Using Differentiation to Find DerivativesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/