§4. Differentiable functions
255
On substituting this result in (3) we find
g(a ~ h) - gta) = e'h [1 - e"h + o(h)] = e'h + o(h)
with c' = -gl(a)/g(a)2, which concludes the proof.
In Newtonian style:
1/(x + xo) = 1/x x 1/(1 + xo/x) = 1/x - xo/x 2 + ... ,
so that for y = 1/x one finds iJ = _X/X2. Again, much quicker!
The three preceding rules of calculus have an obvious consequence for
functions which are everywhere differentiable on an interval I, or of class CP
(p:::; +00): every reasonable algebraic operation (Le. excluding division by 0,
which restricts the set on which the result is defined) transforms functions
of such a type into functions of the same type. For example, every rational
function f(x)/g(x), where f and 9 are polynomials, is Coo on the set where
g(x) -# o. It is also obvious that the derivation f 1-+ f' transforms functions
of class CP into functions of class Cp-l, and in particular the functions of
class Coo into functions of class Coo. If we write CP(I) for the set of functions
of class CP on an interval I of JR., we see that derivation can be considered as
a map
A good example of a function or map defined on a set considerably vaster
than JR, namely a "functional space" .
In this circle of ideas, let us write
for the maps obtained by differentiating an f E CP(1) 2,3, ... times, and
consider an integer n :::; p. Leibniz' formula states that
(15.5)
Dn(fg) = Dn(f).g + nD n - 1 (f).D(g)/1! + ... +
+ n(n - 1) ... (n - k + 1)Dn-k(f)Dk(g)/k! + ... + f.Dn(g)
for f and 9 in CP(1). This bears an analogy to the binomial formula for
(x + y)n, having same coefficients 32 . It is proved in the same way too: one
assumes the formula established for n - 1 and differentiates the two sides to
obtain the formula for n; all one needs is to check the relation
32 On introducing the "divided derivatives" D[n[ = D n In!, Leibniz' formula can
be written in the form D[n1(jg) = 2:D[k1(j)D[n-k1(g). Not yet the everyday
standard.