Math Goodies is a free math help portal for students, teachers, and parents.
|
Interactive Math Goodies Software

testing left nav
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums  Homework Help Forums  Pre-Calculus and Calculus  First Principles New Topic  Topic Locked  Printer Friendly
Author  Topic

uberclay

159 Posts

 Posted - 10/02/2007 :  23:40:37 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)] / ΔxAnd the derivative (dy/dx) can be defined asdy/dx = Δy/Δx lim(Δx->0) [f(x + Δx) - f(x)] / ΔxAm I in the ballpark?

tkhunny

USA
1001 Posts

 Posted - 10/03/2007 :  08:31:46 "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.

uberclay

159 Posts

 Posted - 10/03/2007 :  13:08:04 Thanks!!

uberclay

159 Posts

 Posted - 10/04/2007 :  14:53:51 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 byΔy/Δx = [f(x + Δx) - f(x)] / [(x + Δx) - x] = [f(x + Δx) - f(x)] / ΔxAs 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 aslim(Δx->0) Δy/Δxorlim(Δx->0) [f(x + Δx) - f(x)] / ΔxI 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.

uberclay

159 Posts

 Posted - 10/04/2007 :  18:47:44 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)] / Δxorf'(x) = lim(Δx->0) [f(x + Δx) - f(x)] / ΔxWhere the "y'" and "f'(x)" are the limit notation asked for. Edited by - uberclay on 10/04/2007 18:51:27

uberclay

159 Posts

 Posted - 10/04/2007 :  19:19:34 Much respect for those chat teachers. They have a lot of students eager for answers.I'm all set. What they are looking for isdy/dx = f'(x)f'(x) = ...And an explination.Good night! Edited by - uberclay on 10/04/2007 19:20:35

pka

USA
2731 Posts

 Posted - 10/04/2007 :  19:38:38 quote:Originally posted by uberclayThe slope (m) of the secant AP is given byΔy/Δx = [f(x + Δx) - f(x)] / [(x + Δx) - x] = [f(x + Δx) - f(x)] / ΔxAs 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 aslim(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.

uberclay

159 Posts

 Posted - 10/05/2007 :  00:06:47 Thanks PKA.
Topic
 New Topic  Topic Locked  Printer Friendly Jump To: Select Forum New Visitor Forum       Testing Forum Homework Help Forums       Basic Math and Pre-Algebra       Algebra       Geometry and Trigonometry       Pre-Calculus and Calculus       Probability and Statistics       Standardized Test Preparation Help       Miscellaneous Math Topics Educator Forum       Teacher Talk Parent Forum       Parent's Place  -------------------- Home Active Topics Frequently Asked Questions Member Information Search Page
 Math Forums @ Math Goodies © 2000-2004 Snitz Communications