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Câu hỏi: Cho $f\left( x \right)={{\left( {{e}^{x}}+{{x}^{3}}\cos \text{x} \right)}^{2018}}.$ Giá trị của $f''\left( 0 \right)$ là:
A. 2018
B. $2018.2017$
C. ${{2018}^{2}}$
D. $2018.2017.2016$
A. 2018
B. $2018.2017$
C. ${{2018}^{2}}$
D. $2018.2017.2016$
Sử dụng các công thức tính đạo hàm:
$\begin{aligned}
& \left( {{u}^{n}} \right)'=n.{u}'.u{{n}^{n-1}}(uv)'={u}'v+u{v}' \\
& \left( {{e}^{x}} \right)'={{e}^{x}};\left( \sin x \right)'=\cos x,\left( \cos x \right)'=-\sin x. \\
\end{aligned}$
Ta có:
$\begin{aligned}
& f'\left( x \right)=2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+{{x}^{3}}\cos x \right)' \\
& =2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \\
\end{aligned}$
$f''\left( x \right)=\left( f\left( x \right) \right)'=\left[ 2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \right]'$
$\begin{aligned}
& =2018.2017.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2016}}.\left( {{e}^{x}}+{{x}^{3}}\cos x \right)'\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \\
& +2018.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right)' \\
\end{aligned}$
$\begin{aligned}
& =2018.2017.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2016}}.{{\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right)}^{2}} \\
& +2018.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}\left( {{e}^{x}}+6x\cos x-3{{x}^{2}}\sin x-3{{x}^{2}}\sin x-{{x}^{3}}\cos x \right) \\
\end{aligned}$
Khi $f''\left( 0 \right)=2018.2017.1.1+2018.1.1={{2018}^{2}}.$
$\begin{aligned}
& \left( {{u}^{n}} \right)'=n.{u}'.u{{n}^{n-1}}(uv)'={u}'v+u{v}' \\
& \left( {{e}^{x}} \right)'={{e}^{x}};\left( \sin x \right)'=\cos x,\left( \cos x \right)'=-\sin x. \\
\end{aligned}$
Ta có:
$\begin{aligned}
& f'\left( x \right)=2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+{{x}^{3}}\cos x \right)' \\
& =2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \\
\end{aligned}$
$f''\left( x \right)=\left( f\left( x \right) \right)'=\left[ 2018{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}.\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \right]'$
$\begin{aligned}
& =2018.2017.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2016}}.\left( {{e}^{x}}+{{x}^{3}}\cos x \right)'\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right) \\
& +2018.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right)' \\
\end{aligned}$
$\begin{aligned}
& =2018.2017.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2016}}.{{\left( {{e}^{x}}+3{{x}^{2}}\cos x-{{x}^{3}}\sin x \right)}^{2}} \\
& +2018.{{\left( {{e}^{x}}+{{x}^{3}}\cos x \right)}^{2017}}\left( {{e}^{x}}+6x\cos x-3{{x}^{2}}\sin x-3{{x}^{2}}\sin x-{{x}^{3}}\cos x \right) \\
\end{aligned}$
Khi $f''\left( 0 \right)=2018.2017.1.1+2018.1.1={{2018}^{2}}.$
Đáp án C.