Đạo hàm cấp cao

Ta có ${\rm{d}}y = f'\left( x \right){\rm{d}}x = 2\left( {x - 1} \right){\rm{d}}x$.

Ta có $y = x + \dfrac{1}{{x + 1}}$ $\Rightarrow y' = 1 - \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$ .

$\Rightarrow y'' = \dfrac{2}{{{{\left( {x + 1} \right)}^3}}}$ $\Rightarrow {y^{\left( 3 \right)}} = \dfrac{{ - 6}}{{{{\left( {x + 1} \right)}^4}}}$ $\Rightarrow {y^{\left( 4 \right)}} = \dfrac{{24}}{{{{\left( {x + 1} \right)}^5}}}$ $\Rightarrow {y^{(5)}} =  - \dfrac{{120}}{{{{(x + 1)}^6}}}$.

$\begin{array}{l}f'\left( x \right) = 5{\left( {2x + 5} \right)^4}\left( {2x + 5} \right)' = 10{\left( {2x + 5} \right)^4}\\f''\left( x \right) = 40{\left( {2x + 5} \right)^3}\left( {2x + 5} \right)' = 80{\left( {2x + 5} \right)^3}\\f'''\left( x \right) = 240{\left( {2x + 5} \right)^2}\left( {2x + 5} \right)' = 480{\left( {2x + 5} \right)^2}\end{array}$

$y' = \sqrt {{x^2} + 1}  + x.\dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}$ $= \dfrac{{{x^2} + 1 + {x^2}}}{{\sqrt {{x^2} + 1} }} = \dfrac{{2{x^2} + 1}}{{\sqrt {{x^2} + 1} }}$

$y'' = \dfrac{{4x\sqrt {{x^2} + 1}  - \left( {2{x^2} + 1} \right).\dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}}$ $= \dfrac{{\dfrac{{4x\left( {{x^2} + 1} \right) - x\left( {2{x^2} + 1} \right)}}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}}$ $= \dfrac{{4{x^3} + 4x - 2{x^3} - x}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}$ $= \dfrac{{2{x^3} + 3x}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}$

Ta có $y' =  - 2\sin x\cos x =  - \sin 2x$ $\Rightarrow y'' =  - 2\cos 2x \Rightarrow y''' = 4\sin 2x$

$\Rightarrow y'''\left( {\dfrac{\pi }{3}} \right) = 2\sqrt 3 .$

Ta có $y = \dfrac{1}{2}{x^2} + x + 1\,\, \Rightarrow \,\,y' = x + 1$ và $y'' = 1$.

Khi đó $M = {\left( {y'} \right)^2} - 2y.y'' = {\left( {x + 1} \right)^2} - 2\left( {\dfrac{1}{2}{x^2} + x + 1} \right)$ $= {x^2} + 2x + 1 - {x^2} - 2x - 2 =  - \,1$

Ta có  $y' = 2\cos 2x + 2\sin 2x \Rightarrow y'' =  - 4\cos 2x + 4\sin 2x$

Phương trình $y'' = 0 \Leftrightarrow  - 4\cos 2x + 4\sin 2x = 0 \Leftrightarrow \sin 2x - \cos 2x = 0$

$\Leftrightarrow \sqrt 2 \sin \left( {2x - \dfrac{\pi }{4}} \right) = 0 \Leftrightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = 0 \Leftrightarrow 2x - \dfrac{\pi }{4} = k\pi  \Leftrightarrow x = \dfrac{\pi }{8} + k\dfrac{\pi }{2},{\rm{ }}k \in \mathbb{Z}.$

Ta có $f'\left( x \right) = \sin x + x\cos x$  $\Rightarrow f''\left( x \right) = \cos x + \cos x - x\sin x = 2\cos x - f\left( x \right)$

$\Rightarrow f'''\left( x \right) =  - 2\sin x - f'\left( x \right)$.

Khi đó $f\left( x \right) + f'\left( x \right) + f''\left( x \right) + f'''\left( x \right)$

$= f\left( x \right) + f'\left( x \right) + \left[ {2\cos x - f\left( x \right)} \right] + \left[ { - 2\sin x - f'\left( x \right)} \right] = 2\left( {\cos x - \sin x} \right)$

$\Rightarrow P = f\left( {\dfrac{\pi }{2}} \right) + f'\left( {\dfrac{\pi }{2}} \right) + f''\left( {\dfrac{\pi }{2}} \right) + f'''\left( {\dfrac{\pi }{2}} \right) = F\left( {\dfrac{\pi }{2}} \right) =  - 2$.

Ta có $f'\left( x \right) = \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} \Rightarrow f''\left( x \right) = \dfrac{{ - 2\left( {x + 1} \right).3}}{{{{\left( {x + 1} \right)}^4}}} = \dfrac{{ - 6}}{{{{\left( {x + 1} \right)}^3}}}.$

Phương trình $f'\left( x \right) = f''\left( x \right) \Leftrightarrow \dfrac{3}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{{ - 6}}{{{{\left( {x + 1} \right)}^3}}} \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{ - 2}}{{x + 1}} = 1\\x \ne  - 1\end{array} \right. \Leftrightarrow x =  - 3.$

Ta có $y = {\sin ^3}x\,\, \Rightarrow y' = 3{\sin ^2}x.\cos x$ và $y'' = 6\sin x.{\cos ^2}x - 3{\sin ^3}x.$

Khi đó $M = y'' + 9y = 6\sin x.{\cos ^2}x - 3{\sin ^3}x + 9{\sin ^3}x = 6\sin x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 6\sin x.$

Ta có $\,f'\left( x \right) = 3{x^2} - 4x + 1$ và $f''\left( x \right) = 6x - 4.$

Khi đó $f'\left( {\sqrt 2 } \right) = 7 - 4\sqrt 2$ và $f''\left( {\sqrt 2 } \right) = 6\sqrt 2  - 4$.

Suy ra $M = 7 - 4\sqrt 2  + \dfrac{2}{3}\left( {6\sqrt 2  - 4} \right) = \dfrac{{13}}{3}$.

Ta có $f'\left( x \right) = 4x - 16\sin x + 2\sin 2x$ $\Rightarrow f''\left( x \right) = 4 - 16\cos x + 4\cos 2x$

$\Rightarrow f''\left( \pi  \right) = 24$.

Ta có $y' = \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} \Rightarrow y'' = \dfrac{{ - 2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \dfrac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}.$

Bất phương trình $y'' > 0 \Leftrightarrow \dfrac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}} > 0 \Leftrightarrow {\left( {x - 1} \right)^3} < 0 \Leftrightarrow x < 1$.

Ta có $v\left( t \right)={s}'\left( t \right)=3{{t}^{2}}+8t\Rightarrow a\left( t \right)={v}'\left( t \right)=6t+8.$

Thời điểm vận tốc của vật bằng $11\ {m}/{s}$ $\Rightarrow v\left( t \right)=11\Leftrightarrow 3{{t}^{2}}+8t=11\Leftrightarrow \left[ \begin{align}& t=1 \\  & t=-\dfrac{11}{3}\left( loại \right) \\ \end{align} \right..$

Với $t=1\Rightarrow a\left( 1 \right)=6.1+8=14\ {m}/{{{s}^{2}}}\;.$

Ta có $v\left( t \right)={s}'\left( t \right)=3{{t}^{2}}-6t-9\Rightarrow a\left( t \right)={v}'\left( t \right)=6t-6.$

Thời điểm vận tốc bị triệt tiêu: $v\left( t \right)=0\Leftrightarrow 3{{t}^{2}}-6t-9=0\Leftrightarrow \left[ \begin{align}& t=-1\left( loại \right) \\  & t=3 \\ \end{align} \right..$

Với $t=3\Rightarrow a\left( 3 \right)=6.3-6=12\,\,{m}/{{{s}^{2}}}\;.$

Ta có $y = \sin 5x\cos 2x = \dfrac{1}{2}\left( {\sin 7x + \sin 3x} \right)$.

$\Rightarrow y' = \dfrac{1}{2}\left( {7\cos 7x + 3\cos 3x} \right)$ $\Rightarrow y'' = \dfrac{1}{2}\left( { - 49\sin 7x - 9\sin 3x} \right).$

$\begin{array}{l} y = \frac{{x - 3}}{{x + 4}} = \frac{{x + 4 - 7}}{{x + 4}}\\ = \frac{{x + 4}}{{x + 4}} - \frac{7}{{x + 4}} = 1 - \frac{7}{{x + 4}}\\ y' = \left( {1 - \frac{7}{{x + 4}}} \right)' = - \frac{{ - 7}}{{{{\left( {x + 4} \right)}^2}}} = \frac{7}{{{{\left( {x + 4} \right)}^2}}}\\ y'' = \left[ {\frac{7}{{{{\left( {x + 4} \right)}^2}}}} \right]' = \frac{{ - 7\left[ {{{\left( {x + 4} \right)}^2}} \right]'}}{{{{\left[ {{{\left( {x + 4} \right)}^2}} \right]}^2}}}\\ = \frac{{ - 7.2\left( {x + 4} \right)}}{{{{\left( {x + 4} \right)}^4}}} = - \frac{{14}}{{{{\left( {x + 4} \right)}^3}}}\\ 2{\left( {y'} \right)^2} = 2.{\left[ {\frac{7}{{{{\left( {x + 4} \right)}^2}}}} \right]^2} = \frac{{98}}{{{{\left( {x + 4} \right)}^4}}}\\ \Rightarrow 2{\left( {y'} \right)^2}:y'' = \frac{{98}}{{{{\left( {x + 4} \right)}^4}}}:\left[ { - \frac{{14}}{{{{\left( {x + 4} \right)}^3}}}} \right]\\ = \frac{{98}}{{{{\left( {x + 4} \right)}^4}}}.\frac{{{{\left( {x + 4} \right)}^3}}}{{ - 14}} = \frac{{ - 7}}{{x + 4}} = y - 1\\ \Rightarrow 2{\left( {y'} \right)^2}:y'' = y - 1\\ \Rightarrow 2{\left( {y'} \right)^2} = \left( {y - 1} \right)y'' \end{array}$

Ta có: $y' = \left( {\sin x} \right)' = \cos x,\,\,y'' = \left( {\cos x} \right)' =  - \sin x$.

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

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