Biz - Quatitative.Managment.Method Chapter.05

8/9/2019 Biz - Quatitative.Managment.Method Chapter.05

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Chapter VChapter V

Limits and DerivativesLimits and Derivatives

22

The Difference Quotient:

Suppose y = f (x)

When the variable x changes from the value x0 toa new value x1, the change is measured by thedifference x1 x0. Hence, using the symbol todenote the change, we write

x = x1 x0

For the function f (x) = 5 + x2, we have f (0) = 5+ 02 = 5; and similarly, f (2) = 5+ 22 = 9, etc.


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When x changes from an initial value x0 to anew value (x0 + x), the value of the functiony = f (x) changes from f(x0) to f(x0 + x).

he change in y per unit of change in x can berepresented by the difference quotient

y f(x0 + x) f(x0) = ------------------------ ..(1)x x

44

he above quotient measures the average rateof change of y, which can be calculated if weknow the initial value of x, or x0, and themagnitude of change in x or x . That is,y/x is a function of x0 and x.

Example: Given y = f (x) = 3x2 - 4, we can write:

f (x0) = 3 (x

0)2 4

f(x0 + x) = 3(x0 + x)2 4


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The difference quotient is

y 3 (x0 + x)2 4 (3x0

2 - 4) 6x0 x + 3 (x )2

= ----------------------------------- = ----------------------x x x

= 6x0 +3 x .(2)

Let x0 = 3 and x = 4; then the average rate ofchange of y will be 6(3) + 3(4) = 30.

his means that, on the average, as x changes from3 to 7, the change in y is 30 units per unit change inx.

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The Derivative:

Frequently, we are interested in the rate ofchange of y when x is very small. In such acase, it is possible to obtain an approximationof y/x by dropping all the terms in thedifference quotient involving the expressionx.

In (2), for instance, ifx is very small, wesimply take the term 6x0 on the right as anapproximation ofy/x. The smaller the valueof x, of course, the closer is theapproximation to the true value ofy/x.


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As x approaches zero, (6x0 + 3x) willapproach the value 6x0, and by the sametoken, y/x will approach 6x0 also.

Symbolically, this fact is expressed either bythe statement y/x 6x0 as x 0 or by theequation

ylim = lim (6x0 + 3x) = 6x0x0 x x0

(3) If, as x0, the limit of the difference quotienty/x exists, that limit is identified as thederivative of the function y = f (x).

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First, a derivative is a function; in fact, in thisusage the word derivative really means aderived function. The original function y = f (x)is a primitive function, and the derivative isanother function derived from it. Whereas thedifference quotient is a function of x0 and x.

Secondly, since the derivative is merely a limitof the difference quotient, which measures arate of change of y, the derivative must ofnecessity also be a measure of some rate ofchange.


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Thirdly, derivative functions are commonlydenoted in two ways. Given a primitivefunction y = f (x), one way of denoting its

derivative (if it exists) is to use the symbolf(x), or simply f. The other commonnotation is dy/dx.

Using these notations, we may define thederivative of a given function y = f (x) asfollows:

dy

y f(x) lim dx x0 x

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Problems:1.Given the function y = 4x2 + 9:

(a) Find the difference quotient as a function of

x and x (Use x in lieu of x0).

(b) Find the derivative dy/dx.

(c) Find f(3) and f(4).


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2. Given the function y = 5x2 4x:

(a) Find the difference quotient as a

function of x and x.

(b) Find the derivative dy/dx.

(c) Find f(2) and f(3).

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The Derivative and the Slope of a Curve:The Derivative and the Slope of a Curve:

Given a total cost function C = f (Q), where Cdenotes total cost and Q the output, themarginal cost (MC) is defined as the change intotal cost resulting from a unit increase inoutput; that is MC = C/Q.

It is understood that Q is an extremely smallchange. For the case of a product that hasdiscrete units ( integers only), a change of oneunit is the smallest change possible; but forthe case of a product whose quantity iscontinuous variable, Q will refer to aninfinitesimal change.


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In the second case, it is well known that themarginal cost can be measured by the slope ofthe total cost curve. But the slope of the total

cost curve is nothing but the limit of the ratioC/Q, when Q approaches zero.

In the figure given below, we have drawn atotal cost curve C, which is the graph of the(primitive) function C = f (Q).

Suppose that we consider Q0 as the initial

output level from which an increase in outputis measured, then the relevant point on thecost curve will be A.

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If output is to be raised to Q0

+ Q = Q2

, thetotal cost will be increased from C0 to C0 + C= C2; thus C/Q = (C2 C0) / (Q2 Q0).

Geometrically, this is the ratio of two linesegments, EB/AE, or the slope of the line AB.

his particular ratio measures an average rateof change the average marginal cost for theparticular Q pictured represents adifference quotient. As such, it is a function ofthe initial value Q0 and the amount of changeQ.


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C2

C1

CO

O Q0 Q1 Q2

C

Q

H

G

K

R

A F E

D

B

C= f ( Q)

C

Q

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If a smaller output increment is contemplated (say,

from Q0 to Q1 only), then the average marginal costwill be measured by the slope of the line AD instead.

As we reduce the output increment further andfurther, flatter and flatter lines will result until, in thelimit (as Q 0), we obtain the line KG (which is thetangent line to the cost curve at point A) as therelevant line.

he slope of KG (= HG/KH) measures the slope ofthe total cost curve at point A and represents thelimit ofC/Q, as Q 0, when initial output is at Q= Q0.

herefore, in terms of the derivative, the slope of theC = f (Q) curve at point A corresponds to theparticular derivative value f(Q0).


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The Concept of Limit:The Concept of Limit:

The derivative dy/dx has been defined as the

limit of the difference quotient

y/

x as

x 0. If we adopt the shorthand symbols q y/x

(q for quotient) and v x (v for variation),we have

dy y = lim = lim qdx x0 x v 0

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Left-Side Limit and Right-Side Limit:

Suppose q = g(v). Our immediate interest is infinding the limit of q as v 0, but we may justas easily explore the more general case of vN, where N is any finite real number.

Then lim q will be merely a special case of lim qv0 vN

where N = 0. When we say that v N, the variable v can

approach the number N either from valuesgreater than N, or from values less than N.


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If, as v N from the left side ( from values lessthan N), q approaches a finite number L, we call Lthe left side limit of q.

If L is the number that q tends to as v N fromthe right side (from values greater than N), wecall L the right side limit of q.

The left- and right-side limits may or may not beequal.

The left-side limit of q is symbolized by lim q

vN-

(the minus sign signifies from values less than N)and the right-side limit is written as lim q.

v N+

2020

When and only when the limits have a commonfinite value (say, L), we consider the limit of q to existand write it as

lim q = L.vN

If we have the situation of lim q = (or - ), weshall

vNconsider q to possess no limit, because lim q = means

vN

that q as vN, and q will assume ever-increasingvalues as v tends to N, it would be contradictory tosay that q has a limit.


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As a convenient way of expressing the fact that qas vN.

However, people do indeed write lim q =

vN

and speak of q as having an infinite limit.

In certain cases, only the limit of one side needs to be

considered. In taking the limit of q as v +, forinstance, only the left-side limit of q is relevant,because v can approach + only from the left.

For the case of v , only the right-side limit isrelevant.

2222

L

q

ON

v

L

q

O N v


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L1

L2

q

N

vO

N(d)

M

q

O v

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Whether the limit of q exists in these cases willdepend only on whether q approaches a finitevalue as v +, or as v .

Graphical Illustrations:

Figure 1 shows a smooth curve. As thevariable v tends to the value N from either sideon the horizontal axis, the variable q tends to

the value L. In this case, the left-side limit isidentical with the right-side limit; therefore, wecan write lim q = L.

v N

he curve drawn in Figure 2 is not smooth; ithas a sharp turning point directly above theoint N.


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As v tends N from either side, q again tends toan identical value L. The limit of q again existsand is equal to L.

In Figure 3 as v tends to N, the left-side limitof q is L1, but the right-limit is L2, a differentnumber. Hence, q does not have a limit as vN.

In Figure 4, as v tends to N, the left-side limitof q is - , whereas, the right side limit is +, because the two parts of the (hyperbolic)

curve will fall and rise indefinitely whileapproaching the broken vertical line as anasymptote. Again, lim q does not exist.

v N

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On the other hand, if we are considering a differentsort of limit in diagram d, namely, lim q, then only the

v+left-side limit has relevance, we do find the limit to exist:

lim q = M. Similarly, lim q = M.v+ v

Problem 1:

Given q = 2 + v2, find lim q.v0

Problem 2:

Given q = (1 v2) / (1 v), find lim q.v1


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Example 3:

Given q = (2v + 5) / (v + 1), find lim q.

v

+

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Limit TheoremsLimit Theorems When a single function q = g (v) is involved, the following

theorems are applicable:

Theorem I:

If q = av + b, then lim q = aN +b (a and b are constants).

vN

Example: Given q = 5v + 7, we have lim q = 5(2) + 17

v2Similarly, lim q = 5(0) + 7 = 7.

v0

Theorem II: If q = g (v) = b, then lim q = b.

vN

This theorem, which says that the limit of a constant function isthe constant in that function, is merely a special case of TheoremI, with a = 0.


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eorem : q = v, t en m q =

v N

If q = vk, then lim q = Nk

vN

Example: Given q = v3, we have lim q = (2)3 = 8v2

Theorems Involving Two Functions:

If we have two functions of the same independent variable v, q1 = g (v) and q2 = h (v), and if bothfunctions possess limits as follows:

lim q1 = L1 lim q2 = L2vN vN

where L1 and L2 are two finite numbers, the followingtheorems are applicable.

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Theorem IV (sum-difference limit theorem):lim (q1 q2) = L1 L2

vN

The limit of a sum (difference) of two functions is the sum

(difference) of their respective limits.

Theorem V (product limit theorem):

lim (q1q2) = L1L2

VN

The limit of a product of two functions is the product of their

limits.


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Theorem VI (quotient limit theorem):

q1 L1

lim = (L2 0)vN q2 L2

The limit of a quotient of two functions is the quotient of their

limits. Naturally, the limit L2 is restricted to be nonzero;

otherwise the quotient is undefined.

Example:

Find lim (1 + v) / (2 + v). Since we have here lim (1 + v) = 1

v0 v0and lim (2 + v) = 2, the desired limit is 1 / 2.

v0

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Limit of a Polynomial Function:

Suppose q = g (v) = a0 + a1v + a2v2+ +anv

n is a polynomial

function.

Since the limits of the separate terms are respectively,

lim a0 = a0 lim a1v = a1N lim a2v2 = a2N2 (etc.)

VN VN VN

The limit of the polynomial function is (by the sum limit

theorem)

lim q = a0 + a1N + a2N2 ++ anN

n

vN


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Continuity and Differentiability of a FunctionContinuity and Differentiability of a Function

Continuity of a Function:

When a function q = g (v) possesses a limit as v tends to thepoint N in the domain, and when this limit is equal to g (N)

that is, equal to the value of the function at v = N the function

is said to be continuous at N.

As stated above, the term continuity involves no less than three

requirements:

(1) The point N must be in the domain of the function; i. e.,

g(N) is defined.(2) The function must have a limit as vN; i.e., lim g(v) exists;

vN

(3) That limit must be equal in value to g(N); i.e., lim g(v) = g(N)

vN

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Example: The rational function

q = g(v) = 4v2 / (v2 + 1)

is defined for all finite real numbers; thus its domain consists of

the interval (- ). For any number N in the domain, the limit

of q is (by the quotient limit theorem)lim (4v2)

vN 4N2

lim q = =vN lim (v2 + 1) N2 + 1

vN

which is equal to g (N).

Thus the three requirements of continuity are all met at N. N

can represent any point in the domain of this function;

consequently, this function is continuous in its domain.


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Example: The rational function

v3 + v2 4v 4

q = v2 4

is not defined at v = 2 and at v = -2. Since those two

values of v are not in the domain, the function is

discontinuous at v = -2 and v = 2, despite the fact that a

limit of q exists as v -2 or 2. Graphically, this function

will display a gap at each of these two values of v. But

for other values of v (those which are in the domain), thisfunction is continuous.

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Differentiability of a Function:

Taking the limit of the function y = f(x), we canexamine the whether the function f is continuous at x =x0. The conditions of continuity are:

(1) x = x0 must be in the domain of the function f,

(2) y must have a limit as x x0, and

(3) the said limit must be equal to f(x0).

When these are satisfied, we can write

lim f(x) = f(x0 ) [continuity condition] .(1)x x0


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When the limit concept is applied to the difference

quotient y / x as x 0, on the other hand, we dealinstead with the question of whether the function f is

differentiable at x = x0, i.e., whether the derivative dy/dxexists at x = x0, or whether f(x) exists.

The term differentiable is used because the process ofobtaining the derivative dy/dx is known asdifferentiation (also called derivation).

Since f(x0 ) exists if and only if the limit ofy/x

exists at x = x0 as x 0, the symbolic expression of thedifferentiability of f is

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y f(x0 +x) f(x0)

f(x0) = lim limx 0 x x 0 x

[differentiability condition] .(2)

The properties, continuity and differentiability, areintimately related to each other the continuity of f is anecessary condition for its differentiability. To bedifferentiable at x = x0, the function must first pass the

test of being continuous at x = x0.

We can write (1) and (2) in a more simplified form byreplacing x0 with the symbol N and (2) replacing(x0+x) with the symbol x.


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With the notational change, x now becomes

(x - N), so that the expression x 0 becomes

x N , which is analogous to the expression vNused before in connection with the function q = g(v).

Accordingly, (1) and (2) can now be rewrittenrespectively, as

lim f(x) = f(N) .(3)x N

f(x) f(N)f(N) = lim ..(4)

x N x - N

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y = f(x)

y

x

N x[xo] [x0+x]

xo

y

f(N)

f(x0)

f(x)

f(x0+x)


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Differentiability implies continuity but theconverse is not true.

Continuity is a necessary, but not a sufficient,condition for differentiability.

Let us consider the function

y = f(x) = |x 2| + 1 (see the graph givenin the next slide)

he above function is not differentiable,

though continuous, when x = 2.

First, x = 2 is in the domain of the function.

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Second, the limit of y exists as tends to 2; tobe specific

lim y = lim y = 1

x 2+ x 2-

Third, f(2) is found to be 1.

hus, all the three requirements of continuity

are met. But the function f is not differentiableat x = 2, because, the limit of the differencequotient

lim {f(x) f(2)}/(x 2) = lim (|x 2| + 1 1)/ (x 2) = lim |x 2|/(x 2)x2 x2 x2

does not exist.


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In considering the right-side limit, x mustexceed 2, according to the definition ofabsolute value, we have |x 2| = x 2.

Thus the right-side limit is

lim |x 2|/(x 2) = lim (x 2)/(x 2) = lim 1 = 1

x2+ x2+ x2+

In considering the left-side limit, x must beless than 2, according to the definition ofabsolute value, we have |x 2| = - (x 2).

The left-side limit is

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lim |x 2|/(x 2) = lim { (x 2)}/(x 2) = lim (- 1) = - 1

which is different from the right-side limit.

his shows that continuity does not guaranteedifferentiability.

All differentiable functions are continuous, butnot all continuous functions are differentiable.

The nondifferentiability of the function at x = 2is manifest in the fact that the point (2, 1) hasno tangent line defined, and hence no definiteslope can be assigned to the point.


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o the left of that point, the curve has a slopeof 1, but to the right it has a slope of + 1,and the slopes on the two sides display no

tendency to approach a common magnitude atx = 2.

he point (2, 1) is a special point; it is the onlysharp point on the curve.

At other points of the curve, the derivative isdefined and the function is differentiable.

he function can be divided into two linearfunctions as follows:

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Left part: y = - (x 2) + 1 = 3 x (x2)

Right part: y = (x 2) + 1 = x 1 (x>2)

he left part is differentiable in the interval(-, 2), and the right part is differentiable inthe interval (2, ) in the domain.

Continuity at a point rules out the presence of

a gap, whereas differentiability rules outsharpness.

Differentiability calls for smoothness of thefunction (curve) as well as its continuity.


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(2, 1)

x

y = |x - 2| + 1

o

y

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Rules of Differentiation for aRules of Differentiation for afunction of one Variablefunction of one Variable

Constant-Function Rule:

The derivative of a constant function y = f(x) = k isidentically zero, i.e., is zero for all values of x.

Symbolically, dy / dx = 0 or dk / dx = 0

or f(x) = 0

Power-Function Rule:

dxn / dx = nxn-1 or f(x) = nxn-1


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Problem 1: y = x3, find dy / dx.

Problem 2: y = x9, find dy / dx.

Problem 3: Find the derivative of y = x0.

Problem 4: Find the derivative of y = 1 / x3.

Problem 5: Find the derivative of y = x .

Power-Function Rule Generalized:

(d/dx) cxn

= cnxn-1

or f(x) = cnxn-1

Problem 6: Given y = 2x, find dy / dx.

Problem 7: Given f(x) = 4x3, find f(x).

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Problem 8: Given f(x) = 3x-2, find f(x).

Sum-Difference Rule:

(d/dx) [f(x) g(x)] = (d/dx) f(x) (d/dx) g(x).

Problem 9: Given y = ax2 + bx + c, find dy / dx.

Problem 10: Given y = 7x4 + 2x3 3x +37,

Find dy / dx.

Problem 11: Given the short-run total-cost function

C = Q3 4Q2 + 10Q +75, find marginal-cost function.


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Product Rule:

(d/dx) [f(x) g(x)] = f(x) (d/dx) g(x) + g(x) (d/dx) f(x).

Problem 12: Find the derivative of

y = (2x + 3)(3x2).

Quotient Rule: The derivative of the quotient of twofunctions, f(x) / g(x), is

d f(x) f(x)g(x) f(x)g(x)

=dx g(x) g2(x)

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Problem 13: Find (d/dx) (2x 3/x+1)

Problem 14: Find (d/dx) (5x/x2+1)

Problem 15: Find (d/dx) (ax2 + b/ cx)

Chain Rule:

If we have a function z = f(y), where y in turn afunction of another variable x, y = g(x), then thederivative of z with respect to x is equal to thederivative of z with respect to y, times the derivativeof y with respect to x. Expressed symbolically:

dz/dx = (dz/dy) x (dy/dx) = f(y)g(x)


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If we have z = f(y), y = g(x), and x = h(w), then

dz/dw = dz/dy x dy/dx x dx/dw = f(y)g(x)h(w)

Problem 16: If z = 3y2, where y = 2x + 5. Find dz/dx.

Problem 17: If z = y 3, where y = x3. Find dz/dx.

Problem 18: z = (x2 + 3x 2)17. Find dz/dx.

Inverse-Function Rule:

For inverse functions, the rule of differentiation is

dx/dy = 1 / (dy/dx)

Problem 19: Given y = x5 + x. Find dx/dy.

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Implicit Differentiation:

The function y = f(x) is said to be an explicit function,because y is expressed in terms x.

But if the function is written as g(x, y) = 0, then thefunction is said to be an implicit function. And thedifferentiation of such a function is called implicitdifferentiation.

Here, we treat dy/dx as unknown and differentiate thefunction with respect to x.

Problem 1:

Differentiate xy = 9 implicitly with respect to x.


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Problem 2:

Differentiate x2 + y2 20=0 implicitly with respect to x.

Problem 3:

Differentiate xy y + 4x = 0 implicitly with respect to x.

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Partial Derivatives:

If y = f(x1, x2,..,xn), then f1 = y/ x1, f2 =

y/ x2,., fn = y/ xn.

Problem 20: y = f(x1, x2) = 3x12 + x1x2 + 4x2

2, find thepartial derivatives.

Problem 21: Given y = f(u, v) = (u+4)(3u+2v). Find thepartial derivatives fu and fv.

Problem 22: Given y = (3u 2v) / (u2 +3v). Find thepartial derivatives fu and fv.


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Differentials:

dy (dy/dx) dx or dy f(x)dx

The symbols dy and dx are called differentials of y and xrespectively.

Problem: 23:

Given y = 3x2 + 7x 5, find dy.

The derivative of the function is dy/dx = 6x+7; thus the

desired differential is dy = (6x+7)dx.Problem 24:

Find d if the demand function is Q = 100 2p.

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The marginal function and the average function of thegiven demand curve are

dQ/dP = -2 and Q/P = (100 2P)/P

d = -2 (100 2P)/P = -P/(50-P).

Problem 25:

Find the differential dy, given:

(a)y = -x(x2+3) (b) y = (x-8)(7x+5) (c) y = x/x2+1

Total Differentials:

The concept of differentials can be extended to afunction of two or more independent variables.


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Consider a saving function S = S(Y, i) where S issavings, Y is national income and i interest rate.

The total change in S will then be equal todS = (S/Y) x dy + (S/i) x di

The expression dS, being the sum of the changes fromboth sources, is called the total differential of thesaving function.

The process of finding such a differential is called

total differentiation.

The more general case of a function of n independentvariables can be exemplified by, say, a utility function

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in the general form

U = U(x1, x2,..,xn)

The total differential of this function can be written as

dU = (U/x1)dx1 + (U/x2)dx2 + + (U/xn)dxn

Or dU = U1dx1 + U2 dx2 + + Undxn = Uidxi

Rules of Differentials:

A straightforward way of finding the total differential dy,given a function y = f(x1, x2)

is to find the partial derivatives f1 and f2 substitute theseinto the equation dy = f1dx1 + f2dx2


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Let k be a constant and u and v be two functions of thevariables x1 and x2. Then the following rules are valid.

Rule I. dk = 0 (constant-function rule)Rule II. d(cun) = cnun-1du (power-function rule)

Rule III. d(u v) = du dv (sum-difference rule)

Rule IV. d(uv) = vdu + udv (product rule)

Rule V. d(u/v) = (vdu udv)/v2 (quotient rule)

Rule VI. d(u v w) = du dv dw

Rule VII. d(uvw) = vwdu + uwdv + uvdw

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Problem 1.

Find the total differential dy of the function

y = 5x12 + 3x2

Problem 2.

Find the total differential of the function

y = 3x12 + x1x22

Problem 3.

Find the total differential of the function

y = x1 + x2/2x12


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Total Derivatives:

Given a function of the form Z = f(x, y), we are

interested in finding the rate of change of the functionwith respect to x when x and y are related.

In this kind of differentiation, the independentvariables are actually dependent on other variables. Inconnection with this, three cases are available.

Case I: Given the function Z = f(x, y), where y = g(x),

the total derivative of Z with respect to x isdZ/dx = fx(dx/dx) + fy(dy/dx)

but dx/dx is equal to one, hence dZ/dx= fx + fy(dy/dx),

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where fx and fy are the partial derivatives of the function

with respect to x and y respectively.

fx is the direct effect of the change in x on Z, and fy(dy/dx)

measures the indirect effect of the change in y on Z

change runs from y to x, then to Z.

Problem 1.

Find the total derivative for Z = 3xy where y = 3x2.

Solution:

Zx = fx = 3y, Zy= fy= 3x, and dy/dx = 6x.

Hence, dZ/dx = 3y + 3x(6x) = 3y + 18x2.

(The answer means that the total differential of the functionis divided b dx .


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Example:

For a general function of the form

Z = f(x, y1, y2,yn)

Where y1 = g(x), y2 = s(x), and yn = i(x), the

total derivative would be, dZ/dx =

fx + fy1dy1/dx + fy2dy2/dx + + fyndyn/dx

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Problem 3.

Find the total derivative for Z = 3xy1y2, where y1 = 2x2

and y2 = x.

Solution: Zx = fx = 3y1y2, dy1/dx = 4x, dy2/dx = 1,

Zy1 = fy1 = 3xy2, and Zy2 = fy2 = 3xy1

Hence, dZ/dx = 3y1

y2

+ 12x2y2

+ 3xy1

Case 2: The situation is only slightly more complicatedwhen we have y = f(x1, x2, w) where x1=g(w), x2=h(w).

Now dy/dw = f1(dx1/dw) + f2(dx2/dw) + fw(dw/dw)

= (y/x1)(dx1/dw) + (y/x2)(dx2/dw) + y/w


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Example: Let the production function be

Q = Q (K, L, t) where, aside from the two inputs K and

L, there is a third argument t, denoting time. The presence of the t argument indicates that theproduction function can shift over time in reflection oftechnological changes.

Thus, this is a dynamic rather than a static productionfunction. Since capital and labor, too, can change over

time, we may writeK = K(t) and L = L(t)

Then the rate of change of output with respect to time

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can be expressed, in line with the total-derivative formula

as

dQ/dt = (Q/k)(dk/dt )+ (Q/L)(dL/dt) + Q/t

or, in an alternative symbolism,

dQ/dt = QkK(t) + QLL(t) + Qt


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Problems:

1. Find the total derivative dz/dy, given:

(a) z = f(x, y) = 2x + xy y2, where x = g(y) = 3y2

(b) z = 6x2 3xy + 2y2, where x = 1/y

(c) z = (x + y)(x 2y), where x = 2 7y

2. Find the total derivative dz/dt, given:

(a) z = x2 8xy y3, where x = 3t and y = 1- t

(b) z = 3u + vt, where u = 2t2 and v = t+1

(c) z = f(x, y, t), where x = a + bt and y = c + dt

Biz - Quatitative.Managment.Method Chapter.05

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