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III - Convergence: Continuous variables
methods than contemplating a sketch on a sheet of paper as Newton did,
having had no other choice.
The inverse map 9 : J --+ I, traditionally written g(y) = arcsin y, is
differentiable at all points y = sin x where cos x =1= 0, i.e. for -1 < y < 1
(strict inequalities). Its derivative at the point y = sin x is given by
g'(y) = 1/ cos x = 1/(1 - sin 2 X)1/2 = 1/(1 _ y2)1/2
since cos x > 0 on I. On swapping the letters x and y, we get
(15.7)
arcsin'(x) = 1/(1 - X 2 )1/2 for - 1 < x < 1.
On the other hand, the function is not differentiable at x = -lor + 1 since, at
the corresponding point of I, the derivative of sin x vanishes. The graph of g,
symmetric in the diagonal with that of f, possesses a tangent at these two
points, but they are vertical, which means that the ratio [g(l + k) - g(l)J/k
tends to infinity. One could extend the concept of derivative to cover this
kind of situation, but the slight benefits of such a generalisation would be
very much less than the effort.
Example 3. Consider the function f(x) = expx on I = R, whence f'(x) =
f(x) as observed above; by nO 4, example 2, it maps R onto R+ and has
an inverse g. Since exp' x = exp x is nowhere zero the inverse function 9 is
differentiable and g'(y) = 1/ exp' x if y = expx, which can also be written as
g'(y) = l/y. Surprising?
We conclude these generalities on derivatives with a few more remarks on
notation.
First, the rules of calculus can be put in the form
(15.8) d(f + g) = df + dg, d(fg) = gdf + fdg, d(l/g) = _d9/g 2 .
The second relation, for example, means that at the point x, the differential
h ~ dp(x; h) of the product function p = fg is given by
dp(x; h) = g(x)df(x; h) + f(x)dg(x; h);
for x given this is an identity between linear functions of the auxiliary variable
h and not between functions of x.
The chain rule can similarly be written in a very simple form. Put y = f(x)
and z = g(y) = hex). Then
dz = g'(y)dy and dy = f'(x)dx, whence dz = g'(y)f'(x)dx
and so h'(x) = g'[f(x)]f'(x) since dz = h'(x)dx. Following Leibniz and Co.
one would write this quite simply in the seductive form dz / dx = dz / dy.dy / dx;
III - Convergence: Continuous variables
methods than contemplating a sketch on a sheet of paper as Newton did,
having had no other choice.
The inverse map 9 : J --+ I, traditionally written g(y) = arcsin y, is
differentiable at all points y = sin x where cos x =1= 0, i.e. for -1 < y < 1
(strict inequalities). Its derivative at the point y = sin x is given by
g'(y) = 1/ cos x = 1/(1 - sin 2 X)1/2 = 1/(1 _ y2)1/2
since cos x > 0 on I. On swapping the letters x and y, we get
(15.7)
arcsin'(x) = 1/(1 - X 2 )1/2 for - 1 < x < 1.
On the other hand, the function is not differentiable at x = -lor + 1 since, at
the corresponding point of I, the derivative of sin x vanishes. The graph of g,
symmetric in the diagonal with that of f, possesses a tangent at these two
points, but they are vertical, which means that the ratio [g(l + k) - g(l)J/k
tends to infinity. One could extend the concept of derivative to cover this
kind of situation, but the slight benefits of such a generalisation would be
very much less than the effort.
Example 3. Consider the function f(x) = expx on I = R, whence f'(x) =
f(x) as observed above; by nO 4, example 2, it maps R onto R+ and has
an inverse g. Since exp' x = exp x is nowhere zero the inverse function 9 is
differentiable and g'(y) = 1/ exp' x if y = expx, which can also be written as
g'(y) = l/y. Surprising?
We conclude these generalities on derivatives with a few more remarks on
notation.
First, the rules of calculus can be put in the form
(15.8) d(f + g) = df + dg, d(fg) = gdf + fdg, d(l/g) = _d9/g 2 .
The second relation, for example, means that at the point x, the differential
h ~ dp(x; h) of the product function p = fg is given by
dp(x; h) = g(x)df(x; h) + f(x)dg(x; h);
for x given this is an identity between linear functions of the auxiliary variable
h and not between functions of x.
The chain rule can similarly be written in a very simple form. Put y = f(x)
and z = g(y) = hex). Then
dz = g'(y)dy and dy = f'(x)dx, whence dz = g'(y)f'(x)dx
and so h'(x) = g'[f(x)]f'(x) since dz = h'(x)dx. Following Leibniz and Co.
one would write this quite simply in the seductive form dz / dx = dz / dy.dy / dx;