Giải bài 5.61 - 5.62 trang 209 - SBT Đại số và Giải tích lớp 11

5.61

\(\begin{aligned} & y=\left( 1-x \right){{\left( 1-{{x}^{2}} \right)}^{2}}{{\left( 1-{{x}^{3}} \right)}^{3}} \\ & \Rightarrow y'=-{{\left( 1-{{x}^{2}} \right)}^{2}}{{\left( 1-{{x}^{3}} \right)}^{3}}+\left( 1-x \right)\left[ {{\left( 1-{{x}^{2}} \right)}^{2}}{{\left( 1-{{x}^{3}} \right)}^{3}} \right]' \\ & =-{{\left( 1-{{x}^{2}} \right)}^{2}}{{\left( 1-{{x}^{3}} \right)}^{3}}+\left( 1-x \right)\left[ -4x\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{3}}-9{{x}^{2}}{{\left( 1-{{x}^{3}} \right)}^{2}}.{{\left( 1-{{x}^{2}} \right)}^{2}} \right] \\ & =-{{\left( 1-{{x}^{2}} \right)}^{2}}{{\left( 1-{{x}^{3}} \right)}^{3}}-x\left( 1-x \right)\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{2}}\left[ 4\left( 1-{{x}^{3}} \right)+9\left( 1-{{x}^{2}} \right) \right] \\ & =-{{\left( 1-x \right)}^{2}}\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{2}}\left( 1+x \right)\left( 1+x+{{x}^{2}} \right) \\ &= -x{{\left( 1-x \right)}^{2}}\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{2}}\left[ 4\left( 1+x+{{x}^{2}} \right)+9\left( 1+x \right) \right] \\ & =-{{\left( 1-x \right)}^{2}}\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{2}}\left[ \left( 1+x \right)\left( 1+x+{{x}^{2}} \right)+4x\left( 1+x+{{x}^{2}} \right)+9x\left( 1+x \right) \right] \\ & =-{{\left( 1-x \right)}^{2}}\left( 1-{{x}^{2}} \right){{\left( 1-{{x}^{3}} \right)}^{2}}\left( 1+6x+15{{x}^{2}}+14{{x}^{3}} \right) \\ \end{aligned} \)

5.62

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