Chap 2 A

Representation and Types of Functions Chapter 2A August 1999

1

The tangent problem revisited

Recall that points on the graph of a function f have the form (x, f (x)); that is, if a point (x, y) is on the graph of the function f , then y = f (x). We have defined a secant line for f as a line which passes through two distinct points on the graph of f , say (a, f (a)) and (b, f (b)). The slope of a secant line passing through these points will be m=

f (b) − f (a) . b−a

Of course, we are really interested in computing the slope of tangent lines; if we wish to compute the slope of a tangent line to the graph of f at a, we will compute f (b) − f (a) lim b→a b−a or the equivalent limit f (a + h) − f (a) . lim h→0 h The function f (a + h) − f (a) Fa (h) = h represents the slope of a secant line for the function f passing through the points (a, f (a)) and (a + h, f (a + h)). So, for example, if we define f (x) = x3 and take a = 2, then we can approximate the slope of the tangent line to the graph of y = f (x) by computing the slope of secant lines passing through points on the graph corresponding to points with x coordinate 2 and 2 + h for values of h getting smaller and smaller: h 1/10 1/100 1/1000 1/10000

(2+h)3 −23 h

1261/100 = 12.61 120601/10000 = 12.0601 12006001/1000000 = 12.006001 1200060001/100000000 = 12.00060001

The slopes of the secant lines are getting closer to 12 as h gets smaller. Can we determine if 12 is actually the slope of the tangent line? That is, can we compute (2 + h)3 − 23 ? h→0 h lim

We can expand and simplify to learn that, as long as h is not equal to 0,   2 ((2 + h) − 2) (2 + h) + (2 + h) 2 + 22 (2 + h)3 − 23 = h h   =

2

(2 + h) + (2 + h) 2 + 22 1

But this last expression clearly gets close to 12 as h gets close to 0, and we learn that the slope of the tangent line to the graph of y = x3 at the point (2, 8) is 12. We say that the derivative of f at 2 is 12. Remember that, given the slope m and a point (x0 , y0 ) on a line, the equation of the line is y − y0 = m(x − x0 ), or y = y0 + m(x − x0 ). So, in this case, the equation of the tangent line to the graph of f (x) = x3 at (2, 8) is y = 8 + 12(x − 2). As mentioned, we say that the derivative of f at 2 is 12, the slope of the tangent line to the graph of f at the point corresponding to x = 2. Def: The derivative of f at x, denoted as f 0 (x), is defined as f 0 (x) = lim

h→0

f (x + h) − f (x) . h

It is often possible to compute this limit for all values of x simultaneously, in which case we have constructed a new function, f 0 , which is simply called the derivative of f. For example, since a3 − b3 = (a − b)(a2 + ab + b2 ) we learn that 3

(x + h) − x h

  2 ((x + h) − x) (x + h) + (x + h) x + x2

3

= =

h 2 (x + h) + (x + h) x + x2

So, 3

(x + h) − x3 h→0 h lim

= =

  2 lim (x + h) + (x + h) x + x2

h→0 2

3x .

That is, the slope of the tangent line to the graph of y = x3 at any point (x, x3 ) is 3x2 . Note that this agrees with our computation of the slope at the point (2, 8) . Our goal is to discover a variety of formulas which will allow us to compute derivatives in a relatively simple way. But be sure to remember: the derivative of a function at a particular point, if it exists, is the slope of the tangent line to the graph at that point.

2

Differentiation formulas

We now have a formal definition of the derivative f 0 (x) of the function f at x: f 0 (x) = lim

h→0

f (x + h) − f (x) , h

provided this limit exists. We also know that this limit represents the slope of the tangent line to the graph of f at the point (x, f (x)). However, if we are going to put this idea to use, we need to discover some underlying principles or properties which will allow us to compute derivatives for broad classes of functions without having to resort to the definition. For example, if the function f is linear (i.e. if f (x) = mx + b, for some constants m and b), then the graph of f is a line, and it should be clear that the tangent line at any point is just the original line. So, the derivative of a linear function must be its slope. (Symbolically, if f (x) = mx + b then f 0 (x) = m.) (In the above, we explained why the derivative of a linear function is the slope from a geometric point of view. Use the definition of the derivative to verify that our reasoning was correct, and explain why this yields the following result: If f is a constant function, then f 0 (x) = 0 for all x.)

2

Many basic rules for differentiation follow directly from the properties of the limit. For example, if f (x) = g(x) + h(x), it follows immediately from the definition of the derivative and the properties of the limit that f 0 (x) = g 0 (x)+h0 (x), provided that both of these exist. Similarly, if f (x) = c∗g(x) for some constant c, then f 0 (x) = c∗g 0 (x). Both of these ”rules” allow us to break larger problems into small pieces, and this is just the sort of thing we want to do. We now turn our attention to developing some of the more important rules for differentiation.

2.1

The power rule

One of the most important rules for differentiation tells us how to differentiate a function f (x) = xn ; for some integer n. We have already seen that, for f (x) = x3 , f 0 (x) = 3x2 . Recall that this depended on the formula (x + h)3 = x3 +3x2 h+3xh2 +h3 . For n a positive integer, the binomial theorem tells us that (x+h)n = xn +nxn−1 h+h2 (stuff), where “stuff” is used here to represent other algebraic terms involving non-negative powers of x and h. Using this, we find that if f (x) = xn , then n

f 0 (x)

(x + h) − xn h→0 h n x + nxn−1 h + h2 (stuff) − xn = lim h→0 h nxn−1 h + h2 (stuff) = lim h→0 h
n−1 = lim nx + h (stuff) . =

lim

h→0

But, as h gets close to 0, the expression inside the last limit gets close to nxn−1 . This is the power rule; f (x) = xn =⇒ f 0 (x) = nxn−1 . This of course agrees with our computation of the derivative of the cube function. Since we already know that the derivative of a sum of functions (in this case, we will think of each term of a polynomial as a seperate function) is the sum of the derivatives, and that the derivative of a constant times a function is the constant times the derivative of the function, we can now readily compute the derivative of any polynomial. For example, if p(x) = x4 − 2x3 + 5x2 + 3x + 1, then p0 (x)


d x4 − 2x3 + 5x2 + 3x + 1 dx


d d d d = x4 − 2x3 + 5x2 + (3x) + dx dx dx dx


d d d d = x4 − 2 x3 + 5 x2 + 3 (x) + dx dx dx dx
= 4x3 − 2 3x2 + 5 (2x) + 3 (1) + 0 =

= 2.1.1

d (1) dx d (1) dx

4x3 − 6x2 + 10x + 3.

The power rule for negative and non-integral powers

Though the derivation is more complicated, it turns out that the power rule holds for any power, for whatever values of x the resulting expression makes sense.

3

For example, if f (x) =

√

x = x1/2 ,

then f 0 (x)

= =

1 1/2−1 1 x = x−1/2 2 2 1 √ . 2 x

This makes sense as long as x > 0. Here is another example. Since f (x) =

1 = x−1 , x

we will have f 0 (x)

(−1) x−2 1 = − 2. x =

This makes sense as long as x 6= 0.

2.2

An application: Newton’s method

Before moving on to other procedures for calculating derivatives, let’s illustrate the use of the power formula, as well as an important application of the derivative, by using the tangent line to approximate the root of a function. Here is the idea: given a function f which we know has a root near some number x0 , we will construct the tangent line to the graph at (x0 , f (x0 )) and use the x-coordinate of the point of intersection of the tangent line with the x-axis as a better approximation to the root. 2 For − 2 has a root between 1 and 2. (In fact, we know that the root of f is at √ example, the function f (x) = x √ x = 2, and our calculator tells us that 2 ≈ 1.4. But we want to compute a good approximation, not just write down a symbol representing the number.) We might say that 1.5 is a good starting approximation. (1.52 = 2.25, so f (1.5) = .25) Since f 0 (x) = 2x, the slope of the tangent line to the graph of f at the point (1.5, .25) is 3, so the equation of the tangent line is y = .25 + 3(x − 1.5). Let’s look at a plot of the function and the tangent line.

Figure 1: Approximating a root using the tangent line. You can see from the graph that the tangent line intersects the x-axis very close to the root. We and easily solve for the point of intersection of the tangent line and the x-axis; the y coordinate is 0, so the x-coordinate must satisfy 0 = .25 + 3(x − 1.5). We find x = 1.5 − .25/3 = 1.416666.... This turns out to be a pretty good approximation to √ 2. However, we can obtain an even better approximation by repeating the process, using our new approximation as a starting value. (“Repeating the process” is called “iteration.”) Rounding the result to 2 places, we find f (1.42) = .0164 and f 0 (1.42) = 2.84, so the equation of the tangent line to the graph of f at the point (1.42, .0164) will be 4

y = .0164 + 2.84(x − 1.42). The √ x-coordinate of the point of intersection of this line with the x-axis is approximately 1.414225. How close is this to 2? Let’s examine this process more closely. At the point (x0 , f (x0 )), the slope of the tangent line is f 0 (x0 ) and the equation for the tangent line will be y = f (x0 ) + f 0 (x0 )(x − x0 ). Setting y = 0, we find that the x-coordinate of the point of intersection of the tangent line with the x-axis, call it x1 , will be x1 = x0 − f (x0 )/f 0 (x0 ). Once we have found x1 , we can repeat this process to find a new approximation x2 = x1 − f (x1 )/f 0 (x1 ),... This is Newton’s method. Of course, we can apply this idea to other functions, as well, once we learn how to compute the derivative of the function in question.

2.3

The product rule

A function may be expressed as a product of two functions. For example, the function f (x) = (x2 −2x−2)(x3 +x+1) is a product of the simpler functions g(x) = x2 −2x−2 and h(x) = x3 +x+1. We now know that g 0 (x) = 2x−2 and also that h0 (x) = 3x2 +1. How can we use this information to compute the derivative of the product f (x) = g(x)h(x)? The product rule for differentiation is discovered by a fairly simple, but slightly sneaky, algebraic trick. We manipulate the difference quotient for f as follows: g(x + t)h(x + t) − g(x)h(x) t g(x + t)h(x + t) − g(x)h(x + t) + g(x)h(x + t) − g(x)h(x) = t All we have done is subtract and then add the term g(x)h(x + t) in the numerator of the difference quotient. (Adding and subtracting terms like this may appear sneaky at first, but is really a fairly common algebraic manipulation.) But, we can now break this into 2 pieces: f (x + t) − f (x) t

=

g(x + t) − g(x) h(x + t) − h(x) g(x + t)h(x + t) − g(x)h(x + t) + g(x)h(x + t) − g(x)h(x) = h(x + t) + g(x) t t t Note that the fractions in this last expression are difference quotients for g and h, respectively. As long as the derivatives of both g and h exist at x, we can compute the derivative of f as follows: f 0 (x)

f (x + t) − f (x) t   g(x + t) − g(x) h(x + t) − h(x) = lim h(x + t) + g(x) t→0 t t       g(x + t) − g(x) h(x + t) − h(x) = lim lim h(x + t) + lim g(x) lim t→0 t→0 t→0 t→0 t t 0 0 = g (x)h(x) + g(x)h (x).

=

lim

t→0


In the example f (x) = (x2 −2x−2)(x3 +x+1), we learn that f 0 (x) = (2x − 2) (x3 +x+1)+(x2 −2x−2) 3x2 + 1 .

2.4

The chain rule

Another typical way that two functions are combined to produce a new function is through composition. For example, the function f (x) = (x3 + x2 + x + 1)3 can be written as a composition g(h(x)), where the “outside” function g(x) = x3 and the “inside” function h(x) = x3 + x2 + x + 1. (Throughout this discussion, we’ll assume that values in the range of h are in the domain of g, so that the expressions make sense.) We know how to compute the derivatives of each of these functions separately. How can we use this information to compute the derivative of f ? You will probably not be surprised that the answer lies in the manipulation of the difference quotient for f. In this case, we write f (x + t) − f (x) g(h(x + t)) − g(h(x)) = t t g(h(x + t)) − g(h(x)) h(x + t) − h(x) = h(x + t) − h(x) t 5

which looks very similar to the product of difference quotients for g and h. In fact, while a precise proof requires additional attention to detail, that is essentially what we have, and the end result is that f 0 (x) = g 0 (h(x))h0 (x). So, for our example f (x) = (x3 + x2 + x + 1)3 , we know that the derivative of the “outside” function g(x) = x3 is g (x) = 3x2 and the derivative of the “inside” function h(x) = x3 + x2 + x + 1 is h0 (x) = 3x2 + 2x + 1, so that
2
f 0 (x) = 3 x3 + x2 + x + 1 3x2 + 2x + 1 . 0

2.5

The quotient rule

Quite often, one function may be written as a quotient of two (simpler) functions, f (x) =

g(x) . h(x)

However, since 1 g(x) −1 = g(x) = g(x) (h(x)) , h(x) h(x) we can use a combination of the product and chain rules to compute f 0 (x). From the product rule, f 0 (x) = g 0 (x) (h(x))

−1

+ g(x)

 d  −1 (h(x)) . dx

Now, using the chain rule,  d  −1 −2 (h(x)) = (−1) (h(x)) h0 (x). dx So, f 0 (x)

−1

= g 0 (x) (h(x)) = =

  −2 + g(x) (−1) (h(x)) h0 (x)

g 0 (x) g(x)h0 (x) − 2 h(x) (h(x)) g 0 (x)h(x) − g(x)h0 (x) (h(x))

2

.

This formula is called the quotient rule. Here is an example. Suppose that f (x) =

x2

1 , +1

so that g(x) = 1 and h(x) = x2 + 1. Then g 0 (x) = 0 and h0 (x) = 2x, so
0 x2 + 1 − 1 (2x)) −2x 0 f (x) = = 2 2. 2 2 (x + 1) (x + 1)

6