Cho hàm số $f\left( x \right) = {\left( {ax + b} \right)^5}$ (với $a, b$ là tham số). Tính ${f^{\left( {10} \right)}}\left( 1 \right)$ 

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

${\rm{d}}y = {\left( {x - 1} \right)^2}{\rm{d}}x$.

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

${\rm{d}}y = 2x{\rm{d}}x$.

${y^{(5)}} =  - \dfrac{{120}}{{{{(x + 1)}^6}}}$.

${y^{(5)}} = \dfrac{{120}}{{{{(x + 1)}^6}}}$.

${y^{(5)}} = \dfrac{1}{{{{(x + 1)}^6}}}$.

${y^{(5)}} =  - \dfrac{1}{{{{(x + 1)}^6}}}$.

$f'''\left( x \right) = 80{\left( {2x + 5} \right)^3}$

$f'''\left( x \right) = 480{\left( {2x + 5} \right)^2}$

$f'''\left( x \right) =  - 480{\left( {2x + 5} \right)^2}$

$f'''\left( x \right) =  - 80{\left( {2x + 5} \right)^3}$

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

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

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

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

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

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

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

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

$M = 0.$

$M = 2.$

$M =  - \,1.$

$M = 1.$

$x =  \pm \dfrac{\pi }{4} + k2\pi ,\;k \in \mathbb{Z}.$

$x = \dfrac{\pi }{8} + k\dfrac{\pi }{2},\;k \in \mathbb{Z}.$.

$x = \dfrac{\pi }{8} + k2\pi ,\;k \in \mathbb{Z}.$

$x = \dfrac{\pi }{2} + k\pi ,\;k \in \mathbb{Z}.$