Cho \(f\left(x\right)=\dfrac{1}{x^2-1}.\) Tính \(f'''\left(2\right).\)
\(-\dfrac{80}{27}\).\(-\dfrac{40}{27}\).\(\dfrac{40}{27}\).\(\dfrac{80}{27}\).Hướng dẫn giải:\(f\left(x\right)=\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)=\dfrac{1}{2}\left(\left(x-1\right)^{-1}-\left(x+1\right)^{-1}\right)\), suy ra \(f'\left(x\right)=\dfrac{1}{2}.\left(-1\right)\left(\left(x-1\right)^{-2}-\left(x+1\right)^{-2}\right)\), do đó
\(f"\left(x\right)=\dfrac{1}{2}\left(-1\right)\left(-2\right)\left(\left(x-1\right)^{-3}-\left(x+1\right)^{-3}\right)\)\(\Rightarrow f'''\left(x\right)=\dfrac{1}{2}\left(-1\right)\left(-2\right)\left(-3\right)\left[\left(x-1\right)^{-4}-\left(x+1\right)^{-4}\right]\)
\(f'''\left(2\right)=-3\left[\left(2-1\right)^{-4}-\left(3+1\right)^{-4}\right]=-\dfrac{80}{27}\)