Cho hàm số \(y=\dfrac{x}{x^2-1}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?
A. \(-\dfrac{2019!}{2}\left(\dfrac{1}{\left(x+1\right)^{2019}}+\dfrac{1}{\left(x-1\right)^{2019}}\right)\)
B. \(\dfrac{2019!}{2}\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
C. \(2019!\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
D. \(-\dfrac{2019!}{2}\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
\(y=\dfrac{1}{2}\left(\dfrac{1}{x+1}+\dfrac{1}{x-1}\right)\)
\(y'=\dfrac{1}{2}\left(\dfrac{-1}{\left(x+1\right)^2}+\dfrac{-1}{\left(x-1\right)^2}\right)=\dfrac{1}{2}\left(\dfrac{\left(-1\right)^1.1!}{\left(x+1\right)^2}+\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}\right)\)
\(y''=\dfrac{1}{2}\left(\dfrac{\left(-1\right)^2.2!}{\left(x+1\right)^3}+\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}\right)\)
\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{2}\left(\dfrac{\left(-1\right)^n.n!}{\left(x+1\right)^{n+1}}+\dfrac{\left(-1\right)^n.2!}{\left(x-1\right)^{n+1}}\right)\)
\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{2}\left(\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+1\right)^{2020}}+\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}\right)\)
\(=-\dfrac{2019!}{2}\left(\dfrac{1}{\left(x+1\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)
