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

[daniel_i_l]

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

now I know that the nth derivative of:
<br /> f(x) = x^-1<br />
is:
<br /> (-1)^nn!x^{-1-n}<br />
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.