Cho hàm số \(f\left( x \right) = {\left( {2x + 5} \right)^5}\). Có đạo hàm cấp 3 bằng:

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

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

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

\({\rm{d}}y = 2x{\rm{d}}x\).

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

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

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

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

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

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

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

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

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

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

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

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

$M = 0.$

$M = 2.$

$M =  - \,1.$

$M = 1.$

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

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

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

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