Cho \(f\left(x\right)=\left(2x-3\right)^5.\) Tính \(f'''\left(3\right).\)
\(80\).\(-4320\).\(2160\).\(4320\).Hướng dẫn giải:\(f\left(x\right)=\left(2x-3\right)^5\Rightarrow\)\(f'\left(x\right)=5.2\left(2x-3\right)^4\Rightarrow\)\(f"\left(x\right)=4.5.2^2.\left(2x-3\right)^3\Rightarrow\)\(f'''\left(x\right)=3.4.5.2^3\left(2x-3\right)^2.\) Do đó \(f'''\left(3\right)=3.4.5.2^3.3^2=4320\)
