Chapter III
Differentiation and Integration of the Elementary Functions
3.1. The Simplest Rules for Differentiation and their Applications
In higher analysis and its applications, it is usually the case that the problems of integration are more important than those of differentiation, although differentiation offers less difficulty than integration. Consequently, the natural method of constructing the integral and differential calculus is first to learn to differentiate the widest possible classes of functions and then, by virtue of the fundamental theorem, make the results thus obtained available for the solution of integration problems. In the following sections, it will be our task to carry out this programme. To a certain extent, we shall make a fresh start, since we shall work out the most important differentiations and integrations systematically without referring to the results of the last chapter. In this development of the subject, certain rules for differentiation, with the first of which we are already acquainted , will have an important role.
3.1.1 Rules for Differentiation:
We assume that in the interval under consideration the functions f(x) and g(x) are differentiable; our rules then are:
Rule 1: Multiplication by a constant.
If c is a constant and f(x) = cf(x), then f(x) is differentiable and
This follows immediately from the relation
if we take the limit h ® 0.
Rule 2: Derivative of a sum.
If f (x) = f(x) + g(x), then f (x) is differentiable, and
that is, the processes of differentiation and addition are interchangeable. The same holds for the sum of any finite number n of terms
for which we obtain
We may pass over the proof, which, by 2.3, is fairly obvious.
Rule 3: Derivative of a product.
If f(x) = f(x)g(x), then f(x) is differentiable and
The proof follows from the equation
The passage to the limit h ® 0 can here be carried out directly and yields the result shown.
This formula takes a still more elegant form, if we divide throughout by f(x)=f(x)g(x). We then obtain
where we, of course, assume that f(x) is nowhere equal to zero.
By repeated application of this product formula, we obtain, by induction, for the derivative of a product of n factors an expression consisting of n terms, each of which consists of the derivative of one factor multiplied by all the other factors of the original product. In symbols, this is
or on division by 
, assuming again that f(x)
does not vanish anywhere,
Rule 4: Derivative of a quotient.
We have for a quotient
the rule: The function f(x) is differentiable at every point at which g(x) does not vanish and
If f(x) ¹ 0, this can be rewritten
If we accept the differentiability of f(x) as a hypothesis, we can apply the product rule to f(x)=f(x)g(x) and conclude that
By substituting f(x)/g(x) for f(x) on the right hand side and solving for f '(x), we obtain the above rule. In order to prove the differentiability of f(x) as well as the rule, we use the method: We write
If we now let h tend to 0, we arrive at the result stated, because, by assumption, the two terms obtained by performing the division on the right hand side have definite limits, which are
and 
,
respectively. This proves immediately the existence of the limit on the left hand side as well as the differentiation formula.
3.1.2 Differentiation of the Rational Functions: To begin with, we shall again deduce the differentiation formula
for every positive integer n, basing the proof on the rule for differentiation of a product. We think of xn as a product of n factors, xn = x, ··· . x and hence obtain
If we use this formula and the first rule of differentiation, the second derivative of the function xn becomes:
Continuing this process, we obtain
It is clear from the last of these formulae that the (n+ l)-th derivative of xn vanishes everywhere.
By virtue of our first two rules, a knowledge of the differentiation of powers enables us at once to differentiate any polynomial
We have simply
and
and so on.
The differentiation of any rational function now follows with the help of the quotient rule. In particular, we shall again derive the differentiation formula for the function xn, where n = - m is a negative integer. The application of the quotient rule together with the fact that the derivative of a constant is equal to zero yields
or, if we take m = -n,
which agrees formally with the result for positive values of n and with the results given earlier.
3.1.3 Differentiation of the Trigonometric Functions: For the trigonometric functions sin x and cos x, we have already obtained the differentiation formulae
The quotient rule now enables us to differentiate the functions
According to the rule, the derivative of the first of these functions is
and we obtain the result
Similarly, we obtain
3.2 The corresponding Integral formulae
3.2.1. General Rules for Integration: The fundamental theorem of and the definition of the indefinite integral reveal the possibility of writing down an integration formula corresponding to each differentiation formula. The following rules of integration (of which the first two have already been mentioned) are completely equivalent to the first three rules of differentiation.
Multiplication by a constant: If c is a constant, then
Integration of a sum: It is always true that
To the third rule of differentiation corresponds the rule for the integration of a product or, as it is usually called, the rule for integration by parts. On integration, the product rule yields
The indefinite integral on the left hand side is obviously f'(x)g(x) (except possibly for an additive constant), whence we can write the rule for integration by parts in the form:
This last integration formula, the counterpart of the rule for the differentiation of a product, has been given here only for the sake of completeness; it will not become important for us until 4.4.
3.2.2 Integration of the Simplest Functions:. Corresponding to the differentiation formulae for special functions which we have found above, we now set down the equivalent integration formulae. The formula
expressed as an integration formula, becomes
In fact, this formula merely means that the derivative of the right hand side is equal to the integrand on the left hand side. If we replace n by n +1, we obtain the formula
This formula holds for every integral index n (when n < 0, it holds of course only if x ¹ 0) except when n = -1, when the denominator n + 1 would vanish. In 3.6, this exceptional case will be studied in detail. The fundamental theorem of the integral calculus allows at once to use our integral formulae for the determination of areas, that is, the values of definite integrals. By 2.4.5, we immediately obtain
where, if n is negative, we assume that a and b have the same sign, since otherwise the integrand would be discontinuous in the interval of integration.
There correspond to the differentiation formulae for sin x, cos x the integration formulae:
They yield by way of the fundamental rule of 2.4 the value of the definite integral between any limits, an only restriction being that when the last two formulae are used, the interval of integration must not contain any point of discontinuity of the integrand. For example,
It hardly need be emphasized that, with the help of the first two rules of integration, we are now in a position to integrate any polynomial in x and, in fact, any linear combination with arbitrary constant coefficients of the functions integrated here. However, note that, according to the fundamental theorem, rules of integration and differentiation must be equivalent to each other; it is therefore possible to prove the general integration rules of this section and then to read off the differentiation rules of the preceding section. The reader is well advised to carry out this suggestion.
1. Find the numerical values of all the derivatives of x5 - x4 at x = 1.
2. What is the numerical value of the eleventh derivative of
3. Differentiate the following functions and write down the corresponding integral formulae:
4. Let P(x) = a0
+ a1x + a2x2
+ ··· + anxn.
(a) Calculate the polynomial F(x) from
the equation F(x) - F'(x)
= P(x).
(b)* Calculate F(x) from the equation c0F(x)
+ c1F'(x) + c2F"(x)
= P(x).
6. Differentiate the following functions and write down the corresponding integration formulae:
Recalling that sec x = 1/cos x, cosec x = 1/sin x, find the derivatives of the functions 6. to 9.:
10. Find the limit as n ® ¥ of the absolute value of the n-th derivative of 1/x at the point x = 2.
Evaluate:
3.3 The Inverse Function and its derivative
3.3.1 The General Formula for Differentiation: We have seen earlier (1.2.4 and A1.2.4) that a continuous function y =f(x) has a continuous inverse in every interval in which it is monotonic. More exactly:
If a £ x £ b is an interval in which the continuous function y = f(x) is monotonic, and if f(a) = a and f(b) = b, then x is a function of y which in the interval between a and b is single-valued, continuous and monotonic.
As we have already shown in 2.3.1, the concept of the derivative gives us simple means for recognizing that a function is monotonic and therefore has an inverse. In fact, a differentiable function is certainly always monotonic increasing, if f '(x) is greater than zero throughout the corresponding interval, and similarly is monotonic decreasing, if f'(x) is everywhere less than zero in the interval.
We shall now prove the theorem: If in the interval a < x < b the function y=f(x) is differentiable and in that interval either f'(x) > 0 or else f '(x) < 0 everywhere, then the inverse function x = f(y) also has a derivative at every point of its interval of definition and there exists, for corresponding values of x and y, between the derivative of the given function y = f(x) and that of the inverse function x = f(y) the relationship f '(x) . f '(y) = 1, which we can also write in the form
In this formula, we observe again the flexibility of Leibnitz's notation. It is just as if the symbols dy and dx were quantities which could be operated upon like actual numbers. The proof of this formula is correspondingly simple, if we regard the derivative as the limit of the difference quotient
where x and y =f(x), and x1 and y1 = f(x1), respectively, denote pairs of corresponding values. By assumption, the first of these limiting values is not equal to zero. On account of the continuity of y=f(x) and x=f(y), the equation lim Dx = 0 is equivalent to lim Dy = 0, and consequently the relations y1 ® y and y1 ® x are also equivalent. Therefore the limiting value
exists and is equal to 1/f'(x). On the other hand, the limiting value is by definition the derivative f'(y) of the inverse function f(y), and the formula is proved.
This formula has a simple geometrical meaning,
which is clearly shown in Fig. 1. The tangent to the 
curve y = f(x) or x
= f(y) forms with the positive x-axis an
angle a, with the positive y-axis an angle b and from
the geometrical meaning of the derivative
However, since the sum of the angles a and b is p /2,. tan a tan b=1 and this relationship is exactly equivalent to the differentiation formula.
We have hitherto expressly assumed that either f
'(x) > 0 or f '(x)<0, i.e.,
that f '(x) is never zero. What will happen if f '(x)
= 0? If f '(x)=0 
everywhere in an interval, the function is constant
there, whence it has no inverse, since the same value of y
must correspond to all values of x in the interval. If
the equation f '(x) = 0 is true only at
isolated points and if, for the sake simplicity, f '(x)
is assumed to be continuous, then we must distinguis between
whether on passing through these points f '(x)
changes its sign or not. In the first case, this point separates
a point, where the function is monotonic increasing, from another
point, where it is monotonic decreasing. In the neighbourhood of
such a point, there cannot be a singe-valued inverse function. In
the second case, the vanishing 
of the
derivative does not destroy the monotonic character of the
function y=f(x), so that a single-valued
inverse exists. However, the inverse function will no longer be
differentiable at the corresponding point; in fact, its
derivative will be infinite there. The functions y =
x² and y= x³ at the point x=0 are
examples of the two types. Figs. 2 and 3 illustrate the behaviour
of the two functions, where they pass through the origin, and at
the same time show that one of the functions, namely y = x³,
has a single-valued inverse, but that the other function, y
= x² has not.
3.3.2 The Inverse of the Power Function:. The simplest example of an inverse function is the function y = xn for positive integers n and, as we will at first assume, positive values of x. Under these conditions y' is always positive, so that we can form for all positive values of y a unique positive inverse function
The derivative of this inverse function is immediately obtained in accordance with the general rule above by the calculations
and if we now denote the independent variable by x, we may finally write
which agrees with the result obtained directly in 2.3.3.
The point x = 0 demands special consideration.. If x approaches 0 through positive values, , for n > 1, d(x1/n)/dx will obviously increase beyond all bounds! This corresponds to the fact that, for n > 1, the derivative of the n-th power f(x) = xn vanishes at the origin. Geometrically speaking, this means that the curves y = x1/n, n > 1 touch the y-axis at the origin (Fig. 17, 1.5.7).
For the sake of completeness, it should be noted that, for odd values of n, the assumption that x>0 can be omitted and the function y = xn can be considered for all values of x without loss of its monotonic character or of the uniqueness of its inverse. The differentiation formula d(y1/n)/dy=y1/n-1/n still holds for negative values of y; for x = 0, n > 1, we have d(xn)/dx = 0, which corresponds to an infinite derivative (dy/dx) of the inverse function at the point y = 0.
3.3.3 The Inverse Trigonometric Functions: In order to form the inverses of the trigonometric functions, we once again consider the graphs of sin x, cos x, tan x and cot x. We see at once from Figs. 14 and 15 that it is necessary to select for each of these functions a definite interval, if we are to speak of a unique inverse; in fact, the lines y = c parallel to the x-axis cut the curves, if at all, at an infinite number of points.
For example, for the function y = sin x, the derivative y' = cos x will be positive in the interval -p/2<x<p /2. Hence, in this interval, sin x has an inverse function: We write the inverse function of sin x as
(read arsin y; it is the angle the sine of which has the value y; English texts also use x = sin-1y). This function runs monotonically from -p /2 to +p /2 as y traverses the interval -1 to +1. If we wish to emphasize especially that we are considering the inverse function of the sine for this very interval, we speak of the principal value of the arsine. If we form the inverse function for some other interval in which sin x is monotonic, e.g., the interval +p/2 < x < 3p/2, we obtain another branch of arsin; without an exact statement of the interval, in which the values of the function must lie, arsine is a multi-valued function; in fact, it has an infinite number of values.
In general, the fact that arsin y is multip-valued is expressed by the statement that there corresponds to anyone value y of the sine not only the angle x, but also the angle (2k + 1)p + x, as well as the angle (2k + 1)p - x, where k is any integer (Fig. 4).
The differentiation of the function x = arsin y is performed by our general rule as follows:
where the square root is positive, if we confine ourselves to the first interval above.
If we had chosen instead the interval p/2 < x < 3p/2, corresponding to the substitution of x + p for x, we should have had to use the negative sqare root, since cos x is negative in this interval.
If the independent variable is finally changed back from y to x, the differentiation formula for the function arsin x is
It is assumed here that arsin x lies between -p/2 and +p/2 and the square root sign is chosen positive.
For the inverse function of y= cos x, denoted by arcos x, we obtain the differentiation formula
in exactly the same way. Here we take the positive sign of the root if the value of arcos x is taken in the interval between 0 and p (not as in the case of arc sin x, between -p /2 and + p /2).(Fig. 5).
There remains to say something about the end-points x = -1 and x = +1. The derivatives become infinite as these end-points are approached, corresponding to the fact that the graphs of the inverse sine and inverse cosine must have vertical tangents at these points.
We can deal with the inverse functions of the
tangent and cotangent in an 
analogous
manner. The function y = tan x, the derivative
1/cos²x of which is everywhere positive for x ¹ p/2 + kp, has a
unique inverse in the interval -p/2 < x <
p/2.
We call inverse function x = artan y or (by interchange
of the letters x and y=artan x. We see
at once from Fig. 6 that the original multi-valuedness of the
inverse function - i.e., the multi-valuedness which occurs if the
interval of the values of the function is not fixed - is
expressed by the fact that for each x we could have chosen
instead of y any of the values y + kp (where k is
an integer). For the function y = cot x, the
inverse x = arcot y, (by interchange of x
and y) y = arcot x, is uniquely
determined, if we require that its value shall lie in the
interval from 0 to p ; the many-valuedness of arcot x is otherwise
the same as for artan x.
The differentiation formulae are
or, finally, if we denote the independent variable by x,
3.3.4 The Corresponding Integral Formulae: Expressed in the language of the indefinite integral, the formulas which we have just derived become
There arises no contradiction between the pair of formulae on the left and on the right hand sides, which express each indefinite integral in the form of two functions, which appear to be entirely different. We must remember that, in the case of the indefinite integral, an arbitrary additive constant remains at at our disposal. If we choose these constants so that they differ by p/2 and recall that p/2 - arcos x = arsinx and likewise p/2 - arcot x = artan x, this formal disagreement is immediately removed. The indefiniteness simply depends on the fact that the indefinite integral is not a single definite function, but an entire family of functions which differ from each other by arbitrary additive constants. The equation for an indefinite integral does not specify the value, but only one value of it. As we have already remarked, it would be more correct to express this fact by always including the undetermined constant, thus not writing
but
However, for the sake of convenience, this more detailed form is usually avoided; the reader should therefore be all the more careful to bear in mind the indefiniteness which is always associated with the shorter form.
There follow immediately from the formulae for indefinite integration formulae for definite integration. In particular,
If we set a = 0, b = 1 and recall that tan 0 = 0 and tan p /4 = 1, we obtain the remarkable formula
The number p, which originally arose from the study of the circle, is by this formula brought into a very simple relationship to the rational function 1/(1+x²) and is expressed and defined by the area shown in Fig. 7.
1. If y = x4/4, y = 16 corresponds to x = 8. Find dy/dx for x = 8; solve y=x²/4 for x and find dx/dy for y = 16 and show that the values of these derivatives are consistent with the rule for inverse functions.
2. Prove that
Differentiate the 3. - 10. and write down the corresponding integral formulae:
11. Using graph paper, plot y = 1/(1 + x²)
on a large scale. By counting squares, find 
thus obtaining an
estimate for p /4 (cf. Example 1. in 2.5).
3.4 Differentiation of a function of a function
3.4.1 The Chain Rule: The preceding rules for differentiation enable us to differentiate every function which can be expressed as a rational expression, the terms of which are functions with known derivatives. However, we can take yet another important step forward and differentiate all functions, obtained by compounding functions with known derivatives. Let f (x) be a function which is differentiable in an interval a £ x £ b and assumes all values in the interval £ f £ b. We now wish to consider a second differentiable function g(f) of the independent variable f, in which the variable f ranges over an interval from a to b. We can now regard the function g(f) = g{f(x)} = f(x) as a function of x in the interval a £ f £ b. The function f(x) = g{f(x)} will then be called a compound function of x - compounded from the functions g and f - or a function of a function.
For example, if f(x) = 1- x²
and g(f) = 
f, this compound
function is simply f(x) =
(1 - x²).
For the interval a
£ x £ b, we take here
the interval 0 £ x £ 1. The values of the function f(x)
exactly fill up the interval 0 £ f £ 1; the compound
function f(x) = (1 - x²) is
therefore defined in the interval 0 £ x £1.
Another example of the compounding of functions
is the function f(x) = (1-x²), where
the compounding process may be indicated by the equations
and where the value of the function f(x) runs through all positive numbers ³ 1, so that the function f(x) = g{f(x)} can be formed for all values of x.
When compounding functions in this way, we must
naturally be careful to restrict ourselves to intervals a £ x £ b for
which the compound function is defined. For example, the compound
function (l - x²) is only defined for values of x
in the region -1 £ x £ 1, and not in the region 1 < x £ 2, because, when x
is in this last interval, the values of the function f(x)
consist of negative numbers, for which the function g(f) is not
defined.
Just as we can compound two functions with one another, we can and must consider functions in which the compounding process is performed more than once. Such a function is
which can be built up by the compounding process
For the differentiation of compound functions, we have the fundamental theorem - the chain rule of the differential calculus:
The function f(x) = g{f(x)} is differentiable and its derivative is given by the equation
or, in Leibnitz's notation, by
In words: The derivative of the compound function is the product of the derivatives of the constituent functions.
The proof of this formula follows readily, if we recall the meaning of the derivative. There exist for any arbitrary Dx ¹ 0 and corresponding values of Df and Dg two quantities e and h, tending to 0 with Dx, such that
we have only to calculate h from the second equation and e, where Df ¹ 0, from the first equation, while, if Df = 0, we set e = 0. If we now substitute in the first of these equations the value of Df from the second equation, we obtain
or
However, in this equation, we can let Dx tend to 0 and obtain at once the result stated, since the bracket on the right hand side tends to zero with Dx. Consequently, the left hand side of our equation has a limit f '(x) and this limit is equal to the first term on the right hand side, as stated above.
We could also have proved the rule by carrying
out the passage to the limit Dx ® 0, and consequently Df ® 0, in the equation 
However, the method in the text is to be
preferred. because it avoids the necessity for considering
specially the case f '(x) = 0.
By successive application of our formula, we can immediately extend it to functions which arise from compounding of more than two functions, for example,
we can think of y = f(x) as a function of x' with the derivative given by the rule
The case of a function compounded by an arbitrary number of functions is essentially similar. The proof may be left to the reader.
3.4.2 Examples: As a very simple example, we consider the function y = xa, where we a = p/q, q being a positive integer and p a positive or negative integer, so that a is an arbitrary, positive or negative rational number. Let x be positive. By the chain rule with
we have the formula
so that for arbitrary, rational values of a we obtain the differentiation formula
in agreement with the result already found in another way in 2.3.3. As a second example, we consider
where j = I - x² and - 1 < x < 1. The chain rule yields
Further examples are given in brief calculations:
The chain rule for differentiation can also be expressed in the form of an integration formula in agreement with the fact that there corresponds to each differentiation formula a completely equivalent integration formula. Nevertheless, we will pass over this formula for the present, since we have no immediate need of it here and, moreover, it is discussed in detail lin in 4 .2.
3.4.3 Further Remarks on the Integration and Differentiation of xa, when a is Irrational: In view of the elementary definition of the power xa by the equation
where the numbers rn form a sequence of rational numbers with the limit a, we might be tempted to effect the differentiation of xa by direct passage to the limit in the differentiation formula
We are not entitled to do so unless we have the
might conclude that there follows from the relation 
the relation 
. However,
there is a very serious objection to such a passage to the limit.
In fact, in any 
arbitrarily small
neighbourhood of a given curve one may draw other curves the
direction of which at arbitrarily selected points differs from
the direction of the original curve by any desired amount; for
example, we may approximate to a straight line by a wave lying
arbitrarily near to it, the angle between the wave and the line
reaching a value as large as 45° (Fig. 8). In other words, the
above example shows that from
the fact that two functions differ only very little from each
other, we cannot immediately conclude that their derivatives also
are everywhere nearly equal to one another.
This objection forbids us to perform the apparently obvious
passage to the limit in the absence of further justification .
However, in this respect, the integral behaves quite differently from the derivative. We have already observed that, if two functions differ by less than e throughout the interval from a to b, their integrals must differ by less than e(b-a). We used there this result to establish the validity of the differentiation formula
or, replacing a +1 by a,
Hence, in this indirect way, there is verified
the relation 
given above. This discussion is a characteristic
example of the interrelation of the differential and integral
calculus. Yet, in principle, it is preferable to replace (as we
shall do in 3.6.5) the elementary
definition of xa by another, essentially simpler definition which will
lead us once more to the same result and this time directly,
Differentiate the functions:
Now that we have attained a certain mastery of the problem of differentiating the elementary functions and the functions compounded from them, we are in a position to make a variety of applications. We shall consider here the simplest of these applications - the theory of maxima and minima of a function - in conjunction with a geometrical discussion of the second derivative - and then, in the next section, we shall again take up the general theory.
3.5.1 Convexity or Concavity of
Curves: By definition, the
derivative df(x)/dx of a function
f(x) gives the slope of the curve y=f(x).
This slope itself can be represented by a curve y=df(x)/dx
- the derived curve of the given curve. The slope of this last curve will
be given by df '(x)/dx = d²f(x)/dx²
= f "(x) - the second derivative of f(x)
- and so on. If the second derivative f "(x)
is positive at a point x - so that owing to
continuity (which we here assume) it is positive in a certain
neighbourhood of the point x - then the derivative f '(x)
must increase as it passes this point in the direction of
increasing values of x. Hence, the curve y = f (x)
turns its convex side towards the direction of decreasing values
of y. The opposite is true if f'(x) is negative.
Hence, in the first case, the curve in the neighbourhood of the
point lies above the tangent, in the second case below the
tangent (Figs. 9a and b).
Special consideration is required only in the case of points where f "(x) = 0. As a rule, on passing through such a point, the second derivative f "(x) will change its sign. Such a point will then be a point of transition between the two cases indicated above; that is, the tangent will on one side be above the curve and on the other side below it, so that, besides touching the curve, it will also cross it (Fig. 10). This is a point of inflection of the curve, and the corresponding tangent is called an inflectional tangent.
The
simplest example is given by the function y=a³,
the cubical parabola, for which the x-axis itself is an inflectional tangent at the point x = 0. Another example is given by
the function f(x) = sin x, for
which
f '(x) = d(sin x)/dx = cos x and f "(x)=d²(sin x)/dx = - sin x.
Consequently, f '(0) = 1 and f "(0) = 0; since the sign of f "(x) changes at x = 0, the sine curve has at the origin an inflectional tangent, inclined at an angle of 45° to the x-axis.
However, it must be noted that points can exist where f "(x) = 0, although the tangent does not cut the curve, but remains entirely on one side of it. For example, the curve y = x4 lies entirely above the x-axis, although the second derivative f "(x) vanishes for x = 0.
3.5.2 Maxima and Minima: We say that a continuous function or a curve f(x)
has a maximum (minimum) at a point x, if in at
least some neighbourhood of the point x = x the
values of the function f(x) for x ¹ x are
all less than f(x) (greater than f(x) ).We mean by a neighbourhood of a point an interval a £ x £ b which
contains the point x in its interior. Geometrically speaking, such maxima
and 
minima are the wave-crests and wave-troughs of the curve, respectively. A glance at Fig. 11 tells
us that the value of the maximum at the point P5
may very well be less than the value of the minimum at another
point P2; thus, on account of the restriction
to a certain neighbourhood, the concept of maximum and minimum is
always relative to some extent.
If we wish to focus on the actual greatest or least value of a function, we must employ special means for deciding how this value is to be selected from among the maxima or minima.
Our objective at present is to find the (relative) maxima or minima, or, to use a word that covers both maxima and minima, the relative extreme values (extrema) of a given function or curve. This problem, which is very frequently encountered in geometry, mechanics, and physics and which occurs in many other applications, was in the Seventeenth Century one of the principal incentives for the development of the differential and integral calculus.
The expressions turning value, turning point, are also used. On the other hand, the terms stationary value, stationary point, include inflections as well as maxima and minima.
We see at once that, if a function is assumed to be differentiable, the tangent to its curve at an extreme value x must be horizontal. Hence, the condition
is a necessary condition for an extreme value; by solving this equation for the unknown x, we obtain the points at which an extreme value may possibly occur. Our condition, however, is by no means a sufficient condition for an extreme value; there may be points at which the derivative vanishes, i.e., at which the tangent is horizontal, although the curve has there neither a maximum nor a minimum. This occurs, if at the given point the curve has a horizontal inflectional tangent cutting it, as in the above example of the function y = x³ at the point x = 0.
However, if we have found a point at which f '(x) vanishes, we may immediately conclude that the function has a maximum at that point, if f "(x) < 0, a minimum, if f "(x) > 0. In fact, in the first case, the curve in the neighbourhood of this point lies completely below the tangent, in the second case, completely above the tangent.
Instead of basing the deduction of our necessary condition on intuition, we could, of course, have given an easy proof by purely analytical methods (cf. the exactly analogous considerations for Rolle's theorem. If the function f(x) has a maximum at the point x, then, for all sufficiently small values of h other than 0, the expression f(x) - f(x + h) must be positive, whence the quotient [f(x+h)-f(x)]/h will be positive or negative, according to whether h is negative or positive. Thus, if h tends to zero through negative values, the limit of this quotient cannot be negative, while if h tends to zero through positive values, the limit cannot be positive. However, since we have assumed that the derivative exists, these two limits must be equal to each other and, in fact, to f '(x), which therefore can only have the value zero; we must have f'(x)=0. A similar proof holds for the case of a minimum.
We can also formulate and prove analytically conditions, which are necessary and sufficient for the occurrence of a maximum or a minimum, without involving the second derivative. We assume that the function f(x) is continuous and has a continuous derivative f '(x) which vanishes only at a finite number of points.
Then f(x) has a maximum or a minimum at the point x = x, if and only if the derivative f '(x) changes sign on passing through this point; in particular, the function has a minimum, if the derivative is negative to the left of x and positive to the right of it, while, in the opposite case, it has a maximum.
We prove this by using the mean value theorem. First, we observe that there exist to the left and right of x intervals x1 < x < x and x < x < x2 (extending to the nearest points at which f '(x) = 0) at the each of which f '(x) has only one sign. If the signs of f '(x) in these two intervals differ, then f(x+h)-f(x )=hf '(x+q h) has the same sign for all numerically small values of h, whether h is positive or negative, so that f(x) is an extreme value. If f(x) has the same sign in both intervals, then hf'(x+qh) changes sign with h so that f(x + q h) is greater than f(x) on one side and less than f(x) on the other side, whence there is no extreme value. Our theorem has thus been proved.
At the same time, we see that the value f(x) is a greatest or least value of the function in every interval, containing the point x, in which the only change of sign of f '(x) occurs at x itself.
The mean value theorem, on which this proof is
based, can still be used even if f(x) is not
differentiable at an end-point of the interval to which it is
applied, provided that f(x) is differentiable at
all the other points of the interval; for example, the above
proof still holds if f'(x) does not exist at x = x. This leads us to the following more general result: If the function f(x)
is continuous in an interval containing the point x and everywhere in this interval, with
the possible exception of x
itself, has a derivative f '(x) which
vanishes at not more than a finite number of points, then f(x)
has an extreme value at the point x = x if, and only if, the point x separates two intervals in which f
'(x) has different signs. For
example, the function y = |x| has a
minimum at x = 0, since y' > 0 for x >
0 and y' < 0 for x < 0 (cf. Chapter
II, Fig. 9). The function 
likewise has a minimum at the point x=0,
even though its derivative 2x-1/2/3 is
infinite there (Chapter II. Fig. 12).
In addition, we make the following remark on the general theory of maxima and minima: The finding of maxima and minima is not directly equivalent to the finding of the greatest and least values of a function in a closed interval. In the case of a monotonic function, these greatest and least values will be assumed at the ends of the interval and are therefore not maxima and minima in our sense; in fact, this latter concept refers to a complete neighbourhood of the location in question. For example, the function f(x) = x0 £ x £ 1 in the interval assumes its greatest value at the point x = 1, and its least value at x = 0, and a corresponding statement holds for every monotonic function. The function y=artan x with the derivative l/(l + x²) is monotonic for -¥ £ x £ ¥ and has in that open interval neither a maximum nor a minimum nor a greatest or a least value.
If, after finding the zeros of f '(x), we wish to make sure that we have thereby found the points at which the function has its extreme values, we can often employ the criterion:
A point x , at which f '(x) vanishes, yields the least or greatest value of the function f(x) in an entire interval, if throughout that interval f "(x) > 0 or f "(x) < 0, respectively.
In fact, if both, x and x + h, belong to the interval, by the mean value theorem
Hence, at the point x = x + h, the derivative f '(x) has the same sign as h or the opposite sign, according to whether f "(x) > 0 f "(x) < 0; the statement then follows from the remark following the theorem above.
3.5.3 Examples of Maxima and Minima:
1. Of all rectangles of given area, find that with the least perimeter.
Let a² be the area of the rectangle and x the length of one side (we must consider here x as ranging over the interval 0 < x <¥ ), then the length of the other side is a²/x and half the perimeter is given by
We have
The equation f '(x) = 0 has the single positive root x = a. For this value, f "(x) is positive (as it is for any positive value of x), whence it yields the required least value and we obtain the very plausible result that of all rectangles of given area the square has the smallest perimeter.
2. Of all triangles with given base and area, find that with the least perimeter.
In order to solve this problem, we place the x-axis along the given base AB and the central point of AB as the origin. If C is the vertex of the triangle, h its altitude (which is fixed) and (x,h) are the co-ordinates of the vertex, then the sum of the two sides AC and BC of the triangle, which are to be determined, will be given by
where 2a is the length of the base, whence
We see at once (1) that f '(0) vanishes, (2) that f "(x) is always positive, whence at x = 0 there is a least value. In fact, since f "(x) > 0, the first derivative f '(x) always increases and therefore cannot be equal to zero at any other point, so that the point x = 0 must really yield the least value of f(x). Hence, this least value is given by the isosceles triangle.
Similarly, we find
that of all triangles with given perimeter and given base the
isosceles triangle has the greatest area.
3. Find a point on a given straight line such that the sum of its distances from two given fixed points is a minimum.
Let there be given a straight line and two fixed points A and B on the same side of the line. We wish to find a point P on the straight line such that the distance PA + PB has the least possible value.
We take the x-axis as the given line and use the notation of Fig. 12. Then the distance in question is given by
and we obtain
Hence. the equation f '(x) = 0 yields
which means that the two lines PA and PB must form equal angles with the given line. The positive sign of f "(x) shows us that we really have a least value.
The solution of this problem is closely linked to the optical law of reflection. By an important principle of optics known as Fermat's principle of least time, the path of a light ray is determined by the property that the time taken by the light to go from a point A to at point B under known conditions must be the least possible. If the condition is imposed that a ray of light shall on its way from A to B pass through some point on a given straight line (say, on a mirror), we see that the shortest time will be taken along the ray for which the angle of incidence is equal to the angle of reflection.
Let there be given two points A and B on opposite sides of the x-axis. Which path from A to B corresponds to the shortest possible time, if the velocity on one side of the x-axis is c1 and on the other side c2?
It is clear that the shortest path must lie along two portions of straight lines meeting each other at a point P on the x-axis. Using the notation of Fig. 13, we obtain the two expressions
for the lengths PA, PB, respectively, and we find the time of passage along this path by dividing the lengths of the two segments by the corresponding velocities and adding the results. This yields for the time taken
By differentiation, we obtain
As we readily see from the figure, the equation f '(x) = 0, i.e.,
is equivalent to the condition
We leave it to the reader to prove that there is only one point which satisfies this condition and that this point actually yields the required least value. The physical meaning of our example is again given by the optical principle of least time. A ray of light travelling between two points describes the path of shortest time. If c1 and c2q are the velocities of light on either side of the boundary of two optical media, the path of the light will be that given by our result, which accordingly yields Snell's law of refraction.
1. Find the maxima, minima and points of inflection of the following functions. Graph them, and determine the regions of increase, decrease and of convexity, concavity:
2. Determine the maxima, minima, and points of
inflection of 
Discuss the nature of the roots of 
3. Which point of the hyperbola y² - x²/2 = 1 is nearest to the point x=0, y=3.
4. Let P be a fixed point with coordinates x0,y0 in the first quadrant of a rectangular coordinate system. Find the equation of the line through P such that the length intercepted between the axes is a minimum.
6. A 12 ft high statue stands on a 15 ft high pillar. At what distance must a 6ft high man stand in order that the statue may subtend the greatest possible angle at his eyes?
6. Two sources of light of intensities a and b are at a distance d apart. At which point of the line joining them is the illumination least? (Assume that the illumination is proportional to the intensity and inversely proportional to the square of the distance.)
7. Of all rectangles with a given area, find (a) the one with the smallest perimeter; (b) the one with the shortest diagonal.
8. In the ellipse x²/a² + y²/b² = 1 inscribe the rectangle of greatest area.
9. Two sides of a triangle are a and b. Determine the third side so that the area is a maximum.
10. A circle of radius r is divided into two segments by a line g at a distance h from the centre. Inscribe in the smaller of these segments the rectangle of greatest possible area.
11. Find of all circular cylinders with a given volume the one with the least area.
12. Given the parabola y² = 2px, p > 0, and a point P(x = x, y = h) inside it (h²<2px), find the shortest path (consisting of two line segments) leading from P to a point Q on the parabola and then to the focus F(x = p/2, y = 0) of the parabola. Show that the angle FQP is bisected by the normal to the parabola and that QP is parallel to the axis of the parabola. (Principle of the parabolic mirror)
13.* A prism deflects a beam of light travelling in a plane perpendicular to the edge of the pane. What must the relative position of the prism and the beam be for the deflection to be a minimum?
14. Given n fixed numbers a1,
··· , an, determine x so
that 
is a minimum.
15. Prove that if p > 1 and x > 0, xp - 1 ³ p(x - 1).
16. Prove the inequality 1 ³ sin x/x ³ 2/p, 0 £ x £ p /2.
17. Prove that (a) tan x ³ x, 0 £ x £ p /2, (b) cos x ³ 1 - x²/2.
18.* Given a1 > 0, a2 > 0, ··· , an > 0, determine the minimum of
for x > 0. Use the result to prove by mathematical induction that
