x)
f'(x) = ...

And an explination.

Good night!

Edited by - uberclay on 10/04/2007 19:20:35
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pka
Advanced Member

USA
2731 Posts

Posted - 10/04/2007 :  19:38:38  Show Profile
quote:
Originally posted by uberclayThe slope (m) of the secant AP is given by
Δy/Δx = [f(x + Δx) - f(x)] / [(x + Δx) - x] = [f(x + Δx) - f(x)] / Δx
As P approaches A (and Δx approaches zero), the slope (m) of the secant AP approaches the slope of the tangent to the curve f(x) at A.
lim(P->A) Slope of AP = lim(Δx->0) Δy/Δx
The derivative of f(x) is the instantaneous rate of change of y with respect to x (Δy/Δx) at any particular point on the curve f(x). Therefore the derivative (dy/dx) can be defined as

lim(Δx->0) Δy/Δx or lim(Δx->0) [f(x + Δx) - f(x)] / Δx[/b]
I can't seem to express this definition without limit notation so I am at a standstill.

Oh yes yes you can!
Get your self the book Creative Mathematics by H. S. Wall [University of Texas Press]. Wall gives a totally limit free approach to slope beginning on page 28.
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/05/2007 :  00:06:47  Show Profile
Thanks PKA.
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