Giải bài 5.93 - 5.94 trang 215 - SBT Đại số và Giải tích lớp 11

5.93.

$\begin{aligned} & y=\sin 5x\cos 2x \\ & \Rightarrow y=\dfrac{1}{2}\left( \sin 7x+\sin 3x \right) \\ \end{aligned}$

Ta cos:

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

5.94.

$\begin{aligned} & y=\dfrac{2x+1}{{{x}^{2}}+x-2}=\dfrac{2x+1}{\left( x-1 \right)\left( x+2 \right)} \\ & =\dfrac{x-1+x+2}{\left( x-1 \right)\left( x+2 \right)}=\dfrac{1}{x+2}+\dfrac{1}{x-1} \\ \end{aligned}$

Ta có:

$\begin{aligned} & y'=-\dfrac{1}{{{\left( x+2 \right)}^{2}}}-\dfrac{1}{{{\left( x-1 \right)}^{2}}} \\ & \Rightarrow y''=\dfrac{2}{{{\left( x+2 \right)}^{3}}}+\dfrac{2}{{{\left( x-1 \right)}^{3}}} \\ \end{aligned}$