# Taylor's theorem implies existence of n+1 order derivative?

From my Calculus textbook:

Usually the text is careful to make sure the description above a formula establishes all the necessary preconditions to exist for the formula to be true. I noticed here though that $f^ {(n+1)}$ is referred to, without any mention of a requirement that it exist (f being order n differentiable by itself doesn't imply n+1 differentiability). Does the continuity of all the earlier derivatives imply it or is there an unstated assumption?

The statement says

$f^{(n)}$ is differentiable

so that next derivative exists

• doh, didn't read closely enough Feb 19 '18 at 18:30
• @JosephGarvin No, you didn't. But you did try to, which is more than many students do. So a good question even though it had an easy answer. Feb 19 '18 at 18:31