Giải bài 5.57 - 5.58 trang 208 - SBT Đại số và Giải tích lớp 11

5.57
$\begin{aligned} & y=\dfrac{ax+b}{a+b} \\ & \Rightarrow y'=\dfrac{a}{a+b} \\ \end{aligned}$
5.58

$\begin{aligned} & y=\left( x+1 \right){{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}} \\ & \Rightarrow y'={{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}+\left( x+1 \right)\left[ 2\left( x+2 \right){{\left( x+3 \right)}^{3}}+{{\left( x+2 \right)}^{2}}.3.{{\left( x+3 \right)}^{2}} \right] \\ & ={{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}+\left( x+1 \right)\left( x+2 \right){{\left( x+3 \right)}^{2}}\left[ 2\left( x+3 \right)+3\left( x+2 \right) \right] \\ & ={{\left( x+2 \right)}^{2}}{{\left( x+3 \right)}^{3}}+\left( x+1 \right)\left( x+2 \right){{\left( x+3 \right)}^{2}}\left( 5x+12 \right) \\ & =\left( x+2 \right){{\left( x+3 \right)}^{2}}\left( {{x}^{2}}+5x+6+5{{x}^{2}}+17x+12 \right) \\ & =\left( x+2 \right)\left( x+3 \right)^2\left( 6{{x}^{2}}+22x+18 \right) \\ \end{aligned}$