§4. Differentiable functions
257
the derivative of the function log ( exp x) is 1/ exp x x exp x = 1, which does
not contradict the relation log(expx) = x established in nO 2, example l.
There we first showed that log( exp x) = ax with a mysterious constant a;
the preceding argument now shows that a = 1. Chap. IV will confirm this by
other methods, all roads leading to this fundamental formula; we will even
end up making a theorem out of it ...
An obvious consequence of (D 4), applying it repeatedly, is that if f and
9 are of class CP on I and J, then the composite function c = 9 0 f is again
of class CP on I.
The following rule completes Theorem 7 of nO 4 regarding the inverse of
a. continuous strictly monotone map:
(D 5) Let f be a continuous strictly monotone map of an interval I c lR
onto an interval J c lR and let 9 : J ---+ I be the inverse map of f. Suppose that the function f is differentiable at a point a E I. If so, then 9
possesses a derivative at the point b = f(a) if and only if f'(a) -:I o. And then
g'(b) = 1/ f'(a).
The condition is necessary, since the relation g[f(x)] = x shows that
g'(y)f'(x) = 1
if Y = f(x),
by the preceding rule; this forbids f' from vanishing and even exhibits the
only possible value of g'. It remains to show that, conversely, the condition
I' ( a) -:I 0 assures the existence of g' (b).
We need to show that the ratio [g(b + k) - g(b)]/k tends to 1/ f'(a) as k
tends to o. To do this, put
g(b+k)=a+h i.e. h=g(b+k)-g(b)=p(k),
whence b + k = f(a + h). When k tends to 0 so does h = p(k), since 9 is
continuous. Now the relation
(15.6)
k = (b + k) - b = f(a + h) - f(a) = f'(a)h + o(h)
shows that the ratio k/h = k/p(k) tends to the nonzero limit I'(a) when
k (and so h) tends to O. In consequence, p(k)/k tends to 1/f'(a). But
p(k)/k = [g(b + k) - g(b)]/k, qed.
Example 2. Take I = [-11"/2,11"/2] and f(x) = sinx, a function which the
reader surely knows to be continuous, and strictly increasing on I (though
not on lR), maps I onto J = [-1,1], and everywhere has a derivative given
by f'(x) = cosx; we will justify all this in Chap. IV by much more rigorous
257
the derivative of the function log ( exp x) is 1/ exp x x exp x = 1, which does
not contradict the relation log(expx) = x established in nO 2, example l.
There we first showed that log( exp x) = ax with a mysterious constant a;
the preceding argument now shows that a = 1. Chap. IV will confirm this by
other methods, all roads leading to this fundamental formula; we will even
end up making a theorem out of it ...
An obvious consequence of (D 4), applying it repeatedly, is that if f and
9 are of class CP on I and J, then the composite function c = 9 0 f is again
of class CP on I.
The following rule completes Theorem 7 of nO 4 regarding the inverse of
a. continuous strictly monotone map:
(D 5) Let f be a continuous strictly monotone map of an interval I c lR
onto an interval J c lR and let 9 : J ---+ I be the inverse map of f. Suppose that the function f is differentiable at a point a E I. If so, then 9
possesses a derivative at the point b = f(a) if and only if f'(a) -:I o. And then
g'(b) = 1/ f'(a).
The condition is necessary, since the relation g[f(x)] = x shows that
g'(y)f'(x) = 1
if Y = f(x),
by the preceding rule; this forbids f' from vanishing and even exhibits the
only possible value of g'. It remains to show that, conversely, the condition
I' ( a) -:I 0 assures the existence of g' (b).
We need to show that the ratio [g(b + k) - g(b)]/k tends to 1/ f'(a) as k
tends to o. To do this, put
g(b+k)=a+h i.e. h=g(b+k)-g(b)=p(k),
whence b + k = f(a + h). When k tends to 0 so does h = p(k), since 9 is
continuous. Now the relation
(15.6)
k = (b + k) - b = f(a + h) - f(a) = f'(a)h + o(h)
shows that the ratio k/h = k/p(k) tends to the nonzero limit I'(a) when
k (and so h) tends to O. In consequence, p(k)/k tends to 1/f'(a). But
p(k)/k = [g(b + k) - g(b)]/k, qed.
Example 2. Take I = [-11"/2,11"/2] and f(x) = sinx, a function which the
reader surely knows to be continuous, and strictly increasing on I (though
not on lR), maps I onto J = [-1,1], and everywhere has a derivative given
by f'(x) = cosx; we will justify all this in Chap. IV by much more rigorous