## More on Differentiation

Although this may seem a paradox, all science is dominated by the idea of approximation.

—Bertrand Russell

This chapter presents some extensively used techniques of differentiation. It begins with the generalized power rule and then proceeds to a discussion of the highly useful chain rule. In many economic models, functions are defined implicitly by one or more equations. In some simple but economically relevant cases, we show how to compute derivatives of such functions. Next we consider differentials and linear, quadratic, or higher-order approximations, all of which occur in many applications of mathematics to economics. A discussion of the important economic concept of elasticity ends the chapter.

5.1 The Generalized Power Rule

It is often necessary to differentiate expressions of the form

where g is a differentiate function, and a is a constant. For a = 1. the derivative is just g'(x). For a = 2, we can use the product rule as follows:

>> = [g{x)]2 = g(x) ■ g(x) => >■' = g CO • g(x) + g(x) ■ g\x) = 2g(x) ■ g'(x)

For a = 3, we can combine the previous result with the product rule as follows:

y = [g(jr)]3 = jg(x)]2 • gix) => y = [2g(x) ■ g'(x)] ■ g(x) + [g(x)}2 ■ g\x)

See if you can discern a pattern here. In general, we have the following rule (where a is an arbitrary real number):

### The Generalized Power Rule

Note this important formula. If we put g(x) = x, then g'(x) = 1 and [5.1] reduces to y = xa => y' = axa~\ which is the power of Section 4.5. A generalization of [5.1] is proved in Section 5.2. In the meantime, ambitious students may want to try proving [5.1] by mathematical induction for the case when a is a natural number. (See Problem 10.)

Example 5.1

Differentiate the functions:

Solution The key to applying the generalized power rule is to determine how the given function can be expressed as a power. In the first problem, it is rather obvious:

(a) y = (.x3 + x2)50 = [gix)}50 where gix) = x3 + x2. Differentiating this directly gives g'ix) = 3x2 4- 2x, and so formula [5.1] yields

/ = 50[g(jc)]50-1 • g'ix) = 50U3 + x2)49(3x2 + 2x)

where gix) = (x — l)/(x + 3). In this case, the quotient rule implies that

Hence, [5.1] gives lr..l(I/3)-l „ • \ fx-\\~2ß 4

(c) Here we first notice that y = Jx1 + 1 = (x2 +1)1/2. So >■ = [g(x)]1/2, where g(x) = x2 + 1. Hence, y=\ bw]'"®-' • g\x)=i(*2+1)-'«. 2x =

The generalized power rule can also be formulated in Leibniz's notation.

The Generalized Power Rule (Leibniz's Notation)

When u = g(x) is a function of x, then a_, dy a_, du y = u° — =au° — ax ax

Often we need to combine the generalized power rule with the other rules of differentiation shown earlier. Here is an example from economics.

### Example 5J2

Suppose that the relationship between gross income Y and total income tax T is for taxpayers with incomes between 80,000 and 120,000. The following values for the constants in [*] were estimated, given by the equation

where a, b, c, p, and k are positive constants.

(a) Find an expression for the marginal tax rate. dT/dY.

(b) In an empirical study a = 0.000338. ¿ = 0.81. c = 6467. p = 1.61. k = 0.053

Use these numbers to find the values of T and dT/dY when Y = 100,000.

Solution

(a) Let z = (bY + c)p = up with u = bY + c. Then [5.2] gives dz T du ,

T = 0.000338(0.81 • 100,000 + 6467)161 +0.053 • 100.000 % 35.869.33 and dT

— =0.000338-0.81 • 1.61 • (0.81 -100,000 +6467)0 61 + 0.053 % 0.51

Thus, the marginal tax rate on an income of 100,000 is approximately 51%.

Problems

1. Compute f'(x) when f(x) = (3*2 + I)2 by (a) expanding the square and then differentiating; (b) using [5.1], Compare the answers.

2. Find the derivatives of the functions defined by the following:

d. (x+1) e. (3* - 4)"7 f. (2x2 + 3* - 4)"2

3. Find the derivatives of the functions defined by the following:

4. Find the derivatives of the following functions of t (where a, b, and n are constants):

/at + b\a+l a. (at + 1)~J b. (at + b)n c. i—— I

5. If / is differentiable at x, find expressions for the derivatives of the following functions:

6. Let x — (Ap + B)r and p = at2 + bt + c. Find an expression for dx/dt.

Harder Problems

8. Suppose that [5.1] has already been proved when a is a natural number. Prove that [5.1] is then also valid when a = —n, where n is a natural number. (Hint: Put y = [gOt)]-" = l/jgOt)]", and then use the quotient rule.)

9. Let a, b, m, and n be fixed numbers, where a < b, and m and n are positive. Define the function / for all x by f{x) = (x - a)m ■ (x - b)n. For the equation f'(x) = 0, find a solution xq that lies between a and b.

10. Prove by induction that [5.1] holds when a is a natural number.

11. Prove that

£ [/cor [SCO]" = [mf'(x)g(x) + n/Wg'W] [.f(x)]m~l [g(x)T~]

What do you get if m = n = 1, and if m = — n = 1 ?

### 5.2 Composite Functions and the Chain Rule

If y is a function of u, and u is a function of x, then y is a function of x. In this case, we call y a composite function of x. (In the previous section, we considered the special case where y was given by ua.) Suppose that x changes. This will lead to a change in u and hence a change in y. A change in x, therefore, causes a "chain reaction." If we know the rates of change du/dx and dy/du, then what is the rate of change dy/dxl It turns out that the relationship between these rates of change is simply:

The Chain Rule

dy

dy

du

dx

du

A slightly more detailed formulation of the rule says that if y is a differentiable function of u, and u is a differentiable function of x, then y is a differentiable function of x, and [5.3] holds.

The chain rule is a further generalization of the generalized power rule from the previous section. In the special case where y = ua, we have dy/du = aua~], and substituting this expression into [5.3] gives the formula in [5.2],

It is easy to remember the chain rule when using Leibniz's notation. The left-hand side of [5.3] is exactly what results if we "cancel" the du on the right side. Of course, because dy/du and du/dx are not fractions (but merely symbols for derivatives) and du is not a number, canceling is not defined.

When we interpret the derivatives involved in [5.3] as rates of change, the chain rule becomes rather intuitive, as the next example from economics will indicate.

### Example 53

The demand .r for a commodity depends on price p. Suppose that price p is not constant, but depends on time t. Then x is a composite function of r, and according to the chain rule, dx dx dp

Suppose, for instance, that the demand for butter decreases by 5000 pounds if the price goes up by \$1 per pound. So dx/dp % —5000. Suppose further that the price per pound increases by \$0.05 per week, so dp/dt % 0.05. What is the decrease in demand in pounds per week? Solution: Because the price per pound increases by \$0.05 per week, and the demand decreases by 5000 pounds for every dollar increase in the price, the demand decreases by 5000 • 0.05 % 250 pounds per week. This means that dx/dt % —250 (measured in pounds per week). Note how this argument roughly confirms [*].

The chain rule is very powerful. Facility in applying it comes only from a lot of practice.

Example 5.4

Solution

(a) We can use [5.3] directly. Because dy/du = 5u4 and du/dx = —3x2, we have

dx du dx

(b) In this case, it is not immediately obvious how to apply the chain rule. However, if we rewrite y as y = 10(jc2 + 4x + 5)-7, then y = 10m-7

— = 10(—7)u~7_1 = —70m-8 and ^ = 2x + 4 du dx

So using [5.3] yields

= = —70m-8 • (2x + 4) = — 140(x + 2)/{x2 + 4x + 5)8

dx du dx

Note 1: After a little training, the intermediate steps become unnecessary. For example, to differentiate y = (l-x3)5

we can think of y as y = u5, where u — 1 — x3. We can then differentiate both u5 and 1 — x3 in our heads, and immediately write down y' = 5(1 — x3)4(—3x2).

Note 2: Of course, one could differentiate y = x5/5 using the quotient rule, rather than writing y as y = (l/5)x5 to get y' = (l/5)5x4 = xA. But the latter method is much easier. In the same way, it is unnecessarily cumbersome to apply the quotient rule to the function given in Example 5.4(b). The chain rule is much more effective.

The next example uses the chain rule several times.

Example 5.5

Solution The initial step is easy. Let x(t) = 5u2^, where u = 1 + V/3 + 1, to obtain du -m du x'{t) =5-25u24— = 125m24— [1]

dt dt

The new feature in this example is that we cannot write down du/dt at once. Finding du/dt requires using the chain rule a second time. Let u = 1 +Jv =

1 + t,1/2? where v = r3 + ]. Then d± = 1,1/2-, . dv = 1 1/2 _ , = 1 3 1/2 _ 2 dt 2 dt 2 2

Jc'(r) = 125 (l + \ftr+T)"4 • I(r3 + l)"'/2 ■ 3r2

Suppose, as in the last example, that x is a function of u, u is a function of v, and v is in tum a function of r. Then x is a composite function of r, and the chain rule can be used twice to obtain dx dx du dv dt du dv dt

This is precisely the formula used in the last example. Again the notation is suggestive because the left-hand side is exactly what results if we "cancel" both du and dv on the right-hand side.

### An Alternative Formulation of the Chain Rule

Although Leibniz's notation makes it very easy to remember the chain rule, it suffers from the defect of not specifying where each derivative is evaluated. We remedy this by introducing names for the functions involved. So let y = /(«) and u = g(x). Then >- can be written in the form

Note that when we compute f(g(xj), we first compute g(x), and then second, we apply / to the result. We say that we have a composite function, with g(x) as the the kernel, and / as the exterior function.

Most scientific calculators have several built-in functions. When we punch a number xo and strike the key for the function /, we obtain /(;to). When we compute a composite function given / and g, and try to obtain the value of f(g(x)), we proceed in a similar manner: punch the number xo- then strike the g key to get g(xo), and again strike the / key to get f(g(:Co)). Suppose the machine has the functions \/x and . If we press the number 9, then strike the button

1 ¡x followed by ^fx , we get 1/3 = 0.33____The computation we have performed can be illustrated as follows:

Using function notation, f(x) = *fx and g(x) = l/x, so f(g(x)) = /(1/x) = yTfr = i/V3c. In particular, f(g(9)) = 1/V9 = 1/3.

The Chain Rule

If g is differentiable at xq and / is differentiable at uq = g(xo), then F(x) = f(g(x)) is differentiable at x0 and

In words: to differentiate a composite function, first differentiate the exterior function and substitute in the value of the kernel, then multiply by the derivative of the kernel. It is important to notice that the derivatives /' and g' appearing in formula [5.4] are evaluated at different points; the derivative g' is evaluated at xo, whereas f is evaluated at g(xo).

Example 5.6

Find the derivative of F(x) = f(g(x)) at = —3 if f(u) = u3, and g(x) =2 — x2.

Solution In this case, /'(") = 3u~ and g'(x) = — 2x. So [5.4] gives

Now g(—3) = 2 - (—3)2 = 2-9= -7; g'(-3) = 6; and f(g(-3)) = f'(-7) = 3(—7)2 = 3 - 49 = 147. So F'(-3) = f'{g{-3))g'(-3) = 147 • 6 = 882.

Note: The function that maps * to f{g(x)) is often denoted by fog, and is read as "/ of g" or "/ compounded with g." Correspondingly, go/ denotes the function that maps * to g(/(*)). Thus, we have

Usually, fog and go/ are quite different functions. For instance, the functions used in Example 5.6 give (fog)(x) = (2—x2)3. whereas (gof)(x) = 2—(x3)2 = 2—x6; the two resulting polynomials are not the same.

It is easy to confuse fog with / • g, especially typographically. But these two functions are defined in entirely different ways. When we evaluate / o g at we first compute g(x) and then evaluate / at g(x). On the other hand, the product / - g of / and g is the function whose value at a particular number x is simply the product of f(x) and gU), so (/ • g)(x) = f(x)-g(x).

Proof of the chain rule To find the derivative of F(x) — f[g{x)) at x = we must examine the limit of the following Newton quotient as h tends to 0:

The change in x from xo to xo+h causes the value of g to change by the amount k = g(xo + h)~ g(xQ). As h tends to 0, so k = {[g(x0 + h) - \$(jc0)]//i} • h tends to g'(xo) -0 = 0. Suppose that k ^ 0 whenever h ^ 0 is small enough. Because gixo + h) = g(xo) + k, we can write the Newton quotient as f{g(xo)+k)~ f(g(x0)) k N ~ k ' h f(g{xo) + k)~ f(g{x0)) g(x0 + h)~ g(xo)

As h 0, so k 0. and the last two fractions tend to /'(gUo)) and g'(xo), respectively. This yields the desired formula.

We cannot divide by 0. so the argument fails if g(xQ + h) = g(xo) for arbitrary small values of h. because then k = 0. A more complicated proof takes care of this case as well.

Problems

1. Use the chain rule [5.3] to find dy/dx for the following:

2. Compute the following:

b. dK/dt when K = ALa and L = bt + c (A, a, b. and c are positive constants).

3. Find the derivatives of the following functions, where a, p, q, and b are constants:

0 0