.From time to time, we have made it clear by examples that the general concept of function contains many possibilities foreign to naïve intuition. As a rule, these examples were not given in terms of single analytical expressions. Hence we must show that it is possible to represent various typical discontinuities and abnormal phenomena by means of very simple expressions, built up from the elementary functions. We begin with an example without presence of a discontinuity.
A3.1.1 The Function y = e-1/x²: The function (Fig. 22), which is defined in the first instance only for values of x other than zero, obviously has the limit zero as x®0. In fact, by the transformation l/x² = x , this function becomes y=e-x and
.
Hence, in order to extend our function so that it is continuous for x = 0
we define the value of the function at the point x = 0 by the equation y(0 )= 0.
By the chain rule, the derivative of our function for x ¹ 0 is 
If x
tends to 0, this derivative will also have the limit 0, as we
find immediately from 3.9.5. At the point x =
0 itself, the derivative
is also zero.
If we form the higher derivatives for x ¹ 0, we shall obviously always obtain the product of the function e-1/x² and a polynomial in 1/x, and the passage to the limit x®0 will always yield the limit 0. All the higher derivatives will likewise vanish, like y', at the point x = 0.
Thus, we see that our function is continuous everywhere and differentiable as many times as we please, and yet it vanishes with all its derivatives at the point x = 0. We shall realize later on in A6.1how remarkable this behaviour really is.
A3.1.2. The Function y
= e-1/x: We may readily convince ourselves that, for positive
values of x, this function behaves in the same way as
the preceding function; if x tends to 0 through positive
values, the function tends to 0 and the same is true for all its
derivatives. If we define the value of 
the function at x = 0 as y(0) = 0, all
the right-hand derivatives at the point x = 0 will have
the value 0. It is quite a different matter when x tends
to 0 through negative values; in fact, then the function becomes
infinite and there do not exist left-hand derivatives at the
point Hence, at the point x = 0. the function has a
remarkable kind of discontinuity, quite unlike the infinite
discontinuities of the rational functions, considered in 1.3.1 and 1.8.2.
A3.1.3 The function y = tanh 1/x: We have already seen 1n 1.8.2 and A.1.5 that functions with jump discontinuities can be obtained from simple functions by a passage to the limit. The exponential function, defined on 3.6.4, and the principle of compounding of functions offer us another method for the construction of functions with such discontinuities from elementary functions without any further limiting process. An example of this is the function
with its behaviour at the point x =
0. In the first instance, the function is not defined at this
point. If we approach the point x = 0 through positive
values of x, we obviously obtain the limit 1; on the
other hand, if we approach the point x=0 through negative 
values, we obtain the limit -1. The point x = 0
is therefore a point of discontinuity; as x increases
through 0, the value of the function jumps by 2 (Fig. 24). On the
other hand, the derivative
approaches the limit 0 from both sides, as follows readily from 3.9.5
Another example of the occurrence of a jump discontinuity is given by the function y = artan1/x as x ®0.
A3.1.4 The Function y =x tanh 1/x: In the case of the function
the above discontinuity is removed by the factor x. This function has the limit 0 as x ® 0 from either side, so that we can again appropriately define y(0) as equal to 0. Our function is then continuous at x = 0, its first derivative
has just the same kind of discontinuity as the preceding example. The graph of the function is a curve with a corner (Fig. 25); at the point x = 0, the function has no actual derivative, but a right-hand derivative with the value +1 and a left-hand derivative with the value -1.
A3.1.5 The Function y = x sin 1/x, y(0) = 0: We have already seen that this function is not composed of a finite number of monotonic pieces - as we may say, it is not sectionally monotonic (cf. 10.5.1), but is nevertheless continuous (1.8.3). Indeed, its first derivative
has a discontinuity at x = 0, because, as x tends to 0, this derivative oscillates continually between bounding curves, one positive and one negative, which themselves tend to + ¥ and -¥, respectively. At the actual point x = 0, the difference quotient is [y(h) - y(0)]/h = sin 1/h; since, as h ® 0, this swings backwards and forwards between 1 and -1, an infinite number of times, the function has neither a right-hand nor a left-hand derivative.
A3.2 Remarks on the Differentiability of Functions
The derivative of a function, which is continuous and has a derivative at every point, need not be continuous.
As the simplest example, consider the function
This function is in the first instance not defined at x = 0; we shall define f(0), its value there, as 0, so that the function is now defined and continuous everywhere. For all values of x, other than 0, the derivative is given by the expression
As x tends to 0, f'(x) has no limit. On the other hand, if we form the difference quotient
we see at once that this tends to 0 with h. The derivative therefore exists for x=0 and has the value 0. In order to understand this intuitively - the reason for this paradoxical behaviour - we represent the function graphically (Fig. 26).
It swings backwards and forwards between the curves y = x² and y = -x², which it touches in turn. Thus, the ratio of the heights of the wave-crests of our curve and their distances from the origin steadily becomes smaller. Yet, these waves do not become flatter; in fact, their slope is given by the derivative f '(x) = 2x sin 1/x - cos 1/x; at the points x = 1/(2np), where cos 1/x = 1, this is equal to -1, at the points x - 1/(2n + 1)p , where cos 1/x = -1, it is equal to +1.
In contrast to the possibility
shown here - that a derivative may exist everywhere and yet not
be continuous, we state the following simple theorem, which
throws light on the whole series of earlier examples and
discussions: If we know
that, in a neighbourhood of a point x = a, the function
f(x) is continuous and
has a derivative f '(x)
everywhere, except that we do not know whether f
'(a) exists, and if
moreover the equation 
holds, then also the derivative f
'(x) exists at the point a and f '(a)
= b. The proof follows
immediately from the mean value theorem. In fact, we have [f(a+h)-f(a)]/h=f'(x), where x is a
value intermediate between a and a+h. If now h
tends to 0, by hypothesis, f(x) tends to b,
and our statement follows immediately.
A companion theorem to this theorem may be proved in a similar manner: If the function f(x) is continuous in a £ x £ b and has for a < x < b a derivative, which increases beyond all bounds as a tends to 0, then the right-hand difference quotient [f(a+h)-f(a)]/h also increases beyond all bounds as h tends to 0, so that no finite right-hand derivative exists at x = a. The geometrical meaning of this state of affairs is that at the point with the (finite) co-ordinates (a, f(a)) the curve has a vertical tangent.
A3.3.1 Proof of the Binomial Theorem: Our rules for differentiation enable us to give a simple proof of the binomial theorem; this proof will be introduced here as an example of the method of undetermined parameters which we shall find later on to be important. We wish to expand the quantity (1+x)n in powers of x for all positive integral values of n. We see at once that the function (1+x)n must be a polynomial of degree n, i.e., it must be of the form
and the problem is now to determine the coefficients ak. If we set x = 0, we obtain at once a0 = 1. If we differentiate both sides of the equation once, twice, three times, etc., we obtain the equations
Since these equations hold for all values of x, we can set x = 0 in each of them and thus obtain for the coefficients ak the expressions
We thus obtain the binomial theorem in the form
A3.3.2 Proof of the Binomial Theorem: Leibnitz's Rule: In connection with the above, we ask the reader to prove as an exercise that the successive differentiation of a product may be performed according to Leibnitz's rule:
However. the repeated differentiation of a compound function y = f{f(x)} follows no such easily recalled law. We have from the rules for differentiation (the product and chain rules)
A3.3.3 Further Examples of the Use of the Chain Rule. Differentiation of f(x)g(x). The Generalized Mean Value Theorem: In order to form the derivative of the function xx, we write xx = exlog x, whence the chain rule yields
Similarly, we can carry out the differentiation of the more general expression
by means of the chain rule as follows:
As a further application of the chain rule, we present here a proof of the theorem which we have already called the generalized mean value theorem of the differential calculus, the theorem being established here under less stringent conditions. Let G(x) = u be a function which in the closed interval a £ x £ b is continuous and monotonic, and which in the open interval a < x < b has a derivative which is nowhere equal to 0; moreover, let F(x) be a function which is also continuous for a £ x £ b and differentiable for a < x < b. Introduce by means of the inverse function x = F(u) of G(x) the new independent variable u instead of x into F(x), thus obtaining the compound function f(u) = F(F(u)); by the chain rule,
The ordinary mean value theorem, on application to the function f(u) and to the interval between u1=G(a) and u2=G(b), shows that for an intermediate value w
where x <= F(w) is a value intermediate between a and b.
Exercises: 3.10
1. Find the second derivative of f(g(h(x))).
2. Differentiate the functions: (a)
xsin x, (b) (cos x)tan
x,
(c) logv(x) u(x)
(that is, the logarithm of u(x) to the base v(x),
v(x) > 0).
3. Prove Leibnitz's rule.
4. Find the n-th
derivative of:
5.*Find the n-th derivative of arcsin x at x=0, and then that of (arcsin x)² at x=0.
9. Prove that 