Find nth Derivative of (ax-b)^-1

[daniel_i_l]

The question is to find the nth derivative of:
$$f(x) = (ax - b)^-1$$
the answer in the book is:
$$f^[n](x) = n!(-a)^n(ax-b)^{-1-n}$$

now I know that the nth derivative of:
$$f(x) = x^-1$$
is:
$$(-1)^nn!x^{-1-n}$$
but why would the nth of ax be a^n? Shouldn't it be 0 for every n over 1 which would make the whole term 0?

 

[Tide]

You're not being asked to find the $n^{th}$ derivative of $ax$ nor is it even suggested by the "answer in the book."

 

[daniel_i_l]

but by the chain rule don't I have to multiply by the nth derivative of
(ax + b) ? I thought that the a^n was implied because that is the term that is missing from my answer: a^n * (-1)^n = (-a)^n .

 

[Tide]

Each time you take a derivative you reduce the power of (ax-b) and obtain an additional factor of a.