Lý thuyết Đạo hàm cấp hai

+) ${\left( {\sin x} \right)^{\left( n \right)}} = \sin \left( {x + \dfrac{{n\pi }}{2}} \right)$

+) ${\left( {\cos x} \right)^{\left( n \right)}} = \cos \left( {x + \dfrac{{n\pi }}{2}} \right)$

+) Nếu $n \le m$ thì $\left( {{x^m}} \right)^{\left( n \right)} =m\left( {m - 1} \right)...\left( {m - n + 1} \right).{x^{m - n}}$

+) Nếu $n>m$ thì ${\left( {{x^m}} \right)^{\left( n \right)}} =0$.

$\begin{array}{l} + )y = \sin \left( {ax + b} \right)\\ \Rightarrow {y^{\left( n \right)}} = {a^n}\sin \left( {ax + b + \frac{{n\pi }}{2}} \right)\\ + )y = \cos \left( {ax + b} \right)\\ \Rightarrow {y^{\left( n \right)}} = {a^n}\cos \left( {ax + b + \frac{{n\pi }}{2}} \right)\\ + )y = \frac{1}{{ax + b}}\\ \Rightarrow {y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}.n!.{a^n}}}{{{{\left( {ax + b} \right)}^{n + 1}}}}\\ + )y = \sqrt[m]{{ax + b}}\\ \Rightarrow {y^{\left( n \right)}} = \frac{1}{m}.\left( {\frac{1}{m} - 1} \right)...\left( {\frac{1}{m} - n + 1} \right).{a^n}.{\left( {ax + b} \right)^{\frac{1}{m} - n}} \end{array}$