Chapter 4 The Derivative Function

Introduction

In the last section we saw how, given a function y = f(x) and some particular value of x, we can speak of the derivative of the function at x, denoted by f'(x). In geometrical terms, f'(x) is the slope of the tangent line which grazes the curve at the point (x,f(x)). In this section, instead of taking just one particular value of x (or a few particular values, as in Problem 2 of the last homework ) , we let x vary. That is, we are interested in the rule which for every x in the domain of our function f(x) determines the value of the derivative f'(x). That gives us a new function f', called ``f prime'' or ``the derivative of f.''

Find the derivative function f'(x) for the first function discussed in the last section: .

The point on the upper semicircle above x has coordinates ( ). The radius out to that point has slope = . Since the tangent to the circle at that point is perpendicular to this radius, its slope is the negative reciprocal, i.e., . Thus, f'(x) = . For example, if you plug or -15 into this formula, you get the answers to Problem 2 of the last homework. Of course, the above calculation depends on the clever geometrical method for finding the tangent. We need a more reliable method!

Notation. There are various other ways to denote the derivative f'(x). If ,
we can also write y', or else dy/dx, , or f(x). For example, using the different types of notation in Example 1 , where f(x) = , we can write:

y' = f'(x) = = = - .

Find the derivative function for the second function in the last section: y = .

(distance traveled)/time = .

The limit of this as h approaches zero is . Thus, = . In the case of a distance function, the derivative (the instantaneous velocity ) is often denoted with a dot rather than a prime; so we can write: dot(t) = = .

Find the derivative function for the third function in the last section: y = .

We go through the procedure at the end of the last section, but with any possible x rather than with 2. The difference quotient --- the slope of the chord joining the point ( ) on the reciprocal curve to the nearby point ( ) --- is:

=

= ( = .

As h approaches zero, the denominator approaches , and so we get:

= = = .

.

Sometimes one encounters a point in the domain of a function where there is no derivative, because there is no well defined tangent line. At other times, our limiting process can fail if the tangent is vertical. First we consider the lack of a well defined tangent.

In order for the notion of the tangent line at a point to make sense, the curve must be " smooth '' at that point. This means that if you imagine a particle traveling at some steady speed along the curve, then the particle does not experience an abrupt change of direction. There are two types of situations you should be aware of --- corners and cusps --- where there's a sudden change of direction and hence no derivative.

Discuss the derivative of the absolute value function = . (This is the function that for negative x drops the minus sign and for nonnegative x leaves it unchanged.)

If x is positive, then this is the function , whose derivative is the constant 1. (Recall that when y = , the derivative is the slope m.) If x is negative, then we're dealing with the function , whose derivative is the constant -1. If , then the function has a corner, i.e., there is no tangent line. A tangent line would have to point in the direction of the curve --- but there are two directions of the curve that come together at the origin. Thus,

=

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The function y = pictured below has a derivative for every nonzero value of x (which we will find a formula for later), but it does not have a derivative at the origin, which is a cusp.

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Often in applications one has a function y = f(x) which is not given by a formula at all, but rather by drawing a smooth curve through points corresponding to a table of experimental data. Whenever the function f(x) is given to you as a curve without any formula, you can find the derivative function using a graphical method. Namely, (1) divide up the domain of your function into regular intervals; (2) for each x which is an endpoint of an interval, look at the point over x on the curve; (3) put a small ruler along the curve at that point (x,f(x)) so that it just grazes the curve there (i.e., it is tangent to the curve there); (4) find the slope of the ruler, by measuring distance up divided by distance across (for example, if you are doing this against a piece of graph paper, and if you go over to the right 5 little boxes and up 8 little boxes, then the slope is ); (5) tabulate the slopes obtained for all of the x values; (6) graph these values, drawing a smooth curve between them. The curve you draw is the derivative function .

Use graphical differentiation to find the graph of for the curve pictured below.

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