Tính đạo hàm cấp \(5\) của hàm số \(y=\dfrac{1}{ax+b}\)
\(y^{\left(5\right)}=\dfrac{120a^5}{\left(ax+b\right)^6}\) \(y^{\left(5\right)}=\dfrac{-120a}{\left(ax+b\right)^6}\) \(y^{\left(5\right)}=\dfrac{-24a^5}{\left(ax+b\right)^5}\) \(y^{\left(5\right)}=\dfrac{-120a^5}{\left(ax+b\right)^6}\) Hướng dẫn giải:\(y=\dfrac{1}{ax+b}=\left(ax+b\right)^{-1}\Rightarrow y'=-1.\left(ax+b\right)^{-2}.\left(ax+b\right)'=-1.a.\left(ax+b\right)^{-2}\)
\(\Rightarrow y"=\left(-1\right)\left(-2\right)a\left(ax+b\right)^{-3}\left(ax+b\right)'=\left(-1\right)\left(-2\right)a^2\left(ax+b\right)^{-3}\)
\(\Rightarrow y^{\left(3\right)}=\left(-1\right)\left(-2\right)\left(-3\right)a^3\left(ax+b\right)^{-4}\)\(\Rightarrow y^{\left(4\right)}=\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)a^4\left(ax+b\right)^{-5}\)
\(\Rightarrow y^{\left(5\right)}=\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)\left(-5\right)a^5\left(ax+b\right)^{-6}\)\(=-\dfrac{120a^5}{\left(ax+b\right)^6}\)
Chú ý: Tổng quát ta có \(\left(\dfrac{1}{ax+b}\right)^{\left(n\right)}=\dfrac{\left(-1\right)^nn!a^n}{\left(ax+b\right)^n}\), trong đó \(n!=1.2.3...n.\)