here F(x)= \[(ax+b)^n\]
\[D^n(ax+b)^n=n!a^n\]
This can be proved by taking general case where 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
following are the some particular cases
Case (1) if m=n (positive)
then then equation 1 can be simplified as
\[D^n(ax+b)^n= n(n-1)(n-2).................1*a^n(ax+b)^{n-n}\]
=\[n!a^n\]
