Đạo hàm cấp cao

\(\begin{array}{l}y = \dfrac{{2x + 1}}{{{x^2} - 5x + 6}} = \dfrac{{2x + 1}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{7}{{x - 3}} - \dfrac{5}{{x - 2}}\\ \Rightarrow {y^{\left( 4 \right)}} = 7{\left( {\dfrac{1}{{x - 3}}} \right)^{\left( 4 \right)}} - 5{\left( {\dfrac{1}{{x - 2}}} \right)^{\left( 4 \right)}}\end{array}\)

Xét hàm số \(\dfrac{1}{{ax + b}},\,a \ne 0\) ta có :

\(\begin{array}{l}y' = \dfrac{{ - a}}{{{{\left( {ax + b} \right)}^2}}}\\y'' = \dfrac{{a.2\left( {ax + b} \right).a}}{{{{\left( {ax + b} \right)}^4}}} = \dfrac{{2{a^2}}}{{{{\left( {ax + b} \right)}^3}}}\\y''' = \dfrac{{ - 2{a^2}.3{{\left( {ax + b} \right)}^2}.a}}{{{{\left( {ax + b} \right)}^6}}} = \dfrac{{ - 2.3.{a^3}}}{{{{\left( {ax + b} \right)}^4}}}\\....\\{y^{\left( n \right)}} = \dfrac{{{{\left( { - 1} \right)}^n}.{a^n}.n!}}{{{{\left( {ax + b} \right)}^{n + 1}}}}\\ \Rightarrow {\left( {\dfrac{1}{{x - 3}}} \right)^{\left( 4 \right)}} = \dfrac{{{{\left( { - 1} \right)}^4}{{.1}^4}.4!}}{{{{\left( {x - 3} \right)}^5}}} = \dfrac{{4!}}{{{{\left( {x - 2} \right)}^5}}}\\\,\,\,\,\,{\left( {\dfrac{1}{{x - 2}}} \right)^{\left( 4 \right)}} = \dfrac{{{{\left( { - 1} \right)}^4}{{.1}^4}.4!}}{{{{\left( {x - 2} \right)}^5}}} = \dfrac{{4!}}{{{{\left( {x - 2} \right)}^5}}}\\ \Rightarrow {y^{\left( 4 \right)}} = 7{\left( {\dfrac{1}{{x - 3}}} \right)^{\left( 4 \right)}} - 5{\left( {\dfrac{1}{{x - 2}}} \right)^{\left( 4 \right)}} = \dfrac{{7.4!}}{{{{\left( {x - 3} \right)}^5}}} - \dfrac{{5.4!}}{{{{\left( {x - 2} \right)}^5}}}\end{array}\)

Bước 1:

\(\begin{array}{l}y' = \left( {3{x^2} - 2021x + 2020} \right)'\\ = 3.2.x - 2021\\ = 6x - 2021\end{array}\)

Bước 2:

\( \Rightarrow y'' = \left( {y'} \right)' = 6\)