Hàm số $y{\rm{ }} = \dfrac{{{x^2} + x + 1}}{{x + 1}}$ có đạo hàm cấp 5 bằng:

$y'' = \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}$ 

$y'' = \dfrac{1}{{\sqrt {2x + 5} }}$ 

$y'' =  - \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}$ 

$y''=-\dfrac{1}{\sqrt{2x+5}}$

$y'\left( 1 \right) =  - 4$.

$y'\left( 1 \right) =  - 5$.

$y'\left( 1 \right) =  - 3$.

$y'\left( 1 \right) =  - 2$.

$y' = \dfrac{{ - 13{x^2} - 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}$          

$y' = \dfrac{{ - 13{x^2} + 5x + 11}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}$ 

$y' = \dfrac{{ - 13{x^2} + 5x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}$ 

$y' = \dfrac{{ - 13{x^2} + 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}$

$y' = \sin \left( {x + \dfrac{\pi }{2}} \right)$ 

$y'' = \sin \left( {x + \pi } \right)$ 

$y''' = \sin \left( {x + \dfrac{{3\pi }}{2}} \right)$ 

${y^{\left( 4 \right)}} = \sin \left( {2\pi  - x} \right)$ 

$y = 7x + 2$.

$y = 7x - 2$.

$y =  - 7x + 2$ .

$y =  - 7x - 2$.

$x = \dfrac{\pi }{2}$ 

$x = 0$ hoặc $x = \dfrac{\pi }{6}$ 

$x = 0$ hoặc $x = \dfrac{\pi }{3}$ 

$x = 0$ hoặc $x = \dfrac{\pi }{2}$