Giải bài 54 trang 113 SGK giải tích nâng cao 12

Áp dụng công thức $\left( \ln x \right)'=\dfrac{1}{x},\left( \ln u \right)'=\dfrac{u'}{u}$ ta có

$\begin{align} a)\,& y=\left( 3x-2 \right){{\ln }^{2}}x \\ y'&=3{{\ln }^{2}}x+\dfrac{2\left( 3x-2 \right)\ln x}{x} \\ b)\, & y=\sqrt{{{x}^{2}}+1}{{\ln }^{2}}x \\ y'&=\dfrac{x}{\sqrt{{{x}^{2}}+1}}{{\ln }^{2}}x+\dfrac{2\sqrt{{{x}^{2}}+1}\ln x}{x} \\ \end{align}$

$\begin{align} c)\, & y=x.\ln \dfrac{1}{1+x} \\ y'&=\ln \dfrac{1}{1+x}+x.\dfrac{1}{\dfrac{1}{\left( x+1 \right)}}.\left( \dfrac{1}{x+1} \right)' \\ & =\ln \dfrac{1}{1+x}+x.\left( x+1 \right).\dfrac{-1}{{{\left( x+1 \right)}^{2}}} \\ & =\ln \dfrac{1}{1+x}-\dfrac{x}{x+1} \\ d)\, & y=\dfrac{\ln \left( {{x}^{2}}+1 \right)}{x} \\ y'&=\dfrac{x.\dfrac{\left( {{x}^{2}}+1 \right)'}{{{x}^{2}}+1}-\ln \left( {{x}^{2}}+1 \right)}{{{x}^{2}}} \\ & =\dfrac{2}{{{x}^{2}}+1}-\dfrac{\ln \left( {{x}^{2}}+1 \right)}{{{x}^{2}}} \\ \end{align}$