DerivativesDate: 10/28/96 at 12:53:53 From: Kandice Longhurst Subject: Need help with Calculus! Hello, I am taking calculus this year and I don't understand the concept of the derivative. Could you explain it to me? It would be very helpful. Thank you. Date: 10/29/96 at 1:3:21 From: Doctor Daniel Subject: Re: Need help with Calculus! Hi Kandice, The concept of a derivative is really very important and I hope I can help explain it to you. Unfortunately, mathematicians use it so often that we sometimes assume that this idea, like many other important ones in math, is so obvious that it should be immediately clear. But it's really not. Let's say you were given a function f, which is a function from real numbers to real numbers. (I'm hoping you're pretty clear on the idea of functions here. If you're not, please feel free to ask about that too; it's another extremely important idea which people don't always get.) Anyhow, we have this function f : R -> R. (That notation is just shorthand for f maps from reals to reals.) Maybe, to put a concrete feel to it, f is a function that maps the time, since an experiment began with the amount of radioactive material in a source which is decaying. But the idea is general enough to just work for any function from R -> R. Now, suppose you want to find how fast the function is changing at any given moment. For example, you might be curious how fast the source is emitting radioactive particles. This would be measured in particles/ sec, and it should be fairly easy to compute, all things considered. How would we try doing it? Well, let's say we wanted to find the rate of change of the function at the time = 10 seconds. We could compute the amount of particles left in the sample after 10 seconds, f(10), and the amount of particles left after 20 seconds, f(20). f(20) - f(10) is the amount of particles emitted in 10 seconds. If f(x) is a straight line, then the rate of change at x = 10 seconds will be roughly: f(20)-f(10) ----------- 20-10 This is the slope of the line from (20, f(20)) to (10, f(10)). But suppose it's not a line. Then the rate of change at 10 sec is probably closer to f(11)-f(10), which is the slope of the line from (11, f(11)) to (10, f(10)). In fact, if we get closer and closer to 10, we're computing the slope of a line that is closer and closer to what the function looks like closer and closer to when x = 10. So, for example, 10 * (f(10.1)-f(10)) is closer to our desired rate of change than 1/10 * (f(20)-f(10)). Remember that we're really just computing the slope of lines that look a lot like our curve. What we want is the slope of the curve itself. If you're pretty clear on the definition of a limit, you should see the answer at this point. The slope of the curve ("the rate of change of the function f","the derivative of f") at x = 10 will be this limit: lim (f(10+y)-f(10)) y -> 0 ---------------- y The denominator and numerator are both approaching zero! In practice, however, the value of the numerator will include something with a power of y which will cancel the y on the denominator, and we'll be fine. This is the definition of the derivative of f in general, rather than just at x = 10: f'(x) = lim (f(x+y)-f(x)) y -> 0 ---------------- y What do we know that this must mean? Well, for starters, f has to be a continuous function at the point we're considering. Do you remember what continuity means? That means you could draw the graph of the function without ever lifting a pencil off the paper. In language of limits, that means lim f(x+y) = f(y) y->0 Otherwise, the numerator won't be zero and the limit will be infinite. Another requirement is that the function doesn't change directions suddenly as you get closer to x. Another (though it's obvious) is that the function has to have a value at x; that is, f(x) has to exist. If the derivative has a value at all points, then it's clear that it's also a function. So, for our function f, we can talk about the function f', which is its derivative, where f'(x) is the derivative of f at the given point. You've probably learned formulas by now which help you compute these functions. That's a very long answer to a very important question. The derivative of the function at a given point is the slope of the curve at that point. The way we compute it is by computing the slope of line segments which get closer and closer to the point itself, and eventually taking the limit of this process. It's a function itself, and in practice, rather than computing the limit directly, we compute it using formulas which make our lives much easier. I hope this helps. If you're still confused, please don't hesitate to ask again. It would almost certainly help you to draw some pictures and try out the kind of construction I went through with words above. Good luck, Kandice! Happy math! -Doctor Daniel, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/