Google
Câu hỏi: Đạo hàm của hàm số $y=\left( 3{{x}^{2}}-2 \right){{\log }_{2}}x$ là
A. ${y}'=6x{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x\ln 2}.$
B. ${y}'=6x{{\log }_{2}}x-\dfrac{3{{x}^{2}}-2}{x}.$
C. ${y}'=6x{{\log }_{2}}x-\dfrac{3{{x}^{2}}-2}{x\ln 2}.$
D. ${y}'=6x{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x}.$
A. ${y}'=6x{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x\ln 2}.$
B. ${y}'=6x{{\log }_{2}}x-\dfrac{3{{x}^{2}}-2}{x}.$
C. ${y}'=6x{{\log }_{2}}x-\dfrac{3{{x}^{2}}-2}{x\ln 2}.$
D. ${y}'=6x{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x}.$
$y=\left( 3{{x}^{2}}-2 \right){{\log }_{2}}x$
$\Rightarrow {y}'={{\left( 3{{x}^{2}}-2 \right)}^{\prime }}.{{\log }_{2}}x+\left( 3{{x}^{2}}-2 \right).{{\left( {{\log }_{2}}x \right)}^{\prime }}$
$=6x.{{\log }_{2}}x+\left( 3{{x}^{2}}-2 \right).\dfrac{1}{x\ln 2}$
$=6x.{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x\ln 2}$.
$\Rightarrow {y}'={{\left( 3{{x}^{2}}-2 \right)}^{\prime }}.{{\log }_{2}}x+\left( 3{{x}^{2}}-2 \right).{{\left( {{\log }_{2}}x \right)}^{\prime }}$
$=6x.{{\log }_{2}}x+\left( 3{{x}^{2}}-2 \right).\dfrac{1}{x\ln 2}$
$=6x.{{\log }_{2}}x+\dfrac{3{{x}^{2}}-2}{x\ln 2}$.
Đáp án A.