Question

Cho hàm số \(y=\dfrac{1}{\left(x-1\right)\left(x+3\right)}\). Hỏi đạo hàm cấp 2019 của hàm số bằng biểu thức nào sau đây?

A. \(\dfrac{2019!}{4}\left(\dfrac{1}{\left(x+3\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

B. \(\dfrac{2019!}{4}\left(\dfrac{1}{\left(x+3\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

C. \(-\dfrac{2019!}{4}\left(\dfrac{1}{\left(x+3\right)^{2020}}+\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

D. \(-\dfrac{2019!}{4}\left(\dfrac{1}{\left(x+3\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

Answer 1

C

Answer 2

C

Answer 3

\(y=\dfrac{1}{\left(x-1\right)\left(x+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+3}\right)\)

\(y'=\dfrac{1}{4}\left(\dfrac{-1}{\left(x-1\right)^2}-\dfrac{-1}{\left(x+3\right)^2}\right)=\dfrac{1}{4}\left(\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{\left(-1\right)^1.1!}{\left(x+3\right)^2}\right)\)

\(y''=\dfrac{1}{4}\left(\dfrac{2}{\left(x-1\right)^3}-\dfrac{2}{\left(x+3\right)^3}\right)=\dfrac{1}{4}\left(\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{\left(-1\right)^2.2!}{\left(x+3\right)^3}\right)\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{4}\left(\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{\left(-1\right)^n.n!}{\left(x+3\right)^{n+1}}\right)\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{4}\left(\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{\left(-1\right)^{2019}.2019!}{\left(x+3\right)^{2020}}\right)\)

\(=\dfrac{2019!}{4}\left(\dfrac{1}{\left(x+3\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

Answer 4

C

Answer 5

C

Answer 6

c