Using Differentiation to Find Derivatives

Date: 11/16/2002 at 16:32:22
From: Scott VanHorn
Subject: Derivatives

Dear Dr. Math,

I am a student in a high school Calculus math class. This is our first 
year of calculus.

Recently our class has been working on differentiation and finding 
derivatives. However, we are having a difficult time understanding and 
grasping the concepts of derivatives. We would greatly appreciate it 
if you could simplify the whole idea of using differentiation to find 
derivatives. (It would be extremely helpful if you could possibly 
create for us a real life example or a practical application in which 
derivatives are used.)

Thank you for your time,
Sincerely,
Scott VanHorn and fellow Calculus students

Date: 11/16/2002 at 17:04:07
From: Doctor Jerry
Subject: Re: Derivatives

Hi Scott,

We can't compete with a textbook nor the presence of a teacher, who 
can interact with those listening to him, but I will say a few things 
in the hope that they will help.

First, the idea of a limit is something that is a prerequisite for 
understanding the idea of a derivative. I'll assume that you are 
familiar with the idea of limit.

With certain functions f (whose graphs are "smooth") we may associate 
a second function, which we may designate as f' or, preferred by 
some, Df. What is the value of f'(x) = Df(x)? In terms of the graph 
of f, the value f'(x) is the slope of the tangent line to f at the 
point (x,f(x)). Thus, for example, with the function f(x)=x^2 we 
associate the function f'(x)=2x. If you graph f you will see a 
parabola, opening upwards. Look now at the point (2,f(2))=(2,2^2)=
(2,4). The slope of the line tangent to the graph of f at (2,4) is 
f'(2)=2*2=4. At the point (3,9), the slope of the tangent to the graph 
of f at (3,9) is f'(3)=2*3=6. And so on.

That's one interpretation of f'; it gives the slopes of the tangent 
lines to the graph of f. There are many other interpretations; these 
may depend upon a physical interpretation of x and f(x). If, for 
example, x is time and f(x) is the AMOUNT of radium remaining at time 
x  (radium decays into something else; into, I think, radon, a gas), 
then f(x) has the form

f(x) = C*e^{-k*x},

where C and k are constants depending upon the amount of radium at 
time 0 (x=0) and upon the physical characteristics of radium. We may 
assume, for example, that f(x) gives the amount (in grams) of radium 
present at time x (in years).

f'(x) then gives the rate at which the radium is decaying, in grams 
per year.

You already know that the derivative of f at x is given by

f'(x) = limit as h->0 of the difference quotient [f(x+h)-f(x)]/h. 
  
Some prefer to write this in the equivalent form

f'(x) = limit as x->a of the difference quotient [f(x)-f(a)]/(x-a).

Graphically, the difference quotient [f(x)-f(x)]/(x-a) may be 
interpreted as the slope of the line joining points (a,f(a)) and 
(x,f(x)) on the graph of f.

As x approaches a, the slope of this line approaches the slope of the 
tangent line to the graph of f at (a,f(a)), assuming that it exists.

In the case of f(x) = amount of radium at time x, the difference 
quotient
 
[f(x)-f(a)]/(x-a)

is the change in the amount of radium between times a and x, divided 
by the amount of time x-a between these two times. This gives the 
approximate change in grams per year happening at time a. If x is 
very near a, then the difference quotient will be  very near the 
"instantaneous" rate of change of the radium at time a.

You  might be interested in the fact that in a few months you will be 
able to "differentiate" the expression 

f(x) = C*e^{-k*x},

finding that

f'(x) = (-k)*C*e^{-k*x} .

After understanding the IDEA of derivative, the actual calculation is 
relatively easy.

I hope that this helps a bit.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/