Giải bài 5.14, 5.15 trang 202 - SBT Đại số và Giải tích lớp 11

5.14.
$\begin{align} & y'=\left( 9-2x \right)'\left( 2{{x}^{3}}-9{{x}^{2}}+1 \right)+\left( 9-2x \right)\left( 2{{x}^{3}}-9{{x}^{2}}+1 \right)' \\ & =-2\left( 2{{x}^{3}}-9{{x}^{2}}+1 \right)+\left( 9-2x \right)\left( 6{{x}^{2}}-18x \right) \\ & =-4{{x}^{3}}+18{{x}^{2}}-2-12{{x}^{3}}+90{{x}^{2}}-162x \\ & =-16{{x}^{3}}+108{{x}^{2}}-162x-2 \\ \end{align}$ 
5.15
$\begin{align} & y'=\dfrac{\left( 5-3x-{{x}^{2}} \right)'\left( x-2 \right)-\left( x-2 \right)'\left( 5-3x-{{x}^{2}} \right)}{{{\left( x-2 \right)}^{2}}} \\ & =\dfrac{\left( -2x-3 \right)\left( x-2 \right)-\left( 5-3x-{{x}^{2}} \right)}{{{\left( x-2 \right)}^{2}}} \\ & =\dfrac{-2{{x}^{2}}+x+6-5+3x+{{x}^{2}}}{{{\left( x-2 \right)}^{2}}} \\ & =\dfrac{-{{x}^{2}}+4x+1}{{{\left( x-2 \right)}^{2}}} \\ \end{align}$  

Ghi nhớ:

$(u.v)'=u'v+uv'$

$\left(\dfrac u v\right)'=\dfrac{u'v-uv'}{v^2}$