Cho hàm số \(f\left( x \right) = {\left( {ax + b} \right)^5}\) (với $a, b$ là tham số). Tính \({f^{\left( {10} \right)}}\left( 1 \right)\) 

\(\begin{array}{l}f'\left( x \right) = 5a{\left( {ax + b} \right)^4}\\f''\left( x \right) = 20{a^2}{\left( {ax + b} \right)^3}\\f'''\left( x \right) = 60{a^3}{\left( {ax + b} \right)^2}\\{f^{\left( 4 \right)}}\left( x \right) = 120{a^4}\left( {ax + b} \right)\\{f^{\left( 5 \right)}}\left( x \right) = 120{a^5}\\{f^{\left( 6 \right)}}\left( x \right) = 0\\ \Rightarrow {f^{\left( {10} \right)}}\left( x \right) = 0\,\,\forall x \in R \Rightarrow {f^{\left( {10} \right)}}\left( 1 \right) = 0\end{array}\)