F(x) = \[\frac{1}{(ax+b)}\]
\[D^n\frac{1}{(ax+b)}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}\]
this can be proved by considering function F(x)=\[(ax+b)^m\]
first derivative can be written as \[\frac{\mathrm{d}{y } }{\mathrm{d} x} \]= \[y_1=ma(ax+b)^{m-1}\]
2nd derivative is given by \[y_2=m(m-1)a^2(ax+b)^{m-2}\]
3rd derivative can be given as \[y_3=m(m-1)(m-2)(a^3)(ax+b)^{m-3}\]
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nth derivative is given as \[y_n=m(m-1)(m-2)..........(m-n+1)(a^n)(ax+b)^{m-n}\]..................(1)
This formula is true for all m
when m=-1
\[D^n(ax+b)^{-1}=D^n(\frac{1}{ax+b})\]
= \[(-1)(-2)(-3).....(-n)(a^n)(ax+b)^{-1-n}\]
=\[\frac{(-1)^n n!a^n}{(ax+b)^{n+1}}\]
