Alternate Forms of the Derivative Formula
Date: 10/06/2001 at 23:37:55 From: Dina Subject: Equivalent Forms of the Derivative Formula I am a first-year AP Calculus student and I understand the notion of the derivative and how/why the derivative of f at x is the limit as h approaches 0 of [f(x + h)- f(x)]/h, but I am not as clear about the alternate forms of the derivative and WHY they work. Specifically, why does f'(a) = limit as x approaches a of [f(x)-f(a)]/(x-a), and how do you use this formula to find the derivative? I always get confused about when to plug in x for a or a for x. Also, why does the symmetric difference quotient work (f'(a) = limit as h approaches 0 of [f(a+h)-f(a-h)]/2h)? Why do you even divide by 2h? it looks as if you're taking the average, but I'm still not completely convinced. Any help you could provide to clarify and explain these issues would be greatly appreciated! Thanks a lot.
Date: 10/08/2001 at 11:28:54 From: Doctor Jubal Subject: Re: Equivalent Forms of the Derivative Formula Hi Dina, Thanks for writing to Dr. Math. You say you understand the usual definition of the derivative: f'(x) = lim(h->0) [f(x+h) - f(x)]/h What I'm going to do is start with this form of the derivative definition and derive the alternate forms from it. Let me know whether I succeed in enlightening you in the process. Here's a diagram showing the terms used in the traditional definition of the derivative. x is some point on the x axis. (x+h) is some point to its right. (x+h) - x = h is the distance between them. | | ______ | /| | __ / | | \_/| | | | | | | | | | | +------------------ x->|-|<- (x+h) h The original form is a limit that focuses on the distance h between two points where you evaluate the function. Let's say instead that we want to focus on the position of the points themselves. We'll call the left point a and the right point x. Then the distance between them is x-a. | | ______ | /| | __ / | | \_/| | | | | | | | | | | +---------------- a->|-|<-x x-a The only potentially confusing thing here is that the x we're using now is not the same thing as the x we were using before. It's an unfortunate choice of letters, but you can just look at the two diagrams and see which variables correspond to which other variables in our new naming scheme. Making these substitutions (a for x, x for x+h, x-a for h) gives us: f'(a) = lim((x-a)->0) [f(x) - f(a)]/(x-a) Well, if x-a is approaching zero, then x is approaching a, and we can rewrite this as the second definition you asked about: f'(a) = lim(x->a) [f(x) - f(a)]/(x-a) To get the symmetric formula, we have to try a third naming scheme: Let's call the point on the left a-h, and the point on the right a+h, and then the distance between them has to be (a+h) - (a-h) = 2h. This is where the 2h comes from. It is the distance bewteen (a+h) and (a-h). Here's a diagram to help you keep track of things. | | ______ | /| | __ / | | \_/| | | | | | | | | | | +---------------- (a-h) ->|-|<- (a+h) 2h Then, plugging these new names for the same old quantities back into the derivative definition (a-h for x, a+h for x+h, 2h for h), we get f'(a-h) = lim(2h->0) [f(a+h) - f(a-h)]/2h If 2h is approaching 0, h must be approaching zero, and if h is approaching zero, the a-h on the left-hand side must be approaching a, so we can rewrite: f'(a) = lim(h->0) [f(a+h) - f(a-h)]/2h As I hope you can see, all these definitions of the derivative are really the same definition, just with different names for the variables. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/
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