Đạo hàm cấp cao

Ta có: \(dy = y'dx = \left( {{x^2} + 2x} \right)'dx\).

\(\begin{array}{l}y' = \dfrac{{1.\left( {x - 2} \right) - x.1}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{ - 2}}{{{{\left( {x - 2} \right)}^2}}}\\y'' = \dfrac{{\left( { - 2} \right)'{{\left( {x - 2} \right)}^2} - \left( { - 2} \right).\left( {{{\left( {x - 2} \right)}^2}} \right)'}}{{{{\left( {x - 2} \right)}^4}}} = \dfrac{{4\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^4}}} = \dfrac{4}{{{{\left( {x - 2} \right)}^3}}}\end{array}\)

Cách 1:

\(\begin{array}{l}y' = 3{\left( {{x^2} + 1} \right)^2}\left( {{x^2} + 1} \right)' = 6x{\left( {{x^2} + 1} \right)^2}\\y'' = 6{\left( {{x^2} + 1} \right)^2} + 6x.2\left( {{x^2} + 1} \right).2x\\\,\,\,\,\,\, = 6{\left( {{x^2} + 1} \right)^2} + 24{x^2}\left( {{x^2} + 1} \right)\\y''' = 12\left( {{x^2} + 1} \right).2x + 24.2x.\left( {{x^2} + 1} \right) + 24{x^2}.2x\\\,\,\,\,\,\,\, = 24x\left( {{x^2} + 1} \right) + 48x\left( {{x^2} + 1} \right) + 48{x^3}\\\,\,\,\,\,\, = 24x\left( {{x^2} + 1 + 2\left( {{x^2} + 1} \right) + 2{x^2}} \right) = 24x\left( {5{x^2} + 3} \right)\end{array}\)

Cách 2:

 \(\begin{array}{l}y = {\left( {{x^2} + 1} \right)^3} = {x^6} + 3{x^4} + 3{x^2} + 1\\y' = 6{x^5} + 12{x^3} + 6x\\y'' = 30{x^4} + 36{x^2} + 6\\y''' = 120{x^3} + 72x = 24x\left( {5{x^2} + 3} \right)\end{array}\)

\(\begin{array}{l}y' = \dfrac{{\left( {2x + 5} \right)'}}{{2\sqrt {2x + 5} }} = \dfrac{1}{{\sqrt {2x + 5} }} = {\left( {2x + 5} \right)^{ - \frac{1}{2}}}\\y'' =  - \dfrac{1}{2}.{\left( {2x + 5} \right)^{ - \frac{1}{2} - 1}}.\left( {2x + 5} \right)'\\\,\,\,\,\,\, =  - \dfrac{1}{2}{\left( {2x + 5} \right)^{ - \frac{3}{2}}}.2\\\,\,\,\,\,\, =  - \dfrac{1}{{{{\left( {2x + 5} \right)}^{\frac{3}{2}}}}} =  - \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}\end{array}\)

\(\begin{array}{l}y' = \dfrac{1}{{{{\cos }^2}x}}\\y'' = \dfrac{{ - \left( {{{\cos }^2}x} \right)'}}{{{{\cos }^4}x}} =  - \dfrac{{2\cos x\left( {\cos x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{2\sin x}}{{{{\cos }^3}x}}\end{array}\)

Hàm số viết lại: \(y = {x^4} - 2{x^2} + 1\).

Ta có \(y' = 4{x^3} - 4x\), \(y'' = 12{x^2} - 4\), \(y''' = 24x\), \({y^{\left( 4 \right)}} = 24\).

Khi đó \(M = {y^{\left( 4 \right)}} + 2xy''' - 4y'' = 24 + 2x.24x - 4\left( {12{x^2} - 4} \right) = 40.\)

\(\begin{array}{l}h'\left( x \right) = 15{\left( {x + 1} \right)^2} + 4\\h''\left( x \right) = 30\left( {x + 1} \right) = 0 \Leftrightarrow x =  - 1\end{array}\)

\(y' = \cos x = \sin \left( {x + \dfrac{\pi }{2}} \right) \Rightarrow \) Đáp án A đúng.

\(y'' =  - \sin x = \sin \left( {x + \pi } \right) \Rightarrow \) Đáp án B đúng.

\(y''' =  - \cos x = \sin \left( {x + \dfrac{{3\pi }}{2}} \right) \Rightarrow \) Đáp án C đúng.

$\begin{array}{l}f'\left( x \right) =  - 2\sin \left( {2x - \dfrac{\pi }{3}} \right)\\f''\left( x \right) =  - 4\cos \left( {2x - \dfrac{\pi }{3}} \right)\\f'''\left( x \right) = 8\sin \left( {2x - \dfrac{\pi }{3}} \right)\\{f^{\left( 4 \right)}}\left( x \right) = 16\cos \left( {2x - \dfrac{\pi }{3}} \right)\\{f^{\left( 4 \right)}}\left( x \right) =  - 8 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) =  - \dfrac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l}2x - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} + k2\pi \\2x - \dfrac{\pi }{3} =  - \dfrac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x =  - \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\\x \in \left[ {0;\dfrac{\pi }{2}} \right] \Rightarrow x = \dfrac{\pi }{2}\end{array}$

\(y' = 2\cos 2x;\,\,y'' =  - 4\sin 2x =  - 4y \Leftrightarrow 4y + y'' = 0\)

\(\begin{array}{l}y' = \dfrac{1}{{{x^2}}}\\y'' =  - \dfrac{{\left( {{x^2}} \right)'}}{{{x^4}}} =  - \dfrac{{2x}}{{{x^4}}} =  - \dfrac{2}{{{x^3}}}\\y''' =  - 2.\dfrac{{ - \left( {{x^3}} \right)'}}{{{x^6}}} = \dfrac{{2.3{x^2}}}{{{x^6}}} = \dfrac{6}{{{x^4}}}\end{array}\)

$\begin{array}{l}f'\left( x \right) = 3{\sin ^2}x\left( {\sin x} \right)' + 2x = 3{\sin ^2}x\cos x + 2x\\f''\left( x \right) = 3.\left( {{{\sin }^2}x} \right)'.\cos x + 3{\sin ^2}x.\left( {\cos x} \right)' + 2\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x\left( {\sin x} \right)'\cos x - 3{\sin ^2}x.\sin x + 2\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x{\cos ^2}x - 3{\sin ^3}x + 2\\f''\left( { - \dfrac{\pi }{2}} \right) = 6\sin \left( { - \dfrac{\pi }{2}} \right){\cos ^2}\left( { - \dfrac{\pi }{2}} \right) - 3{\sin ^3}\left( { - \dfrac{\pi }{2}} \right) + 2 = 3 + 2 = 5.\end{array}$

Ta có \(y' = 15{x^4} - 20{x^3} + 3 \Rightarrow y'' = 60{x^3} - 60{x^2}\).

Bất phương trình \(y'' < 0 \Leftrightarrow 60{x^3} - 60{x^2} < 0\)\( \Leftrightarrow {x^2}\left( {x - 1} \right) < 0 \Leftrightarrow \left\{ \begin{array}{l}x < 1\\x \ne 0\end{array} \right.\).

Đáp án A:

\(\begin{array}{l}y = \dfrac{1}{{\cos x}}\\y' = \dfrac{{ - \left( {\cos x} \right)'}}{{{{\cos }^2}x}} = \dfrac{{\sin x}}{{{{\cos }^2}x}}\\y'' = \dfrac{{\cos x.{{\cos }^2}x - \sin x.2\cos x\left( {\cos x} \right)'}}{{{{\left( {{{\cos }^2}x} \right)}^2}}} = \dfrac{{{{\cos }^3}x + 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} = \dfrac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^3}x}}\end{array}\). 

Đáp án B:

\(\begin{array}{l}y =  - \dfrac{1}{{\cos x}}\\y' = \dfrac{{\left( {\cos x} \right)'}}{{{{\cos }^2}x}} =  - \dfrac{{\sin x}}{{{{\cos }^2}x}}\\y'' =  - \dfrac{{\cos x.{{\cos }^2}x - \sin x.2\cos x\left( {\cos x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{ - {{\cos }^3}x - 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} =  - \dfrac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^4}x}}\end{array}\)

Đáp án C:

\(\begin{array}{l}y = \cot x\\y' =  - \dfrac{1}{{{{\sin }^2}x}}\\y' = \dfrac{{2\sin x\left( {\sin x} \right)'}}{{{{\sin }^4}x}} = \dfrac{{2\sin x\cos x}}{{{{\sin }^4}x}} = \dfrac{{2\cos x}}{{{{\sin }^3}x}}\end{array}\)

Đáp án D:

\(\begin{array}{l}y = \tan x\\y' = \dfrac{1}{{{{\cos }^2}x}}\\y'' = \dfrac{{ - 2\cos x\left( {\cos x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{2\sin x\cos x}}{{{{\cos }^4}x}} = \dfrac{{2\sin x}}{{{{\cos }^3}x}}\end{array}\)

\(\begin{array}{l}f'\left( x \right) = 5a{\left( {ax + b} \right)^4}\\f''\left( x \right) = 20{a^2}{\left( {ax + b} \right)^3}\\f'''\left( x \right) = 60{a^3}{\left( {ax + b} \right)^2}\\{f^{\left( 4 \right)}}\left( x \right) = 120{a^4}\left( {ax + b} \right)\\{f^{\left( 5 \right)}}\left( x \right) = 120{a^5}\\{f^{\left( 6 \right)}}\left( x \right) = 0\\ \Rightarrow {f^{\left( {10} \right)}}\left( x \right) = 0\,\,\forall x \in R \Rightarrow {f^{\left( {10} \right)}}\left( 1 \right) = 0\end{array}\)

\(\begin{array}{l}y'\left( x \right) =  - \sin x\\y''\left( x \right) =  - \cos x\\y'''\left( x \right) = \sin x\\{y^{\left( 4 \right)}}\left( x \right) = \cos x = y\\{y^{\left( 5 \right)}}\left( x \right) =  - \sin x = y'\\{y^{\left( 6 \right)}}\left( x \right) =  - \cos x = y''\\{y^{\left( 7 \right)}}\left( x \right) = \sin x = y'''\\....\end{array}\)

Ta có: \(2018 = 504.4 + 2 \Rightarrow {y^{\left( {2018} \right)}}\left( x \right) = y''\left( x \right) =  - \cos x\)

Ta có :

\(\begin{array}{l}a = v' = \left( {s'} \right)' = s''\\s' = 3{t^2} - 4t + 4\\s'' = 6t - 4 = a\\a\left( 2 \right) = 6.2 - 4 = 8\,\,\left( {m/{s^2}} \right)\end{array}\)

\(\begin{array}{l}y = \sin 5x.\sin 3x =  - \dfrac{1}{2}\left( {\cos 8x - \cos 2x} \right)\\ \Rightarrow y' =  - \dfrac{1}{2}\left( { - 8\sin 8x + 2\sin 2x} \right) = 4\sin 8x - \sin 2x\\\,\,\,\,\,\,y'' = 32\cos 8x - 2\cos 2x\\\,\,\,\,\,\,y''' =  - 256\sin 8x + 4\sin 2x\\\,\,\,\,\,\,{y^{\left( 4 \right)}} =  - 2048\cos 8x + 8\cos 2x\end{array}\)

Ta có :

$\begin{array}{l}y' = \dfrac{{\left( {2x - {x^2}} \right)'}}{{2\sqrt {2x - {x^2}} }} = \dfrac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }} = \dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\y'' = \dfrac{{ - \sqrt {2x - {x^2}}  - \left( {1 - x} \right).\dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}}}{{2x - {x^2}}} \end{array}$

$= \dfrac{{ - \left( {2x - {x^2}} \right) - {{\left( {1 - x} \right)}^2}}}{{\sqrt {2x - {x^2}} \left( {2x - {x^2}} \right)}} $ $= \dfrac{{ - 2x + {x^2} - 1 + 2x - {x^2}}}{{\sqrt {2x - {x^2}} \left( {2x - {x^2}} \right)}} $ $= \dfrac{{ - 1}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }}$

Thay vào \({y^3}.y'' + 1 \) \(= {\left( {\sqrt {2x - {x^2}} } \right)^3}.\dfrac{{ - 1}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }} + 1 \) \(=  - 1 + 1 = 0\)

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