Then
is differentiable on
and
.
Let
. We prove that
is differentiable on
and we will compute the first
derivative of
.
As the sine function is continuous on
, the function
is well-defined on
. For any real
, we have:
![]() |
![]() |
The function
is differentiable on
, by Theorem 8.1; we have:
![]() |
Thus
is differentiable on
and
![]() |
Actually, in Example 8.2, we could have computed an explicit formula for
. In the
Example 8.3, this should be impossible.
Let
. The function
defined by
is
continuous on
, therefore by Theorem 8.1, the function
is differentiable on
and we have:
![]() |
Noah Dana-Picard 2007-12-28