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Date: 10/28/96 at 12:53:53
From: Kandice Longhurst
Subject: Need help with Calculus!


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 

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:


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 

      lim     (f(10+y)-f(10))
     y -> 0  ----------------    

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 

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  ----------------

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)  

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 

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!   
Associated Topics:
High School Calculus
High School Definitions

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